2026 Cambridge IAL Biology (9700) 模拟试题及答案详解

1 division on the stage micrometer \(= 0.01\text{ mm} = 10\ \mu\text{m}\).
6 stage micrometer divisions \(= 6 \times 10\ \mu\text{m} = 60\ \mu\text{m}\).
Since 50 eyepiece graticule units (epu) align with 6 stage micrometer divisions, 50 epu \(= 60\ \mu\text{m}\).
Therefore, 1 epu \(= \frac{60\ \mu\text{m}}{50} = 1.2\ \mu\text{m}\).
The chloroplast measures 5 epu, so its actual length is \(5 \times 1.2\ \mu\text{m} = 6.0\ \mu\text{m}\).

B is correct [1 mark].
A is incorrect as it is the value of 1 epu (1.2 µm).
C is incorrect as it results from dividing 50 by 6 (8.33 µm).
D is incorrect as it represents the total stage micrometer span (60 µm).

Amylopectin is a branched polysaccharide component of starch made exclusively of \(\alpha\)-glucose monomers. It contains \(\alpha\)-1,4-glycosidic bonds along its linear chains and \(\alpha\)-1,6-glycosidic bonds at branch points.

C is correct [1 mark].
A is incorrect because amylose, not amylopectin, has only \(\alpha\)-1,4 bonds and is unbranched.
B is incorrect because it specifies \(\beta\)-glucose monomers.
D is incorrect because it specifies \(\beta\)-glucose and \(\beta\)-glycosidic bonds (cellulose features).

An unchanged \(V_{\text{max}}\) with an increased \(K_m\) is characteristic of competitive inhibition. Competitive inhibitors have a shape similar to the substrate and bind reversibly to the active site (statement 1). Because they compete with substrate molecules for the active site, increasing substrate concentration outcompetes the inhibitor and restores the maximum rate \(V_{\text{max}}\) (statement 2). Statements 3 describes non-competitive inhibition, and statement 4 is incorrect because an increased \(K_m\) indicates a lower affinity.

A is correct (1 and 2 only) [1 mark].
B, C, and D are incorrect because statements 3 and 4 are false for competitive inhibition.

Cholesterol molecules interact with phospholipid fatty acid tails to regulate membrane fluidity, preventing the membrane from becoming too fluid at high temperatures and too rigid at low temperatures.

D is correct [1 mark].
A is incorrect because glycoproteins act as cell surface receptors and antigens, not as barriers to non-polar molecules.
B is incorrect because phospholipid bilayers allow small non-polar molecules through, while blocking polar molecules and ions.
C is incorrect because channel proteins facilitate diffusion down a gradient without hydrolysing ATP (ATP hydrolysis is performed by active transport pumps).

In a diploid cell with \(2n = 8\), during metaphase of mitosis, DNA replication (in S phase) has already occurred. There are 8 chromosomes, each composed of two sister chromatids joined at a single centromere, giving 8 centromeres. Each sister chromatid has two telomeres (one at each end). Thus, each duplicated chromosome contains \(2 \times 2 = 4\) telomeres. For 8 duplicated chromosomes, there are \(8 \times 4 = 32\) telomeres.

B is correct [1 mark].
A is incorrect as 16 telomeres assumes 2 telomeres per duplicated chromosome instead of 4.
C and D are incorrect because each duplicated chromosome has only 1 centromere holding sister chromatids together, not 2.

DNA polymerase can only synthesise DNA in the 5' to 3' direction by adding nucleotides to the exposed 3' -OH end of the growing strand (statement 4). Because the lagging strand template runs 5' to 3' towards the replication fork, synthesis is discontinuous, forming Okazaki fragments (statement 2). DNA ligase joins these Okazaki fragments by forming phosphodiester bonds (statement 3). Statement 1 is false because synthesis is discontinuous and proceeds 5' to 3'.

C is correct (2, 3 and 4) [1 mark].
A and B are incorrect because statement 1 is incorrect.
D is incorrect because statement 2 is also a correct statement regarding lagging strand synthesis.

Active loading involves: (1) proton pumps generating a steep electrochemical \(\text{H}^+\) gradient by pumping \(\text{H}^+\) out into the cell wall apoplast; (2) \(\text{H}^+\) diffusing back via cotransporters, carrying sucrose into companion cells against its concentration gradient; (3) sucrose diffusing down its concentration gradient into sieve tube elements via plasmodesmata; and (4) the accumulation of sucrose lowering the water potential in the sieve tube element, drawing water in by osmosis and generating high hydrostatic pressure. All 4 statements are correct.

D is correct (1, 2, 3 and 4) [1 mark].
A, B, and C are incomplete combinations.

Event 5 describes \(\text{H}^+\) ions (produced from the dissociation of carbonic acid) acting as an allosteric effector that binds to oxyhaemoglobin to form haemoglobinic acid (HHb), lowering the affinity of haemoglobin for oxygen and causing oxygen release. This shift of the oxygen dissociation curve to the right in the presence of increased carbon dioxide and \(\text{H}^+\) is the Bohr effect. (Events 3 and 4 describe the chloride shift).

A is correct [1 mark].
B is incorrect because the chloride shift is described by events 3 and 4.
C is incorrect because endocytosis is unrelated to gas transport.
D is incorrect because active transport is not involved in this passive ion exchange and buffering process.

1. Calculate the value of 1 eyepiece graticule unit (epu) at \(\times 100\) magnification:
- 10 stage micrometer divisions = \(10 \times 0.01\text{ mm} = 0.10\text{ mm} = 100\,\mu\text{m}\).
- 40 epu = \(100\,\mu\text{m}\).
- \(1\text{ epu} = \frac{100}{40} = 2.5\,\mu\text{m}\) at \(\times 100\).

2. Determine the calibration factor at \(\times 400\) magnification:
- The total magnification increased by a factor of \(\frac{400}{100} = 4\).
- Therefore, each epu at \(\times 400\) covers \(\frac{2.5\,\mu\text{m}}{4} = 0.625\,\mu\text{m}\).

3. Calculate the actual size of the cell:
- \(\text{Actual diameter} = 28 \times 0.625\,\mu\text{m} = 17.5\,\mu\text{m}\).

Award 1 mark for correct option B.
- A is incorrect: calculates \(28 \times 0.25\,\mu\text{m}\) (incorrect magnification scaling).
- B is correct: \(28 \times 0.625\,\mu\text{m} = 17.5\,\mu\text{m}\).
- C is incorrect: calculation error by a factor of 4 (omitted objective magnification adjustment: \(28 \times 2.5\,\mu\text{m} = 70.0\,\mu\text{m}\)).
- D is incorrect: multiplied by the magnification factor instead of dividing.

- Statement 1 is correct: glycine is the smallest amino acid (R-group is \(-\text{H}\)), allowing tight packing of the three chains into a triple helix.
- Statement 2 is correct: collagen is formed from three helical polypeptide chains wound around one another.
- Statement 3 is incorrect: adjacent collagen molecules in a fibril are held together by covalent cross-links formed between lysine residues, not disulfide bridges.
- Statement 4 is correct: hydrogen bonds between the peptide bonds of adjacent chains stabilise the triple helix.

Therefore, statements 1, 2, and 4 are correct.

Award 1 mark for correct option C (1, 2, and 4 only).
- A, B, and D are incorrect as they include statement 3 (covalent cross-links in collagen fibrils involve lysine side chains, not disulfide bonds).

- Inhibitor P increases \(K_m\) from \(0.4\) to \(1.2\text{ mmol dm}^{-3}\) while \(V_{\max}\) remains unchanged at \(100\text{ a.u.}\). This is characteristic of a competitive inhibitor that binds reversibly to the active site. Because \(V_{\max}\) is unchanged, high substrate concentrations can outcompete the inhibitor. Thus, statements 1 and 3 are correct.
- Inhibitor Q decreases \(V_{\max}\) from \(100\) to \(45\text{ a.u.}\) while leaving \(K_m\) unchanged at \(0.4\text{ mmol dm}^{-3}\). This is characteristic of a non-competitive inhibitor that binds to an allosteric site (statement 4 is correct). Because \(K_m\) is unchanged, the enzyme's affinity for substrate is unaffected; thus, statement 2 is incorrect.

Statements 1, 3, and 4 only are supported.

Award 1 mark for correct option B (1, 3, and 4 only).
- A, C, and D are incorrect because statement 2 is false (unchanged \(K_m\) indicates unchanged affinity).

- Statement 1 is correct: at \(0.0\%\) mass change (\(0.3\text{ mol dm}^{-3}\)), dynamic equilibrium is reached, meaning no net movement of water occurs.
- Statement 2 is incorrect: plant cells possess a rigid cellulose cell wall that prevents osmotic lysis (bursting); cells become turgid instead.
- Statement 3 is correct: the incipient net flux is zero when cell water potential equals external solution water potential (at \(0.3\text{ mol dm}^{-3}\)).
- Statement 4 is correct: at \(0.6\text{ mol dm}^{-3}\), mass decreases because water moves out of the cells by osmosis down a water potential gradient.

Therefore, statements 1, 3, and 4 are correct.

Award 1 mark for correct option D (1, 3, and 4 only).
- A, B, and C are incorrect because plant cells do not burst (statement 2 is false due to the cell wall).

1. Diploid number \(2n = 12\) means there are 12 chromosomes.
2. In S phase prior to mitosis, DNA replicates, so at late prophase, each of the 12 chromosomes consists of 2 identical sister chromatids attached at 1 centromere.
- Number of centromeres = 12.
- Total sister chromatids = \(12 \times 2 = 24\).
3. Each sister chromatid contains one double-stranded DNA molecule.
- Number of double-stranded DNA molecules = 24.
4. Each sister chromatid has 2 ends, each protected by a telomere.
- Number of telomeres = \(24 \times 2 = 48\).

Award 1 mark for correct option B.
- A is incorrect: confuses unreplicated chromosomes with replicated chromosomes (DNA = 12, telomeres = 24).
- C is incorrect: centromeres do not duplicate until anaphase (there are 12 centromeres, not 24).
- D is incorrect: confuses haploid and diploid counts.

1. Total bases in double-stranded DNA = \(1200 \times 2 = 2400\) bases.
2. Cytosine (C) = 32%, so Guanine (G) = 32% (by complementary base pairing).
- Total C + G = \(32\% + 32\% = 64\%\).
- Total A + T = \(100\% - 64\% = 36\%\).
- Adenine (A) = \(\frac{36\%}{2} = 18\%\).
- Number of adenine bases = \(0.18 \times 2400 = 432\).
3. The mRNA transcribed from one strand will be 1200 nucleotides in length.
4. Since each amino acid is encoded by a triplet codon (3 nucleotides):
- Maximum number of amino acids = \(\frac{1200}{3} = 400\).

Award 1 mark for correct option B.
- A is incorrect: calculates 18% of 1200 instead of 2400 bases, giving 216.
- C is incorrect: assumes 1 nucleotide per amino acid instead of a triplet codon.
- D is incorrect: calculates 36% of 2400 (the combined A+T count) instead of adenine alone.

- 1 decreases rate: proton pumps (\(\text{H}^+\)-ATPase) require ATP to pump \(\text{H}^+\) into the cell wall.
- 2 decreases rate: an increase in apoplast pH means a lower concentration of \(\text{H}^+\) ions in the cell wall, which reduces the proton concentration gradient needed to drive sucrose cotransport into companion cells.
- 3 decreases rate: lower mesophyll sucrose concentration decreases the supply of sucrose diffusing toward companion cells.
- 4 decreases rate: lower temperature decreases kinetic energy, reducing the rate of enzyme-mediated ATP production and carrier protein activity.

All four changes (1, 2, 3, and 4) will decrease the rate of sucrose loading.

Award 1 mark for correct option D (1, 2, 3, and 4).
- A, B, and C are incomplete as all four listed factors directly reduce the rate of proton pumping, cotransport, or sucrose availability.

- Statement 1 is correct: carbonic anhydrase in red blood cells catalyses \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\).
- Statement 2 is correct: carbonic acid dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\); haemoglobin buffers \(\text{H}^+\) by forming haemoglobinic acid (\(\text{HHb}\)), preventing a drop in cytoplasmic pH.
- Statement 3 is incorrect: hydrogencarbonate ions (\(\text{HCO}_3^-\)) diffuse out of the RBC into plasma, and chloride ions (\(\text{Cl}^-\)) diffuse INTO the RBC to maintain electrical neutrality (the chloride shift).
- Statement 4 is correct: higher \(\text{H}^+\) concentration causes haemoglobin to release oxygen more readily (Bohr effect), shifting the curve to the right.

Thus, statements 1, 2, and 4 only are correct.

Award 1 mark for correct option B (1, 2, and 4 only).
- A, C, and D are incorrect because statement 3 incorrectly states the direction of the chloride shift (chloride ions enter red blood cells, they do not leave).

1. Calculate the value of 1 eyepiece graticule unit (epu) at \(\times 100\):
- \(16\text{ stage micrometer divisions} = 16 \times 0.01\text{ mm} = 0.16\text{ mm} = 160\,\mu\text{m}\).
- \(1\text{ epu at } \times 100 = \frac{160\,\mu\text{m}}{40} = 4.0\,\mu\text{m}\).

2. Adjust calibration for \(\times 400\) magnification:
- When the objective lens changes from \(\times 10\) to \(\times 40\), magnification increases by a factor of \(4\).
- Therefore, \(1\text{ epu at } \times 400 = \frac{4.0\,\mu\text{m}}{4} = 1.0\,\mu\text{m}\).

3. Calculate actual length:
- \(\text{Actual length} = 6\text{ epu} \times 1.0\,\mu\text{m} = 6.0\,\mu\text{m}\).

B is correct (6.0 \(\mu\text{m}\)).
[1 mark for correct selection]

Amylose is an unbranched helical polymer of \(\alpha\)-glucose linked only by \(\alpha\)-1,4-glycosidic bonds.
Amylopectin is a branched polymer of \(\alpha\)-glucose containing both \(\alpha\)-1,4- and \(\alpha\)-1,6-glycosidic bonds.
Cellulose is an unbranched, straight chain polymer of \(\beta\)-glucose containing only \(\beta\)-1,4-glycosidic bonds, where alternate glucose monomers are rotated \(180^\circ\).

C is correct.
[1 mark for correct identification of monomer type, bond types, and branching status]

A non-competitive inhibitor binds to an allosteric site on the enzyme (away from the active site), altering the tertiary structure and shape of the active site so catalytic activity is reduced. Increasing substrate concentration cannot overcome this inhibition, so the maximum rate \(V_{\max}\) decreases. However, the affinity of the unaffected active sites for substrate remains unaltered, meaning the Michaelis-Menten constant \(K_{\text{m}}\) (the substrate concentration at \(\frac{1}{2}V_{\max}\)) remains unchanged.

A is correct (\(V_{\max}\) decreases and \(K_{\text{m}}\) remains unchanged).
[1 mark for correct deduction of enzyme kinetic parameters under non-competitive inhibition]

Cholesterol regulates membrane fluidity. At physiological body temperatures (around \(37\,^\circ\text{C}\)), cholesterol interacts with fatty acid hydrocarbon tails of phospholipids, reducing their sideways movement and packing them moderately tightly, which decreases membrane fluidity and prevents the membrane from becoming overly permeable to small water-soluble solutes. At low temperatures, it prevents phospholipids from packing closely and solidifying.

D is correct (It decreases membrane fluidity and reduces permeability to small hydrophilic molecules).
[1 mark]

In double-stranded DNA, adenine (\(22\%\)) pairs with thymine (\(22\%\)), meaning the remaining \(56\%\) consists of guanine (\(28\%\)) and cytosine (\(28\%\)). However, mRNA is transcribed from only ONE of the two DNA strands (the template strand). The proportions of bases on a single strand can deviate entirely from the overall double-stranded averages, provided the complementary strand has the complementary base counts. Therefore, the exact percentage of cytosine (or any individual base) in the mRNA cannot be determined without individual strand data.

D is correct (The percentage of cytosine in the mRNA cannot be determined from this information alone).
[1 mark for understanding base complementarity in double-stranded DNA vs single-stranded mRNA]

The correct sequence for phloem loading at a source is:
1st (Step 3): \(\text{H}^+\) ions are actively pumped out of companion cells into the apoplast by proton pumps consuming ATP.
2nd (Step 1): \(\text{H}^+\) ions flow down their electrochemical gradient back into the companion cell through cotransporter proteins, bringing sucrose along with them.
3rd (Step 4): Sucrose accumulates in high concentrations in companion cells and diffuses via plasmodesmata into adjacent sieve tube elements.
4th (Step 2): The lowered water potential in the sieve tube element causes water to enter by osmosis, increasing hydrostatic pressure.

B is correct (3 \(\rightarrow\) 1 \(\rightarrow\) 4 \(\rightarrow\) 2).
[1 mark for logical mechanistic order of active loading in phloem]

At the onset of ventricular systole (isovolumetric ventricular contraction), ventricular muscles contract, causing left ventricular pressure to rise rapidly above left atrial pressure. This causes the left atrioventricular (bicuspid/mitral) valve to snap shut. However, until the ventricular pressure exceeds the resting diastolic pressure in the aorta (around \(80\text{ mmHg}\)), the aortic semilunar valve remains closed. During this brief window, left ventricular pressure is higher than atrial pressure but lower than aortic pressure.

C is correct (early ventricular systole / isovolumetric contraction phase).
[1 mark for accurate pressure relation across heart chambers and vessels]

Artificial active immunity involves the injection of an attenuated/inactivated pathogen or antigen (vaccine), which stimulates the host's own immune system to produce antibodies and generates long-lived memory cells. Artificial passive immunity involves the injection of pre-formed antibodies (e.g., antivenom or antitoxin), providing immediate short-term protection without stimulating the host immune system, meaning no memory cells are produced.

A is correct.
[1 mark for identifying correct mechanism and presence of memory cells for active vs passive artificial immunity]

1. Calculate the distance represented by 10 stage micrometer divisions:
\(10 \times 0.01\text{ mm} = 0.10\text{ mm} = 100\,\mu\text{m}\).
2. Calculate the distance represented by 1 eyepiece graticule unit:
\(100\,\mu\text{m} / 25 = 4.0\,\mu\text{m}\).
3. Calculate the actual width of the cell:
\(18 \times 4.0\,\mu\text{m} = 72\,\mu\text{m}\).

B is correct (72 µm).
- A (45 µm) results from incorrectly inverting the calibration ratio (25 / 10 = 2.5 µm/unit; 18 × 2.5 = 45 µm).
- C (144 µm) results from a factor of 2 calculation error.
- D (180 µm) assumes 1 graticule unit = 10 µm without proper calibration.

Collagen is a fibrous structural protein consisting of three helical polypeptide chains wound into a triple helix (tropocollagen), with every third amino acid being glycine to allow close packing. It does not contain prosthetic groups. In contrast, haemoglobin is a globular protein consisting of four polypeptide chains (two alpha and two beta chains), each associated with a non-polypeptide haem prosthetic group containing an iron ion.

C is correct: Haemoglobin has a prosthetic group (haem) and quaternary structure with four subunits, while collagen consists of a triple helix without a prosthetic group.
- A is incorrect because collagen has a quaternary structure (triple helix).
- B is incorrect because collagen is insoluble in water and haemoglobin is soluble.
- D is incorrect because glycine is the predominant repeating residue in collagen, not haemoglobin.

A competitive inhibitor has a molecular shape complementary to the active site of the enzyme, competing directly with the substrate for binding. Because competitive inhibition can be overcome by increasing the substrate concentration, the maximum velocity (\(V_{\max}\)) remains unchanged. However, a higher substrate concentration is required to reach half \(V_{\max}\), which means the value of \(K_{\text{m}}\) increases (indicating a lower apparent affinity of the enzyme for its substrate).

A is correct: Competitive inhibitors bind to the active site, leaving Vmax unchanged and increasing Km.
- B is incorrect because non-competitive inhibitors lower Vmax and do not change Km.
- C is incorrect because allosteric inhibitors act non-competitively and lower Vmax.
- D is incorrect because covalent, irreversible inhibitors reduce the number of functional active sites and decrease Vmax.

Water moves by osmosis from a region of higher (less negative) water potential to a region of lower (more negative) water potential down a water potential gradient.
- Plant tissue water potential = \(-650\text{ kPa}\).
- Solution \(\mathbf{W}\) (\(-300\text{ kPa}\)): Water enters the tissue \(\rightarrow\) mass increases.
- Solution \(\mathbf{X}\) (\(-650\text{ kPa}\)): No net water movement \(\rightarrow\) mass remains constant.
- Solution \(\mathbf{Y}\) (\(-900\text{ kPa}\)): Water leaves the tissue (gradient = \(250\text{ kPa}\)) \(\rightarrow\) mass decreases.
- Solution \(\mathbf{Z}\) (\(-1200\text{ kPa}\)): Water leaves the tissue with the steepest gradient (gradient = \(550\text{ kPa}\)) \(\rightarrow\) greatest net loss of water and greatest decrease in mass.

D is correct: Solution Z has the lowest water potential (-1200 kPa), generating the largest water potential gradient for water to move out of the cells by osmosis.
- A causes an increase in tissue mass.
- B results in no net change in mass.
- C causes a smaller decrease in mass than Z.

1. Total number of nucleotides in the double-stranded DNA = \(1600 \times 2 = 3200\text{ bases}\).
2. According to Chargaff's rules:
- Percentage of Adenine (\(\text{A}\)) = \(22\%\)
- Percentage of Thymine (\(\text{T}\)) = \(22\%\)
- Combined \(\text{A} + \text{T} = 44\%\)
3. Combined Guanine (\(\text{G}\)) + Cytosine (\(\text{C}\)) = \(100\% - 44\% = 56\%\).
4. Since \(\%\text{G} = \%\text{C}\), the percentage of guanine = \(56\% / 2 = 28\%\).
5. Number of guanine bases = \(0.28 \times 3200 = 896\).

C is correct (896 guanine bases).
- A (352) is the number of guanine bases if calculated out of 1600 single nucleotides instead of base pairs.
- B (448) is 28% of 1600.
- D (704) is the number of adenine bases (22% of 3200 = 704).

- Statement 1 is correct: Proton pumps in the companion cell surface membrane use ATP to pump \(\text{H}^+\) out into the cell wall.
- Statement 2 is correct: \(\text{H}^+\) ions diffuse back into the companion cell down their concentration gradient through cotransporter proteins, carrying sucrose molecules against their concentration gradient.
- Statement 3 is correct: The high concentration of sucrose lowers the water potential inside the sieve tube element, drawing water in from adjacent xylem by osmosis, which raises hydrostatic pressure.
- Statement 4 is incorrect: Sucrose enters companion cells via \(\text{H}^+\)/sucrose cotransporters, not through aquaporins (which transport water).

B is correct (1, 2, and 3 only).
- Statement 4 is false because aquaporins are water-channel proteins and do not transport sucrose.

Elevated carbon dioxide and lowered pH (from carbonic acid and lactic acid) increase the concentration of hydrogen ions (\(\text{H}^+\)). Hydrogen ions bind to haemoglobin to form haemoglobinic acid (\(\text{HHb}\)), causing a conformational change that lowers haemoglobin's affinity for oxygen. This shifts the oxygen dissociation curve to the right (the Bohr shift), causing more oxygen to be released (unloaded) at respiring tissues at any given partial pressure of oxygen.

A is correct: High CO2 and low pH cause the Bohr shift, shifting the dissociation curve to the right and promoting oxygen unloading.
- B is incorrect because shifting to the right decreases affinity, not increases it.
- C is incorrect because the curve shifts to the right, not the left.
- D is incorrect because the curve shifts to the right and oxygen release is increased.

In the hybridoma method:
1. A mouse or other small mammal is injected with the target antigen to stimulate an immune response.
2. B-lymphocytes (plasma cells) producing specific antibodies against the antigen are isolated from the spleen.
3. These B-lymphocytes are fused with myeloma (cancerous plasma) cells using a fusing agent (such as polyethylene glycol, PEG) to form hybridoma cells.
4. The hybridoma cells are cultured, screened for production of the desired antibody, and cloned to produce large quantities of identical (monoclonal) antibodies.

D is correct: B-lymphocytes from the spleen of an immunised animal are fused with myeloma cells to produce hybridoma cells.
- A is incorrect because T-lymphocytes do not secrete antibodies.
- B is incorrect because red blood cells lack nuclei and do not secrete antibodies.
- C is incorrect because bacterial cells are not fused with B-lymphocytes in this technique.

1. Determine the distance represented by 20 stage micrometer divisions:
\(20 \times 0.01\text{ mm} = 0.20\text{ mm} = 200\,\mu\text{m}\).

2. Calculate the value of 1 eyepiece graticule unit (epu):
\(1\text{ epu} = \frac{200\,\mu\text{m}}{40} = 5.0\,\mu\text{m}\).

3. Calculate the actual length of the cell:
\(14\text{ epu} \times 5.0\,\mu\text{m} = 70\,\mu\text{m}\).

B is correct [1 mark].
A is incorrect: calculated as \(14 \times 2.5\,\mu\text{m} = 35\,\mu\text{m}\).
C is incorrect: calculated without converting units correctly (\(140\,\mu\text{m}\)).
D is incorrect: calculated as \(280\,\mu\text{m}\).

Glycogen is a branched polymer composed exclusively of \(\alpha\)-glucose monomers linked by \(\alpha\)-1,4-glycosidic bonds along linear chains and \(\alpha\)-1,6-glycosidic bonds at branch points. It does not contain \(\beta\)-glucose, and its 1,6-bonds are at branch points, not throughout the unbranched segments.

C is correct [1 mark].
A is incorrect: glycogen consists of \(\alpha\)-glucose, not \(\beta\)-glucose.
B is incorrect: hydrogen bonding between parallel chains forming microfibrils is a feature of cellulose.
D is incorrect: 1,6-glycosidic bonds form the branches, whereas 1,4-glycosidic bonds form the main chains.

A competitive inhibitor competes directly with the substrate for the active site. At very high substrate concentrations, the substrate outcompetes the inhibitor, so the maximum velocity (\(V_{\max}\)) remains unchanged. However, a higher concentration of substrate is required to achieve half of \(V_{\max}\), which means the enzyme's affinity appears lower and \(K_{\mathrm{m}}\) increases.

A is correct [1 mark].
B is incorrect: \(K_{\mathrm{m}}\) increases, not decreases.
C is incorrect: \(V_{\max}\) is unchanged in competitive inhibition (it decreases in non-competitive inhibition).
D is incorrect: \(V_{\max}\) does not decrease in competitive inhibition.

When the water potential of the external sucrose solution matches the internal water potential of the potato cells, there is no net movement of water by osmosis into or out of the tissue. Consequently, the potato tissue neither gains nor loses mass, corresponding to a percentage change in mass of \(0\%\).

C is correct [1 mark].
A is incorrect: highest percentage gain in mass occurs in pure water (highest external water potential).
B is incorrect: maximum rate of change does not signify isotonicity.
D is incorrect: greatest loss in mass occurs when external water potential is much lower than tissue water potential.

1. In double-stranded DNA, complementary base pairing dictates that the percentage of cytosine (C) equals the percentage of guanine (G):
\(\%\text{C} = \%\text{G} = 34\%\).

2. The total percentage of C and G is:
\(34\% + 34\% = 68\%\).

3. The remaining percentage consists of adenine (A) and thymine (T):
\(100\% - 68\% = 32\%\).

4. Since \(\%\text{A} = \%\text{T}\):
\(\%\text{A} = \frac{32\%}{2} = 16\%\).

A is correct [1 mark].
B is incorrect: \(32\%\) is the sum of adenine and thymine.
C is incorrect: \(34\%\) is the percentage of guanine.
D is incorrect: \(68\%\) is the sum of cytosine and guanine.

The Casparian strip is a band of suberin embedded in the radial and transverse cell walls of the root endodermis. Suberin is waterproof, which blocks the apoplast pathway (water movement through cell walls and intercellular spaces) and forces water and dissolved mineral ions into the symplast pathway across the selectively permeable cell surface membrane of endodermal cells.

D is correct [1 mark].
A is incorrect: the Casparian strip is in the endodermis, not the pericycle or cortex.
B is incorrect: the Casparian strip does not prevent backflow from xylem to phloem.
C is incorrect: suberin blocks the apoplast pathway, not the symplast pathway.

A shift to the right of the oxygen dissociation curve (the Bohr effect / Bohr shift) occurs when haemoglobin has a lower affinity for oxygen, promoting oxygen unloading at actively respiring tissues. This shift is caused by an increase in partial pressure of carbon dioxide (\(p\text{CO}_2\)), an increase in hydrogen ion concentration (a decrease in pH), or an increase in temperature.

B is correct [1 mark].
A is incorrect: an increase in pH (lower \([\text{H}^+]\)) shifts the curve to the left.
C is incorrect: a decrease in \(p\text{CO}_2\) shifts the curve to the left.
D is incorrect: a decrease in temperature shifts the curve to the left.

Penicillin functions by inhibiting transpeptidase enzymes that form peptide cross-links between peptidoglycan molecules during bacterial cell wall synthesis. Viruses are non-cellular entities containing a nucleic acid core surrounded by a protein coat (capsid), and they completely lack peptidoglycan cell walls as well as metabolic enzymes targeted by antibiotics.

A is correct [1 mark].
B is incorrect: penicillin does not target 70S ribosomes (aminoglycosides or tetracyclines do, but viruses do not have ribosomes anyway).
C is incorrect: viruses are not eukaryotes.
D is incorrect: viruses do not produce penicillinase as a natural mechanism of cellular wall resistance; they inherently lack peptidoglycan cell walls.

(a) Structural adaptations of companion cells include:
1. Many mitochondria to supply ATP required for active proton pumping.
2. Infolded cell surface membrane (transfer cell characteristics) to provide a large surface area for carrier/cotransporter proteins.
3. Numerous plasmodesmata connecting companion cells to sieve tube elements to facilitate rapid symplastic diffusion of sucrose.

(b) Active loading mechanism:
1. ATP-dependent proton pumps ( ext{H}^+ pumps) in the companion cell surface membrane actively transport ext{H}^+ ions out into the cell wall / apoplast.
2. This builds up a high concentration / electrochemical gradient of ext{H}^+ ions outside the cell.
3. ext{H}^+ ions diffuse back down their electrochemical gradient into the companion cell through sucrose- ext{H}^+ cotransporter proteins.
4. The movement of ext{H}^+ provides the energy to transport sucrose into the companion cell against its concentration gradient (secondary active transport/symport).
5. Sucrose then diffuses into the sieve tube element via plasmodesmata.

(c) Mass flow mechanism:
1. The high concentration of sucrose in the sieve tube element lowers its solute/water potential.
2. Water moves into the sieve tube element from nearby xylem vessels down a water potential gradient by osmosis.
3. The entry of water increases the volume and turgor/hydrostatic pressure at the source.
4. At the sink, sucrose is unloaded and used/stored, raising the water potential; water moves out into xylem/cells, reducing hydrostatic pressure.
5. The resulting hydrostatic pressure gradient between source and sink forces phloem sap to move by mass flow.

(a) [max 2]
1. Many / dense mitochondria to synthesize ATP (for active transport) ;
2. Infoldings of cell surface membrane / large surface area (of membrane) for transport / cotransporter proteins ;
3. Numerous plasmodesmata connecting to sieve tube element (for symplastic movement) ;
4. Presence of (specific) ext{H}^+ pumps / sucrose-cotransporters in membrane ;

(b) [max 4]
1. ext{H}^+ ions / protons actively pumped out of companion cell into cell wall / apoplast ;
2. (using ATP from) ATP synthase / hydrolysis of ATP / proton pump ;
3. sets up a proton / ext{H}^+ / electrochemical concentration gradient ;
4. ext{H}^+ ions move back into companion cell by facilitated diffusion via cotransporter / symporter protein ;
5. sucrose is transported simultaneously against its concentration gradient (into companion cell) ;
6. sucrose diffuses into sieve tube element via plasmodesmata ;

(c) [max 4]
1. entry / loading of sucrose lowers the water potential ( ext{AW}) in sieve tube element ;
2. water enters sieve tube element from xylem (vessels) by osmosis / down a water potential gradient ;
3. generates high hydrostatic / turgor pressure (at source) ;
4. at sink, sucrose is removed / unloaded AND water leaves sieve tube (by osmosis) ;
5. lowering hydrostatic pressure at sink ;
6. (phloem sap moves) down a hydrostatic pressure gradient from source to sink (by mass flow) ;

(a) In erythrocytes:
1. ext{CO}_2 combines with ext{H}_2 ext{O} to form carbonic acid ( ext{H}_2 ext{CO}_3).
2. This reaction is catalysed by the enzyme carbonic anhydrase.
3. Carbonic acid rapidly dissociates into ext{H}^+ and ext{HCO}_3^- ions.

(b) Chloride shift:
1. ext{HCO}_3^- diffuses out of erythrocytes into the plasma down its concentration gradient.
2. To balance the loss of negative charge and maintain electrical neutrality across the erythrocyte membrane, chloride ions ( ext{Cl}^-) move from the plasma into the erythrocyte.

(c) Buffering and Bohr effect:
1. Free ext{H}^+ ions bind to oxyhaemoglobin / haemoglobin, acting as a buffer to maintain a physiological pH.
2. This forms haemoglobinic acid ( ext{HHb}).
3. The binding of protons causes an allosteric conformational change in the globin polypeptide chains / quaternary structure.
4. This lowers haemoglobin's affinity for oxygen.
5. Consequently, oxygen is released more readily to actively respiring tissues where ext{CO}_2 / ext{H}^+ concentrations are high, shifting the oxygen dissociation curve to the right.

(a) [max 3]
1. carbon dioxide reacts / combines with water to form carbonic acid / ext{H}_2 ext{CO}_3 ;
2. catalysed by carbonic anhydrase ;
3. carbonic acid dissociates into hydrogen ions / ext{H}^+ AND hydrogen carbonate ions / ext{HCO}_3^- ;

(b) [max 2]
1. ext{HCO}_3^- diffuses out of red blood cell into plasma ;
2. chloride ions / ext{Cl}^- diffuse / move into red blood cell ;
3. to maintain electrical neutrality / preserve membrane potential / balance charge ;

(c) [max 5]
1. ext{H}^+ ions (from carbonic acid) bind to haemoglobin / oxyhaemoglobin ;
2. forming haemoglobinic acid / ext{HHb} ;
3. haemoglobin acts as a buffer / prevents blood pH decreasing / maintains neutral pH ;
4. binding of ext{H}^+ alters tertiary / quaternary structure / conformation of haemoglobin ;
5. reduces affinity of haemoglobin for oxygen ;
6. promotes release / unloading / dissociation of oxygen (from oxyhaemoglobin) ;
7. provides more oxygen to actively respiring tissues (at higher partial pressures of ext{CO}_2) / shifts dissociation curve to the right ;

(a) Catalytic mechanism:
1. The substrate enters the active site, which is not an exact rigid fit initially.
2. The active site moulds/changes conformation around the substrate to achieve a tight fit, forming an enzyme-substrate (ES) complex (induced fit).
3. This induces strain/distortion on specific chemical bonds in the substrate.
4. Consequently, the activation energy ( ext{E}_ ext{a})—the minimum energy required to start the reaction—is lowered, enabling the substrate to reach the transition state rapidly at physiological temperatures.

(b) Michaelis-Menten constant:
1. ext{K}_ ext{m} is the substrate concentration required to achieve half of the maximum rate of reaction (rac{1}{2} ext{V}_ ext{max}).
2. ext{K}_ ext{m} is inversely proportional to enzyme affinity: a lower ext{K}_ ext{m} indicates higher affinity.

(c) Comparison of inhibitors:
1. Competitive inhibitor: Has a complementary shape to the active site; competes directly with substrate; ext{K}_ ext{m} increases because higher substrate concentration is needed to achieve rac{1}{2} ext{V}_ ext{max}; ext{V}_ ext{max} is unchanged because high [S] outcompetes inhibitor.
2. Non-competitive inhibitor: Binds to an allosteric site (site other than the active site); causes a conformational change in the active site; prevents catalysis regardless of substrate concentration; ext{V}_ ext{max} decreases; ext{K}_ ext{m} remains unchanged as remaining functional active sites have unaltered affinity.

(a) [max 4]
1. active site is not fully complementary to substrate (initially) ;
2. active site changes shape / conformation as substrate binds (forming ES complex) ;
3. strains / distorts bonds in substrate / stabilizes transition state ;
4. lowers activation energy / ext{E}_ ext{a} ;
5. provides alternative reaction pathway (so reaction occurs at lower temperature) ;

(b) [max 2]
1. substrate concentration at half maximum velocity / rac{1}{2} ext{V}_ ext{max} ;
2. inverse relationship with affinity / lower ext{K}_ ext{m} = higher affinity (for substrate) ;

(c) [max 4]
1. competitive inhibitor has similar shape to substrate AND binds to active site ;
2. non-competitive inhibitor binds to allosteric site / site other than active site ;
3. non-competitive changes shape / conformation of active site (preventing ES complex formation) ;
4. competitive inhibitor increases ext{K}_ ext{m} AND ext{V}_ ext{max} is unchanged ;
5. non-competitive inhibitor decreases ext{V}_ ext{max} AND ext{K}_ ext{m} is unchanged ;
6. competitive inhibition can be overcome by increasing substrate concentration (non-competitive cannot) ;

(a) Enzyme functions:
1. DNA polymerase aligns free deoxyribonucleoside triphosphates with complementary bases on the template strand.
2. DNA polymerase catalyses the condensation reaction forming phosphodiester bonds between the 3'-OH of the existing strand and the 5'-phosphate of the incoming nucleotide.
3. DNA ligase seals single-stranded nicks on the lagging strand by catalysing phosphodiester bond formation between adjacent Okazaki fragments.

(b) Continuous vs discontinuous synthesis:
1. DNA strands in a double helix are antiparallel (one is 5' to 3', the other 3' to 5').
2. DNA polymerase can only extend a strand from the 3' end (synthesizing 5' to 3').
3. The leading strand template exposes the 3' end towards the replication fork, allowing uninterrupted synthesis.
4. The lagging strand template exposes the 5' end towards the fork, so synthesis occurs in discontinuous sections (Okazaki fragments) moving away from the fork.

(c) Translation process:
1. Ribosome attaches to mRNA at the start codon (AUG).
2. tRNA molecules with complementary anticodons bring specific amino acids to the ribosome.
3. Complementary base pairing occurs between mRNA codon and tRNA anticodon.
4. Peptide bonds are formed between adjacent amino acids (catalysed by peptidyl transferase in the large ribosomal subunit).
5. Ribosome moves along mRNA until a stop codon is encountered, releasing the completed polypeptide.

(a) [max 4]
1. DNA polymerase adds free (activated) nucleotides to exposed template strand ;
2. via complementary base pairing (A-T, C-G) ;
3. catalyses formation of phosphodiester bonds (between adjacent nucleotides) ;
4. synthesizes in 5' to 3' direction ;
5. DNA ligase joins Okazaki fragments / discontinuous segments (on lagging strand) ;
6. by forming phosphodiester bonds between sugar-phosphate backbones ;

(b) [max 3]
1. DNA strands are antiparallel / run in opposite directions (5' to 3' and 3' to 5') ;
2. DNA polymerase active site is specific to the 3' (OH) end of nucleotide / can only synthesize in 5' to 3' direction ;
3. leading strand synthesized continuously towards replication fork / unwinding site ;
4. lagging strand synthesized discontinuously (in Okazaki fragments) away from replication fork ;

(c) [max 3]
1. mRNA binds to small subunit of ribosome ;
2. tRNA carrying specific amino acid binds to mRNA codon via complementary anticodon ;
3. peptide bond forms between adjacent amino acids (catalysed by peptidyl transferase) ;
4. ribosome moves / translocates along mRNA (until stop codon reached) ;

(a) Antibody structure and function:
1. Quaternary structure: 4 polypeptide chains (2 heavy, 2 light) connected by covalent disulfide bridges.
2. Variable region ( ext{V}_ ext{H} and ext{V}_ ext{L}): Unique amino acid sequence forming two identical antigen-binding sites complementary to a specific epitope on an antigen.
3. Constant region: Identical within an antibody class, binds to Fc receptors on macrophages/neutrophils to facilitate opsonisation.
4. Hinge region: Provides flexibility, enabling the antibody to bind two distinct antigen molecules at different angles/distances to cause agglutination.

(b) Clonal selection and expansion:
1. Foreign antigen enters the body and encounters specific B-cells whose membrane-bound antibodies have complementary binding sites.
2. Antigen binds to the specific B-cell receptor (clonal selection).
3. T-helper cells release chemical signals (cytokines / interleukins) that stimulate the activated B-cell.
4. The selected B-cell undergoes repeated mitotic divisions (clonal expansion).
5. Resulting clone differentiates into plasma cells (which synthesize and secrete large quantities of soluble antibodies) and long-lived memory B-cells.

(c) Differences:
1. Natural active: Triggered by natural infection; individual's own plasma cells produce antibodies; memory cells are formed; long-lasting protection.
2. Artificial passive: Pre-synthesised antibodies are injected (e.g., antivenom/antitoxin); no immune response is mounted by recipient; no memory cells produced; short-term protection as antibodies are broken down.

(a) [max 4]
1. quaternary structure / 4 polypeptide chains (2 heavy and 2 light) joined by disulfide bonds / bridges ;
2. variable region / antigen-binding site has specific / complementary shape to one antigen ;
3. two antigen-binding sites allow binding to two antigens / agglutination / cross-linking ;
4. constant region binds to receptors on phagocytes / macrophages (facilitating opsonisation / engulfment) ;
5. hinge region provides flexibility (for binding at variable distances) ;

(b) [max 4]
1. specific B-cell has membrane receptor / antibody complementary to antigen ;
2. antigen binds to B-cell receptor (clonal selection) ;
3. T-helper cell secretes cytokines / interleukins to activate B-cell ;
4. selected B-cell divides by mitosis (clonal expansion) ;
5. differentiates into plasma cells AND memory cells ;
6. plasma cells secrete specific / monoclonal antibodies ;

(c) [max 2]
1. natural active: antigen enters body naturally via infection AND individual produces own antibodies / memory cells (long-term) ;
2. artificial passive: pre-formed / foreign antibodies injected / introduced AND no memory cells formed (short-term) ;

(a) Phospholipid bilayer:
1. Phospholipids are amphipathic: hydrophilic glycerol-phosphate heads interact with water outside and inside the cell; hydrophobic fatty acid tails orient inwards away from water.
2. This forms a continuous double-layered hydrophobic interior.
3. Non-polar, lipid-soluble molecules (e.g., steroid hormones, ext{O}_2, ext{CO}_2) dissolve easily in the hydrophobic fatty acid core and pass across by simple diffusion.
4. Polar, hydrophilic, or charged particles (e.g., ext{Na}^+, glucose) are repelled by the hydrophobic core and cannot pass without channel or carrier proteins.

(b) Facilitated diffusion vs Active transport:
1. Energy requirement: Facilitated diffusion is a passive process relying on kinetic energy of particles (no ATP); active transport requires ATP hydrolysis.
2. Direction: Facilitated diffusion occurs down a concentration gradient (high to low concentration); active transport moves solutes against their concentration gradient (low to high concentration).
3. Membrane proteins: Facilitated diffusion utilizes either protein channels or carrier proteins; active transport relies exclusively on specific carrier protein pumps (which undergo ATP-dependent conformational changes).

(c) Role of cholesterol:
1. At high temperatures: Cholesterol restricts the lateral diffusion/movement of phospholipid fatty acid tails, reducing membrane fluidity and preventing excessive permeability.
2. At low temperatures: Cholesterol disrupts regular packing of fatty acid hydrocarbon tails, preventing the bilayer from solidifying/freezing, thus preserving membrane flexibility and function.

(a) [max 4]
1. phospholipids arranged with hydrophilic heads facing aqueous environments (exterior and cytoplasm) ;
2. hydrophobic (fatty acid) tails orient inwards / face each other forming hydrophobic core ;
3. lipid-soluble / non-polar molecules dissolve in hydrophobic core / diffuse directly through bilayer ;
4. polar molecules / ions are repelled by hydrophobic core / cannot pass through non-polar fatty acids ;
5. require protein channels / carrier proteins for transport ;

(b) [max 4]
1. facilitated diffusion is passive / does not require ATP ;
2. active transport requires energy / ATP (hydrolysis) ;
3. facilitated diffusion moves substances down a concentration gradient ;
4. active transport moves substances against a concentration gradient ;
5. facilitated diffusion uses channel proteins OR carrier proteins ;
6. active transport uses (specific) carrier proteins / protein pumps only ;

(c) [max 2]
1. at high temperatures: binds to / restricts movement of phospholipids, decreasing fluidity / preventing membrane becoming too permeable ;
2. at low temperatures: disrupts close packing of fatty acid tails / prevents crystallization, maintaining fluidity / preventing solidification ;

### Detailed Worked Solution

(a) (i) Dilution Table Calculations:
Using the dilution formula \(C_1 V_1 = C_2 V_2\) where \(C_1 = 2.0\%\) and \(V_2 = 20.0\text{ cm}^3\):
- For \(1.6\%\): \(V_1 = \frac{1.6 \times 20.0}{2.0} = 16.0\text{ cm}^3\) of H; volume of W \(= 20.0 - 16.0 = 4.0\text{ cm}^3\).
- For \(1.2\%\): \(V_1 = \frac{1.2 \times 20.0}{2.0} = 12.0\text{ cm}^3\) of H; volume of W \(= 20.0 - 12.0 = 8.0\text{ cm}^3\).
- For \(0.8\%\): \(V_1 = \frac{0.8 \times 20.0}{2.0} = 8.0\text{ cm}^3\) of H; volume of W \(= 20.0 - 8.0 = 12.0\text{ cm}^3\).
- For \(0.4\%\): \(V_1 = \frac{0.4 \times 20.0}{2.0} = 4.0\text{ cm}^3\) of H; volume of W \(= 20.0 - 4.0 = 16.0\text{ cm}^3\).
- For \(2.0\%\): \(20.0\text{ cm}^3\) of H; volume of W \(= 0.0\text{ cm}^3\).

(a) (ii) Results Table:
- Table contains headings with correct slash units: `Concentration of hydrogen peroxide / %`, `Time taken for disc to rise, t / s` (Trials 1, 2, 3, and Mean), `Rate of reaction / s^-1`.
- Independent variable values ordered logically.
- All raw times recorded to whole seconds or \(0.1\text{ s}\).
- Mean time calculated correctly from repeats.
- Rate calculated accurately as \(1000 / t_{\text{mean}}\) and presented to a consistent number of significant figures (typically 2 or 3 s.f.).

(a) (iii) Uncertainty:
- Reaction time / human delay in starting or stopping the stopwatch upon observing the disc reach the surface.
- Smallest division of digital stopwatch: \(0.01\text{ s}\) (or \(0.1\text{ s}\) / \(1\text{ s}\) for analogue).

(b) (i) Graph:
- A (Axes): \(x\)-axis labelled `Concentration of hydrogen peroxide / %`, \(y\)-axis labelled `Mean rate of reaction / s^-1`.
- S (Scale): Linear scales where data occupies greater than \(50\%\) of the grid along both axes (e.g., \(x\)-axis: 0 to 2.2 with 0.2 or 0.4 per large grid unit; \(y\)-axis: 0 to 90 with 10 or 20 per large grid unit).
- P (Plotting): Points \((0.2, 14.5), (0.6, 38.0), (1.0, 58.5), (1.4, 72.0), (1.8, 80.0), (2.0, 81.5)\) plotted with small neat crosses or circled dots precisely to within half a small square.
- L (Line): Smooth best-fit curve or clean point-to-point ruled lines without double lines or excessive thickness.

(b) (ii) Unknown Determination:
- A horizontal construction line drawn from \(65.0\text{ s}^{-1}\) on the \(y\)-axis to intersect the plotted line/curve, and a vertical line dropped to the \(x\)-axis.
- Reading falls within the range \(1.16\%\) to \(1.24\%\) (e.g., \(1.20\%\)).

(b) (iii) Explanation of kinetics:
- From \(0.2\%\) to \(1.2\%\), substrate concentration is limiting: as concentration increases, substrate-active site collisions increase per unit time, forming more enzyme-substrate complexes (ESCs), causing the rate to increase proportionally.
- Above \(1.4\%\), the rate begins to plateau / level off: substrate concentration is no longer limiting; active sites of catalase become saturated, and enzyme concentration becomes the limiting factor (approaching \(V_{\text{max}}\)).

(c) (i) Uncontrolled Variables & Standardisation:
1. Temperature of solution / surroundings: Submerge reaction test tubes in a thermostatically controlled water bath set at \(25\text{ }^\circ\text{C}\).
2. Volume of enzyme absorbed by disc: Use an automatic micropipette to dispense a fixed volume (e.g. \(20\text{ }\mu\text{L}\)) of enzyme extract directly onto dry paper discs.

(c) (ii) Modification for pH:
- Prepare at least five different buffer solutions across a range of pH (e.g., pH 5, 6, 7, 8, 9).
- Keep hydrogen peroxide concentration constant (e.g. \(1.0\%\)) and mix equal volumes of buffer with substrate before placing the disc.

### Marking Scheme

(a) (i) [3 marks]
- mp1: Correct volumes of H calculated and stated for all concentrations (16.0, 12.0, 8.0, 4.0 $\text{cm}^3$).
- mp2: Correct volumes of W calculated to give a total volume of $20.0\text{ cm}^3$ for each concentration (4.0, 8.0, 12.0, 16.0 $\text{cm}^3$).
- mp3: Inclusion of $2.0\%$ ($20.0\text{ cm}^3$ H, $0.0\text{ cm}^3$ W) OR correct table format with clear headings and slash units (`/ cm^3` and `/ %`).

(a) (ii) [5 marks]
- mp1: Table drawn with ruled borders and complete headings with units: `Concentration of hydrogen peroxide / %` (or `Concentration / %`), `Time / s` (raw trials and mean), `Rate / s^-1` (or `1000/t / s^-1`).
- mp2: Data recorded for all five concentrations plus unknown U.
- mp3: All raw times recorded to the same degree of precision (whole seconds or to 1 decimal place).
- mp4: Means calculated correctly from concordant raw repeats for each concentration.
- mp5: Rate values calculated correctly ($1000 / t_{\text{mean}}$) and recorded to consistent significant figures (2 or 3 s.f.).

(a) (iii) [1 mark]
- mp1: Identifying human reaction time error in judging when disc reaches the surface AND stating the resolution/smallest division of the stopwatch (e.g. $0.01\text{ s}$ or $0.1\text{ s}$). (Do not accept just 'stopwatch error').

(b) (i) [4 marks]
- mp1 (Axes): Independent variable on $x$-axis (`Concentration of hydrogen peroxide / %`) and dependent variable on $y$-axis (`Mean rate of reaction / s^-1` or `/ s^-1`). Both axes must have clear labels and units.
- mp2 (Scale): Linear scale along both axes; plotted points occupy $>50\%$ of grid width and height; origin (0,0) or clear non-awkward increments.
- mp3 (Plots): All 6 points from Table 1.2 plotted accurately to within $\pm 0.5$ small square using small neat crosses ($+$ or $\times$) or circled dots.
- mp4 (Line): Either points joined with straight ruled lines from point to point OR a single smooth line of best fit; line does not extrapolate beyond the data range; line width $<0.5\text{ mm}$ without breaks or feathery edges.

(b) (ii) [1 mark]
- mp1: Correct reading from candidate's graph corresponding to rate $65.0\text{ s}^{-1}$ (expected $1.16\% - 1.24\%$) with construction line shown.

(b) (iii) [2 marks]
- mp1: (At lower concentrations / $<1.2\%$) Rate increases because substrate is limiting; more collisions between active sites and substrate molecules per unit time, resulting in more enzyme-substrate complexes (ESCs) formed.
- mp2: (At higher concentrations / $>1.4\%$) Rate levels off / plateaus because enzyme concentration / active sites become limiting / all active sites are saturated / $V_{\text{max}}$ is approached.

(c) (i) [3 marks]
- mp1: Identifying uncontrolled variable 1: Temperature / volume of enzyme absorbed / depth of disc placement.
- mp2: Standardisation 1: Thermostatically controlled water bath / micropipetting measured volume of enzyme onto disc.
- mp3: Standardisation 2: Matching a second distinct uncontrolled variable with its appropriate method of control.

(c) (ii) [2 marks]
- mp1: Use of at least 5 different pH buffers (e.g., pH 4, 5, 6, 7, 8).
- mp2: Keep hydrogen peroxide concentration, enzyme concentration, and temperature constant.

(a) Plan Drawing:
1. Sharp, clear, unbroken lines throughout; no shading, stippling, or cross-hatching.
2. Sector outlines one-quarter of the stem circumference, showing outer epidermis/cuticle, distinct cortex layer, vascular bundle ring, and central parenchymatous pith.
3. Vascular bundles arranged in a distinct ring with correct proportional depth (cortex, vascular bundle, and pith realistically proportioned; drawing occupies more than 50% of the allocated space).
4. Sclerenchyma caps correctly placed immediately external to the phloem.
5. Three correct label lines touching the target tissues: Xylem (innermost tissue of bundle), Phloem (between xylem and sclerenchyma cap), Sclerenchyma cap (fibrous tissue capping the outer margin of vascular bundle).

(b) High-Power Cellular Drawing:
1. Sharp, continuous lines depicting four touching cells (no gaps between adjacent cell walls).
2. Double lines representing cell walls drawn for all cells, with lumens left completely unshaded.
3. Cell walls of the two xylem vessels drawn noticeably thicker than the walls of the two adjacent xylem parenchyma cells.
4. Vessels drawn significantly larger than parenchyma cells with angular/polygonal/rounded realistic outlines.
5. Correct label with a straight ruled leader line pointing directly to one xylem vessel element.

(c)(i) Calibration calculation:
- Distance on stage micrometer = \(40 \times 0.01\text{ mm} = 0.40\text{ mm} = 400\,\mu\text{m}\).
- Value of 1 epu = \(\frac{400\,\mu\text{m}}{25\text{ epu}} = 16.0\,\mu\text{m}\).

(c)(ii) Cortex Thickness calculation:
- Mean measurement in epu = \(\frac{18.5 + 19.0 + 18.0}{3} = \frac{55.5}{3} = 18.5\text{ epu}\).
- Mean actual thickness = \(18.5\text{ epu} \times 16.0\,\mu\text{m/epu} = 296\,\mu\text{m}\).
- Answer to 3 significant figures: \(296\,\mu\text{m}\).

(d) Comparison Table:

| Feature / Observable Structure | Stem on slide K1 | Stem L1 (Fig. 1.1) |
| :--- | :--- | :--- |
| Vascular bundle arrangement | Arranged in a distinct ring / periphery | Scattered throughout the ground tissue / stem |
| Central pith / cavity | Solid parenchymatous pith present | Hollow central canal / aerenchyma present / no distinct pith |
| Sclerenchyma bundle sheath / cap | Distinct cap on outer edge of phloem | Complete sclerenchyma sheath encircling each bundle / absent cap |
| Outer stem surface / Epidermis | Smooth / continuous circular outline | Ridged / fluted stem with epidermal hairs (trichomes) |

(a) Plan drawing [max 5]:
- mp1 Sharp, continuous, unbroken lines with no sketchy overlapping lines AND strictly no shading or stippling anywhere [1]
- mp2 Size rule: Drawing occupies at least 50% of the available space AND represents a sector showing ~\(\frac{1}{4}\) of the stem outline [1]
- mp3 Correct tissue organisation: Continuous epidermis, discrete cortex, vascular bundles positioned in a ring/arc, and central pith [1]
- mp4 Proportions: Radial width of vascular bundles relative to the total cortex and pith radius is anatomically realistic (bundles neither ridiculously large nor minute) [1]
- mp5 Correct labels with ruled lines (no arrowheads) touching: xylem, phloem, AND sclerenchyma (cap) [1]

(b) High-power drawing [max 5]:
- mp1 Quality of drawing: clean, single lines, no feathery lines, drawing large enough to show detail [1]
- mp2 Group of exactly 4 touching/adjacent cells drawn (2 larger vessel elements + 2 smaller parenchyma cells) [1]
- mp3 Cell walls shown as continuous double lines with distinct middle lamella/junctions between cells [1]
- mp4 Relative wall thickness: xylem vessel walls drawn appreciably thicker than parenchyma walls [1]
- mp5 Accurate label with ruled leader line to a xylem vessel element [1]

(c)(i) Graticule calibration [max 2]:
- mp1 Correct conversion of stage micrometer distance to micrometres (\(40 \times 0.01\text{ mm} = 0.40\text{ mm} = 400\,\mu\text{m}\)) [1]
- mp2 Correct final value: \(16.0\,\mu\text{m}\) or \(16\,\mu\text{m}\) [1]

(c)(ii) Cortex measurement and actual size [max 3]:
- mp1 Calculation of mean epu = \(18.5\text{ epu}\) [1]
- mp2 Multiplication of candidate's mean epu by calibration factor from (c)(i) (e.g. \(18.5 \times 16.0\)) [1] (Allow ECF from (c)(i))
- mp3 Correct answer rounded to 3 significant figures: \(296\,\mu\text{m}\) [1]

(d) Comparison table [max 4]:
- mp1 Table drawn with ruled lines, clear column headers identifying Feature, Stem K1, and Stem L1 [1]
- mp2 First valid comparative point contrasting an observable anatomical feature (e.g., vascular bundles in a ring in K1 vs scattered in L1) [1]
- mp3 Second valid comparative point (e.g., solid cellular pith in K1 vs hollow lumen/air spaces in L1) [1]
- mp4 Third valid comparative point (e.g., sclerenchyma capped bundles in K1 vs complete fibrous sheath/no caps in L1, OR smooth epidermis in K1 vs trichomes/ridges in L1) [1]
(Do not credit non-observable physiological/biochemical features such as transpiration rate or water potential)

(a)(i)
1. Aerenchyma tissue contains large, continuous internal air spaces / channels.
2. Allows oxygen to diffuse from aerial parts / leaves down to submerged roots.
3. Rapid upward stem elongation (internode elongation) ensures leaves remain above water surface to capture oxygen and light.

(a)(ii)
1. Glycolysis breaks down glucose to pyruvate, generating a net yield of 2 ATP (via substrate-linked phosphorylation) and reduced NAD.
2. Pyruvate is decarboxylated by pyruvate decarboxylase to produce ethanal (releasing \(\text{CO}_2\)).
3. Ethanal is reduced to ethanol by the enzyme alcohol dehydrogenase / ethanol dehydrogenase.
4. This reaction oxidises reduced NAD back to \(\text{NAD}^+\), allowing glycolysis to continue.
5. Cannot continue indefinitely because ethanol is toxic / damages cell membranes / denatures proteins, and energy yield is low (only 2 ATP per glucose).

(b)(i)
\(\text{RQ} = \frac{\text{volume of } \text{CO}_2 \text{ produced}}{\text{volume of } \text{O}_2 \text{ consumed}} = \frac{10.1}{14.4} = 0.70138... \approx 0.70\)

(b)(ii)
Lipid / fatty acid (or triglyceride).

(a)(i) [max 3]
1. aerenchyma / specialised air spaces in stem and roots ;
2. allows diffusion / transport of oxygen / gases from aerial parts / leaves to roots ;
3. rapid stem / internode elongation / growth ;
4. keeps leaves / photosynthetic tissue above water level ;

(a)(ii) [max 4]
1. glycolysis produces pyruvate and reduced NAD / NADH ;
2. pyruvate is converted / decarboxylated to ethanal AND carbon dioxide / \(\text{CO}_2\) released ;
3. ethanal is reduced to ethanol ;
4. by alcohol / ethanol dehydrogenase ;
5. reduced NAD / NADH is oxidised / \(\text{NAD}^+\) is regenerated (to allow glycolysis to continue) ;
6. ethanol is toxic / poisonous / damages membranes / cell death OR low ATP yield / inefficient ;

(b)(i) [2]
1. correct formula or substitution: \(10.1 / 14.4\) ;
2. \(0.70\) (must be to 2 decimal places) ;

(b)(ii) [1]
1. lipid / fatty acid / triglyceride ;

(a)
1. Photosystems contain accessory pigments (chlorophyll b, carotenoids) that absorb photons and pass excitation energy by resonance transfer to the primary pigment / chlorophyll a reaction centre.
2. In PSII (P680), light absorption leads to photoactivation / excitation of electrons, which leave the reaction centre and enter the electron transport chain.
3. The oxygen-evolving complex associated with PSII catalyses the photolysis of water (\(2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + 4\text{e}^- + \text{O}_2\)) to replace electrons lost by P680.
4. In PSI (P700), excited electrons reduce \(\text{NADP}^+\) to reduced NADP via ferredoxin-NADP reductase in non-cyclic photophosphorylation, or cycle back via the electron transport chain in cyclic photophosphorylation generating ATP only.

(b)(i)
Enzyme X: Rubisco (Ribulose bisphosphate carboxylase / ribulose-1,5-bisphosphate carboxylase-oxygenase).
Compound Y: GP (Glycerate 3-phosphate).

(b)(ii)
1. TP is converted to glycerol, and GP is converted to fatty acids (via acetyl-CoA), which condense to form lipids / triglycerides / phospholipids.
2. GP / TP is converted to keto acids which, by the addition of ammonium / amino groups (transamination), form amino acids for protein synthesis.

(c)
1. At low light intensity, light energy is the limiting factor; light-dependent reactions are photochemical and independent of temperature, so changing temperature has no effect.
2. At high light intensity, light is in excess and temperature becomes the limiting factor because the light-independent Calvin cycle is controlled by enzymes (e.g. rubisco); at \(25\,^\circ\text{C}\) enzymes and substrates have greater kinetic energy, forming more enzyme-substrate complexes per unit time.

(a) [max 4]
1. accessory pigments / light-harvesting complex absorb light / photons and pass energy to primary pigment / reaction centre / chlorophyll a ;
2. PSII / P680 involves photolysis of water / splitting of water into \(\text{H}^+\), \(\text{e}^-\), and \(\text{O}_2\) (by oxygen-evolving complex) ;
3. excited electrons from PSII pass down ETC to generate proton gradient / ATP ;
4. PSI / P700 absorbs light to excite electrons that reduce \(\text{NADP}^+\) to reduced NADP / NADPH ;
5. PSI can participate in cyclic photophosphorylation (producing ATP only) ;

(b)(i) [2]
1. Enzyme X: rubisco / ribulose bisphosphate carboxylase (-oxygenase) ;
2. Compound Y: glycerate 3-phosphate / GP ;

(b)(ii) [2]
1. TP / GP used to produce glycerol and fatty acids to make lipids / triglycerides / phospholipids ;
2. TP / GP used (with nitrates / amino groups) to make amino acids / proteins ;

(c) [2]
1. at low light intensity, light is the limiting factor / photochemical reactions are temperature-independent ;
2. at high light intensity, temperature is limiting because Calvin cycle / enzyme-catalysed reactions / rubisco have more kinetic energy / more frequent successful collisions / more ESCs at \(25\,^\circ\text{C}\) ;

(a)
1. High hydrostatic pressure is generated in the glomerulus because the afferent arteriole has a wider diameter/lumen than the narrower efferent arteriole.
2. Fluid and small dissolved molecules are forced out through the fenestrations / pores in the capillary endothelium.
3. The basement membrane acts as a selective molecular sieve / primary filter, preventing large plasma proteins (molecular mass > 68,000) and blood cells from passing through.
4. Podocytes with interdigitating foot processes / pedicels form filtration slits through which the filtrate passes into the lumen of the Bowman's capsule.

(b)(i)
Epithelial cells of the collecting duct (or distal convoluted tubule).

(b)(ii)
1. ADH binds to complementary G-protein coupled receptors on the basal / plasma membrane of the target cell.
2. Activates the enzyme adenylyl cyclase, which catalyses the synthesis of cyclic AMP (cAMP) from ATP.
3. cAMP acts as a second messenger, activating protein kinase A (PKA), which triggers the movement of vesicles carrying aquaporins (water channel proteins) to fuse with the luminal / apical membrane, increasing membrane permeability to water.

(b)(iii)
1. Drinking water increases the water potential of blood / body fluids.
2. Hypothalamic osmoreceptors detect the increase (water enters them by osmosis causing swelling) and reduce electrical impulse transmission to the posterior pituitary gland, reducing / stopping ADH release.

(a) [max 4]
1. high hydrostatic pressure in glomerulus due to afferent arteriole being wider / larger diameter than efferent arteriole ;
2. fluid passes through fenestrations / pores in capillary endothelium ;
3. basement membrane acts as main / selective filter (preventing plasma proteins / RBCs passing) ;
4. filtration slits / gaps between podocyte foot processes / pedicels ;
5. small molecules (glucose, urea, ions, water) enter Bowman's capsule space / lumen ;

(b)(i) [1]
1. (epithelial / principal) cells of collecting duct / distal convoluted tubule ;

(b)(ii) [3]
1. binding of ADH activates G-protein AND adenylyl cyclase ;
2. adenylyl cyclase produces cyclic AMP / cAMP (from ATP) / activates protein kinase A ;
3. vesicles containing aquaporins / water channels move to and fuse with apical / luminal membrane ;
4. increases permeability of cell surface membrane to water (by osmosis) ;

(b)(iii) [2]
1. blood / plasma water potential increases / detected by osmoreceptors in hypothalamus ;
2. osmoreceptors swell / absorb water AND decrease nerve impulses to posterior pituitary ;
3. less / no ADH released into blood (resulting in less water reabsorption / dilute urine) ;

(a)
1. Action potential arrives at the presynaptic axon terminal, causing depolarisation and opening of voltage-gated \(\text{Ca}^{2+}\) channels.
2. \(\text{Ca}^{2+}\) ions diffuse into the presynaptic neurone down their electrochemical gradient.
3. Influx of \(\text{Ca}^{2+}\) causes synaptic vesicles containing acetylcholine (ACh) to move to and fuse with the presynaptic membrane (exocytosis).
4. ACh diffuses across the synaptic cleft and binds to nicotinic ligand-gated receptors on the motor end plate / sarcolemma.
5. Ligand-gated \(\text{Na}^+\) channels open, \(\text{Na}^+\) ions diffuse into the muscle fibre, generating an end-plate potential that reaches threshold to trigger an action potential along the sarcolemma and down T-tubules.

(b)(i)
I-band, H-zone, and the sarcomere itself (all three decrease; A-band remains constant).

(b)(ii)
1. \(\text{Ca}^{2+}\) ions bind to troponin, causing it to change shape.
2. This pulls tropomyosin away, exposing the myosin-binding sites on the actin filament.
3. Myosin heads bind to the exposed sites, forming actin-myosin cross-bridges.
4. The release of ADP and Pi causes the power stroke, tilting the myosin head and pulling the actin filament towards the centre of the sarcomere.
5. ATP binds to the myosin head, causing the cross-bridge to detach / break.
6. Hydrolysis of ATP by ATPase in the myosin head provides energy to recock the head back to its energized resting conformation.

(a) [max 4]
1. action potential depolarises presynaptic membrane AND voltage-gated \(\text{Ca}^{2+}\) channels open ;
2. \(\text{Ca}^{2+}\) enters presynaptic terminal by diffusion ;
3. synaptic vesicles containing acetylcholine / ACh fuse with presynaptic membrane / release ACh by exocytosis ;
4. ACh diffuses across cleft AND binds to receptors on sarcolemma / motor end plate ;
5. \(\text{Na}^+\) channels open / \(\text{Na}^+\) enters muscle cell, causing depolarisation / action potential ;

(b)(i) [2]
1. I-band AND H-zone ;
2. sarcomere ;
(Reject if A-band is included as decreasing in length)

(b)(ii) [max 4]
1. \(\text{Ca}^{2+}\) binds to troponin ;
2. troponin changes shape / conformation, displacing / moving tropomyosin ;
3. exposes myosin-binding sites on actin ;
4. myosin heads bind to actin forming cross-bridges ;
5. ATP binding causes myosin head to detach from actin ;
6. ATP hydrolysis (by ATPase) resets / recocks myosin head (to high-energy state) ;

(a)(i)
Complementary epistasis / epistasis.

(a)(ii)
9 purple : 7 white.

(a)(iii)
Total offspring = 320.
Expected ratio = 9 : 7 (total parts = 16).
Expected purple: \(\frac{9}{16} \times 320 = 180\).
Expected white: \(\frac{7}{16} \times 320 = 140\).

Calculation:
\(\chi^2 = \frac{(192 - 180)^2}{180} + \frac{(128 - 140)^2}{140} = \frac{12^2}{180} + \frac{(-12)^2}{140} = \frac{144}{180} + \frac{144}{140} = 0.800 + 1.029 = 1.829 \approx 1.83\).

(a)(iv)
1. Degrees of freedom \(= 2 - 1 = 1\), giving a critical value of 3.84 at \(p = 0.05\).
2. Calculated \(\chi^2\) (1.83) is less than the critical value (3.84), meaning \(p > 0.05\); there is no significant difference between observed and expected ratios (deviations are due to chance), so the hypothesis/ratio 9:7 is supported.

(b)
1. The regulatory gene (lacI) is constitutively expressed to synthesize an active repressor protein.
2. In the absence of lactose, the repressor protein binds specifically to the operator locus.
3. This physically blocks RNA polymerase from binding to the promoter / traversing the operon.
4. Therefore, transcription of the structural genes (lacZ, lacY, lacA) is switched off / prevented.

(a)(i) [1]
1. (complementary) epistasis ;

(a)(ii) [1]
1. 9 purple : 7 white (accept 9:7) ;

(a)(iii) [2]
1. correct expected numbers: 180 purple and 140 white ;
2. correct calculation of \(\chi^2 = 1.83\) (allow 1.828 to 1.83) ;

(a)(iv) [2]
1. degrees of freedom = 1 AND critical value is 3.84 ;
2. calculated value is less than critical value / \(p > 0.05\) AND difference between observed and expected is not significant / due to chance / hypothesis accepted ;

(b) [max 4]
1. lacI / regulatory gene is transcribed / translated to produce repressor protein ;
2. repressor protein is active / in correct conformation in absence of lactose / allolactose ;
3. repressor binds to the operator (region / DNA sequence) ;
4. prevents / blocks RNA polymerase from binding to promoter / moving along DNA ;
5. transcription / expression of structural genes / lacZ, lacY, lacA is switched off / prevented ;

(a)(i)
Frequency of homozygous recessive genotype (\(q^2\)) \(= \frac{108}{1200} = 0.09\).
Recessive allele frequency, \(q = \sqrt{0.09} = 0.30\).
Dominant allele frequency, \(p = 1 - q = 1 - 0.30 = 0.70\).

(a)(ii)
Frequency of heterozygotes \(= 2pq = 2 \times 0.70 \times 0.30 = 0.42\).
Expected number of heterozygous plants \(= 0.42 \times 1200 = 504\).

(a)(iii)
Three conditions: large population size; random mating; no gene flow / migration into or out of the population; no mutations; no selection pressure / all genotypes have equal reproductive success.

(b)
1. Sympatric speciation occurs within the same geographic location / without physical geographic barrier.
2. Reproductive isolation arises via mechanisms such as polyploidy / autopolyploidy / allopolyploidy, ecological niche differentiation, or disruptive selection for different flowering times / pollinator preferences.
3. Disruptive selection favours extreme phenotypes, preventing gene flow between sub-groups, leading to divergence of gene pools until the two groups can no longer interbreed to produce fertile offspring.

(a)(i) [2]
1. \(q^2 = 108 / 1200 = 0.09\) so \(q = 0.30\) ;
2. \(p = 1 - 0.30 = 0.70\) ;

(a)(ii) [2]
1. \(2pq = 2 \times 0.70 \times 0.30 = 0.42\) ;
2. \(0.42 \times 1200 = 504\) (allow ECF from (a)(i)) ;

(a)(iii) [3]
Any three from:
1. large population / no genetic drift ;
2. random mating ;
3. no mutation ;
4. no migration / no gene flow / closed population ;
5. no selection / all phenotypes equally fertile / viable ;

(b) [max 3]
1. speciation occurs in the same geographic area / without geographic barrier ;
2. reproductive isolation occurs (e.g. polyploidy / change in flowering times / different pollinators / behavioural isolation) ;
3. disruptive selection favours different phenotypes / adaptations to distinct ecological niches ;
4. prevents gene flow / leads to divergence of gene pools / cannot interbreed to produce fertile offspring ;

(a)
1. Membrane lipids: Archaea possess ether bonds connecting branched isoprenoid hydrocarbon chains to glycerol (sometimes forming monolayers), whereas Bacteria have ester bonds linking unbranched fatty acid chains to glycerol.
2. Cell wall: Archaea cell walls lack peptidoglycan (they have pseudopeptidoglycan / glycoproteins), whereas Bacteria cell walls contain peptidoglycan (murein).
3. Protein synthesis / genetics: Archaea have RNA polymerase with complex multi-subunit structure similar to eukaryotes and use methionine as the initiating amino acid, whereas Bacteria have simple RNA polymerase (4 subunits) and use formylmethionine (fMet).

(b)(i)
For Forest B, \(N = 100\):
- Beetle P: \((80/100)^2 = 0.80^2 = 0.6400\)
- Beetle Q: \((5/100)^2 = 0.05^2 = 0.0025\)
- Beetle R: \((5/100)^2 = 0.05^2 = 0.0025\)
- Beetle S: \((10/100)^2 = 0.10^2 = 0.0100\)

\(\sum (n/N)^2 = 0.6400 + 0.0025 + 0.0025 + 0.0100 = 0.6550\)
\(D = 1 - 0.6550 = 0.345\) (or using \(n(n-1)/N(N-1)\): \((80\times 79 + 5\times 4 + 5\times 4 + 10\times 9)/(100\times 99) = (6320+20+20+90)/9900 = 6450/9900 = 0.6515\), \(D = 1 - 0.6515 = 0.348\)).
Using the provided syllabus formula \(D = 1 - \sum (n/N)^2\):
\(D = 1 - 0.655 = 0.345\).

(b)(ii)
1. Forest A has higher biodiversity because \(D\) is closer to 1.0 (species are much more evenly distributed / high species evenness).
2. A higher diversity index means Forest A is more stable / more resilient to environmental stress / disease because it possesses more complex food webs and no single species dominates.

(c)
1. Captive breeding programmes (maintaining genetic diversity via studbooks, assisted reproduction, IVF, artificial insemination).
2. Reintroduction / release programmes to re-establish wild populations in protected habitats.
3. Research and education / raising public awareness and funding for in situ conservation.

(a) [3]
Any three differences:
1. membrane lipids: ether links in Archaea VS ester links in Bacteria / branched (isoprene) chains in Archaea VS unbranched fatty acids in Bacteria ;
2. cell wall: peptidoglycan absent in Archaea VS peptidoglycan present in Bacteria ;
3. RNA polymerase: complex (similar to eukaryotes) in Archaea VS simple in Bacteria ;
4. initiating amino acid: methionine in Archaea VS formylmethionine / fMet in Bacteria ;
5. sensitivity to antibiotics: Archaea not sensitive to chloramphenicol / penicillin VS Bacteria sensitive ;

(b)(i) [3]
1. calculated \((n/N)^2\) values: \(0.64\), \(0.0025\), \(0.0025\), \(0.01\) ;
2. sum of \((n/N)^2 = 0.655\) ;
3. \(D = 1 - 0.655 = 0.345\) ;
(If alternate formula \(\sum n(n-1)/N(N-1)\) used: allow \(D = 0.348\))

(b)(ii) [2]
1. Forest A has higher biodiversity / higher evenness / not dominated by one species ;
2. higher \(D\) indicates a more stable ecosystem / greater resilience to environmental change / complex food webs ;

(c) [2]
Any two from:
1. captive breeding programmes / maintain genetic diversity (using studbooks) ;
2. use of assisted reproduction / IVF / artificial insemination / surrogacy ;
3. reintroduction of animals into restored / protected wild habitats ;
4. education / raising public awareness / funds for conservation ;
5. scientific research into physiology / behaviour / disease treatment ;

(a)
1. Reverse transcriptase uses isolated mRNA as a template and synthesises a single strand of complementary DNA (cDNA) using DNA nucleotides.
2. The mRNA strand is digested / removed using an enzyme (e.g. RNase H) or alkali treatment.
3. DNA polymerase uses the single-stranded cDNA as a template to synthesise the second, complementary DNA strand by adding complementary nucleotides, producing a double-stranded cDNA molecule.

(b)
1. A promoter is a specific non-coding DNA sequence that allows RNA polymerase and transcription factors to bind.
2. Without a prokaryotic promoter, bacterial RNA polymerase would not recognise the human gene and transcription would not occur (ensures efficient transcription / high expression).

(c)
1. The gene for GFP is included in the plasmid vector; host bacteria transformed with the plasmid produce GFP and emit green light / fluoresce when viewed under ultraviolet (UV) light.
2. Recombinant colonies can be identified directly and non-destructively without plating on antibiotic-containing media / killing non-resistant cells.
3. Eliminates the risk of horizontal gene transfer of antibiotic resistance genes to pathogenic bacterial strains.

(d)
1. Stem cells / T-lymphocytes are removed from the bone marrow or blood of the SCID patient.
2. A normal, functional copy of the human ADA allele is inserted into a viral vector (e.g. retrovirus / lentivirus) or plasmid, which infects the cells and integrates the normal allele into the host genome.
3. The genetically modified cells expressing functional adenosine deaminase are cultured and infused back into the patient's body.

(a) [3]
1. reverse transcriptase uses (isolated) mRNA as a template (with primers) ;
2. joins (free DNA) nucleotides to form single-stranded cDNA ;
3. DNA polymerase synthesises the complementary second strand of DNA to form double-stranded cDNA ;

(b) [2]
1. provides binding site for (host) RNA polymerase / transcription factors ;
2. allows / initiates transcription (of the inserted cDNA) / ensures high level of protein expression ;

(c) [3]
1. GFP gene included in plasmid / vector AND cells that take up plasmid fluoresce (green) under UV light ;
2. easy / fast visual screening / non-destructive / does not require killing bacteria with antibiotics ;
3. prevents risk of spreading antibiotic resistance to pathogenic bacteria ;

(d) [2]
1. normal / functional ADA allele inserted into vector / retrovirus / lentivirus / plasmid ;
2. T-lymphocytes / bone marrow stem cells removed from patient AND transduced / infected with vector ;
3. modified cells producing functional ADA enzyme are transfused / injected back into patient ;

(a)
- Substrate-linked phosphorylation: Direct transfer of a phosphate group from a phosphorylated metabolic intermediate / substrate to ADP to produce ATP, without the direct involvement of an electron transport chain / oxygen.
- Oxidative phosphorylation: Formation of ATP from ADP and inorganic phosphate (\(\text{P}_\text{i}\)) driven by energy released during the transfer of electrons along the electron transport chain to oxygen, using a proton gradient and ATP synthase (chemiosmosis).

(b)(i)
\[\text{RQ} = \frac{\text{Volume of } \text{CO}_2 \text{ produced}}{\text{Volume of } \text{O}_2 \text{ absorbed}} = \frac{39.2}{56.0} = 0.70\]

(ii)
- Species A: Carbohydrate / glucose (\(\text{RQ} = 1.0\))
- Species B: Lipid / triglyceride / fatty acid (\(\text{RQ} = 0.7\))

(c)
- Contains electron transport chains / carriers (cytochromes and proteins) that transfer electrons released from reduced NAD / reduced FAD.
- Protons (\(\text{H}^+\)) are pumped from the mitochondrial matrix across the inner membrane into the intermembrane space.
- Inner mitochondrial membrane is impermeable to protons, establishing and maintaining a steep electrochemical / proton gradient.
- Protons flow back into the matrix down their electrochemical gradient through ATP synthase, driving the phosphorylation of ADP to ATP.
- Folding into cristae provides a large surface area for housing numerous electron transport chains and ATP synthase complexes.

(d)
- In ethanol fermentation, glucose only undergoes glycolysis where substrate-linked phosphorylation yields a net of only \(2\) ATP per glucose molecule.
- Glucose is only partially oxidised (pyruvate is reduced to ethanol via ethanal), leaving potential chemical energy trapped in ethanol.
- Without oxygen acting as the terminal electron acceptor, the link reaction, Krebs cycle, and oxidative phosphorylation / electron transport chain cannot function, meaning no further ATP is generated by chemiosmosis.

(a) [max 2]
- mp1: Substrate-linked is the direct transfer of a phosphate group from a phosphorylated substrate / intermediate to ADP (to form ATP);
- mp2: Oxidative phosphorylation involves electron transfer along an electron transport chain / respiratory chain / chemiosmosis / ATP synthase coupled to electron flow to oxygen;
- mp3: Substrate-linked occurs in cytoplasm / matrix whereas oxidative occurs at inner mitochondrial membrane;

(b)(i) [1]
- mp1: \(0.70\) or \(0.7\) (correct calculation \(39.2 / 56.0\));

(b)(ii) [2]
- mp1: Species A: Carbohydrate / glucose / hexose sugar;
- mp2: Species B: Lipid / fatty acid / triglyceride;

(c) [max 3]
- mp1: Contains electron carriers / electron transport chain (ETC) / cytochromes;
- mp2: Energy from electron flow used to pump \(\text{H}^+\) / protons into intermembrane space;
- mp3: (Inner membrane) is impermeable to \(\text{H}^+\) / maintains proton gradient / proton motive force;
- mp4: Protons diffuse / move back down gradient through ATP synthase (driving ATP synthesis / chemiosmosis);
- mp5: Cristae / folding provides large surface area for more ETC proteins / ATP synthase enzymes;

(d) [max 2]
- mp1: Only glycolysis operates / only substrate-linked phosphorylation occurs (giving net \(2\) ATP per glucose);
- mp2: Glucose is only partially broken down / ethanol still contains much chemical energy / incomplete oxidation;
- mp3: No oxygen to act as final electron acceptor, so link reaction / Krebs cycle / oxidative phosphorylation / ETC cannot occur;

(a)
- Arrival of an action potential causes depolarisation of the presynaptic membrane.
- This depolarisation stimulates voltage-gated calcium ion (\(\text{Ca}^{2+}\)) channels to open in the presynaptic membrane.
- Calcium ions diffuse rapidly into the presynaptic knob down their electrochemical concentration gradient.
- The influx of \(\text{Ca}^{2+}\) causes synaptic vesicles containing acetylcholine to move towards the presynaptic membrane and fuse with it, releasing acetylcholine into the synaptic cleft by exocytosis.

(b)
- The action potential travels along the T-tubules (transverse tubules) deep into the muscle fibre close to the sarcoplasmic reticulum.
- Depolarisation causes voltage-gated calcium channels in the sarcoplasmic reticulum membrane to open, allowing stored \(\text{Ca}^{2+}\) to diffuse out into the sarcoplasm / around the myofibrils.
- Calcium ions bind to troponin molecules on the actin / thin filaments.
- This causes troponin to change shape, pulling tropomyosin away from the myosin-binding sites on the actin filament, exposing these binding sites for myosin heads to form cross-bridges.

(c)
- Binding of an ATP molecule to the myosin head causes the detachment of the myosin head from the actin filament (breaking the cross-bridge).
- Hydrolysis of ATP to ADP and inorganic phosphate (\(\text{P}_\text{i}\)) by ATPase located on the myosin head releases energy, causing the myosin head to be 'cocked' / return to its upright, high-energy resting conformation.

(d)
- Deflection / bending of a sensory hair opens mechanosensitive / calcium channels, creating a receptor potential.
- If two hairs are touched within a short timeframe (approx. \(20\text{--}35\) seconds), action potentials are generated that travel across the leaf lobes.
- Action potentials stimulate proton pumps (\(\text{H}^+\) extrusion) into the cell walls of outer lobe cells / rapid entry of water / loss of turgor in hinge cells, causing rapid change in lobe curvature / convex to concave snapping closed.

(a) [max 3]
- mp1: Depolarisation of presynaptic membrane / arrival of action potential;
- mp2: Voltage-gated \(\text{Ca}^{2+}\) / calcium ion channels open;
- mp3: Calcium ions / \(\text{Ca}^{2+}\) enter / diffuse into presynaptic knob / neurone down concentration gradient;
- mp4: Synaptic vesicles fuse with presynaptic membrane to release acetylcholine via exocytosis;

(b) [max 3]
- mp1: Action potential / depolarisation passes down T-tubules / transverse system;
- mp2: Opens calcium channels in sarcoplasmic reticulum, releasing \(\text{Ca}^{2+}\) into sarcoplasm;
- mp3: \(\text{Ca}^{2+}\) binds to troponin;
- mp4: Troponin changes shape / conformation and pulls / shifts tropomyosin;
- mp5: Exposes myosin-binding sites on actin (allowing cross-bridge formation);

(c) [max 2]
- mp1: Binding of ATP causes myosin head to detach from actin / breaks cross-bridge;
- mp2: Hydrolysis of ATP (to ADP and \(\text{P}_\text{i}\)) / ATPase activity provides energy to cock / re-orient the myosin head (into high-energy state);

(d) [max 2]
- mp1: Deflection / touching of sensory hair stimulates receptor potential / opens ion channels;
- mp2: Action potential spreads across leaf lobes if stimulated twice (within approx. \(20\text{--}35\) s) / two hairs stimulated;
- mp3: Causes proton pumping / \(\text{H}^+\) entry into cell wall / acid growth hypothesis / change in turgor / water movement causing rapid elastic snap from convex to concave;

(a) Experimental design:
- Preparation of concentrations: Dilute \(0.10\text{ mol dm}^{-3}\) (\(100\text{ mmol dm}^{-3}\)) sodium malonate stock solution with distilled water using simple or serial dilution to produce at least 5 different concentrations (e.g., 0, 10, 20, 40, 80 \(\text{mmol dm}^{-3}\)).
- Standardised volumes: Use fixed volumes of yeast suspension (e.g., \(2\text{ cm}^3\)), succinate solution (e.g., \(1\text{ cm}^3\)), and DCPIP (e.g., \(1\text{ cm}^3\)) across all tubes.
- Controlled variables: Equilibrate all reaction components in a thermostatically controlled water bath at a constant temperature (e.g., \(30^\circ\text{C}\) to \(35^\circ\text{C}\)) for 5 minutes before mixing. Ensure pH is buffered if buffer is provided.
- Measurement: Zero/calibrate the colorimeter using a blank cuvette (distilled water or yeast suspension without DCPIP) using a red filter (wavelength 600–660 nm). Transfer reaction mixture to cuvette and record absorbance every 30 seconds for 5 minutes.
- Repetition: Repeat the complete procedure 3 times at each malonate concentration to calculate mean initial rates and identify anomalies.
- Safety: Wear safety goggles and gloves when handling chemical reagents (DCPIP and malonate) to prevent skin/eye contact.

(b) Control experiment: Set up a tube replacing the yeast suspension with boiled/denatured yeast suspension (or distilled water) with succinate, DCPIP, and malonate. Purpose: to verify that the reduction of DCPIP is caused by dehydrogenase activity in living yeast cells rather than non-enzymatic/abiotic reduction.

(c)(i) A red filter (600–660 nm) is used because blue solutions absorb light in the complementary red region of the spectrum, providing maximum sensitivity to changes in DCPIP concentration.

(c)(ii) Plot absorbance on the y-axis against time on the x-axis. Draw a tangent to the curve at \(t = 0\text{ s}\). Calculate the gradient of this tangent: \(\text{gradient} = \frac{\Delta \text{absorbance}}{\Delta \text{time}}\).

(d) Calculation:
- Initial mean rate at \(0.0\text{ mmol dm}^{-3} = 0.84\text{ a.u. min}^{-1}\)
- Initial mean rate at \(80.0\text{ mmol dm}^{-3} = 0.16\text{ a.u. min}^{-1}\)
- Decrease \(= 0.84 - 0.16 = 0.68\text{ a.u. min}^{-1}\)
- Percentage decrease \(= \frac{0.68}{0.84} \times 100\% = 80.952\% \approx 81.0\%\) (or \(81\%\)).

(a) Planning [Max 8 marks]:
- mp1 (Independent variable): Description of preparing at least 5 different concentrations of sodium malonate by serial or proportional dilution from stock solution with distilled water.
- mp2 (Preparation detail): Correct dilution table or explicit stated volumes of stock and water (e.g., \(8\text{ cm}^3\) stock + \(2\text{ cm}^3\) water = \(80\text{ mmol dm}^{-3}\)).
- mp3 (Dependent variable): Measurement of absorbance over time (at stated regular intervals, e.g. every 15/30 s) using a colorimeter OR time taken to reach a specific absorbance endpoint.
- mp4 (Controlled variable 1): Temperature controlled using a thermostatically controlled water bath (with stated temperature between \(25^\circ\text{C}\) and \(40^\circ\text{C}\)) AND pre-equilibrating reagents before mixing.
- mp5 (Controlled variable 2): Constant volume and concentration of yeast suspension AND constant volume/concentration of succinate solution.
- mp6 (Colorimeter procedure): Reference to zeroing/calibrating the colorimeter using a blank (cuvette with water/yeast extract) AND using a red filter/wavelength ~600–660 nm.
- mp7 (Reliability): Replicating the investigation at each concentration at least 3 times (or 3 trials per concentration) to calculate a mean.
- mp8 (Safety / Risk assessment): Identifies a specific hazard and low-risk mitigation, e.g., DCPIP/malonate is an irritant/stain, wear eye protection / gloves.

(b) Control experiment [2 marks]:
- mp1: Tube with boiled / heat-denatured yeast suspension OR distilled water in place of yeast suspension (all other reagents identical) [1].
- mp2: To confirm that the decolourisation/reduction of DCPIP is due to enzyme/dehydrogenase activity (and not non-biological reduction) [1].

(c)(i) Filter selection [1 mark]:
- Red filter / 600–660 nm because oxidised DCPIP is blue and absorbs red light (or transmits blue light) [1].

(c)(ii) Initial rate determination [2 marks]:
- mp1: Draw a tangent to the curve at time = 0 [1].
- mp2: Calculate the gradient of the tangent (\(\Delta \text{absorbance} / \Delta \text{time}\)) [1].

(d) Calculation [2 marks]:
- mp1: Correct formula/working: \(\frac{0.84 - 0.16}{0.84} \times 100\) OR \(\frac{0.68}{0.84} \times 100\) [1].
- mp2: 81.0% or 81% (allow 80.95%) [1].

(a) Null hypothesis: There is no significant difference between the mean shell aspect ratio of Littorina littorea on the exposed rocky shore and the sheltered rocky shore (any observed difference is due to chance alone).

(b) Sampling procedure:
- Establish a grid system over a defined area of the intertidal zone using two tape measures at right angles.
- Generate random coordinates using a random number table or computer random number generator.
- Place a frame quadrat at each set of coordinates and select adult L. littorea found within the quadrat, or use systematic sampling along a belt transect down the tidal gradient.

(c)(i) Validity of \(t\)-test:
1. The data are continuous (ratio of lengths/widths measured in mm).
2. Comparing two separate/independent sample means.
3. The data are assumed to be normally distributed and have similar sample variances/standard deviations.

(c)(ii) Calculation:
- \(|\bar{x}_A - \bar{x}_B| = |1.216 - 1.485| = 0.269\)
- \(s_A^2 = (0.043)^2 = 0.001849\)
- \(s_B^2 = (0.049)^2 = 0.002401\)
- \(\frac{s_A^2}{n_A} = \frac{0.001849}{15} = 0.00012327\)
- \(\frac{s_B^2}{n_B} = \frac{0.002401}{15} = 0.00016007\)
- Sum of variance terms \(= 0.00012327 + 0.00016007 = 0.00028334\)
- Square root \(= \sqrt{0.00028334} = 0.016833\)
- \(t = \frac{0.269}{0.016833} = 15.98\) (using unrounded numbers: \(t = 15.96\) to \(16.0\))

(c)(iii) Degrees of freedom: \(v = (n_A - 1) + (n_B - 1) = 14 + 14 = 28\).

(d) Interpretation:
- At \(v = 28\) and \(p = 0.05\), critical value is \(2.048\).
- Since calculated \(t\) (\(16.0\)) is greater than the critical value \(2.048\) (and also greater than critical value at \(p = 0.001\), \(3.674\)), the probability that the difference is due to chance is less than 0.05 (\(p < 0.001\)).
- Reject the null hypothesis: there is a statistically significant difference between the mean aspect ratios of the two populations.

(e) Uncontrolled variables and modification:
- Variables: Temperature fluctuations / desiccation stress, differences in predator pressure (e.g. shore crabs), food availability (algal cover), age/size distribution of snails.
- Modification: Raise juvenile snails from both sites under identical controlled laboratory conditions (common garden experiment) with constant water flow and diet to determine if differences are phenotypic plasticity or genetically determined.

(a) Null hypothesis [1 mark]:
- States that there is no significant difference between the mean shell aspect ratio of L. littorea from the exposed shore and the sheltered shore [1].
(Reject: 'no correlation' or 'no effect on growth').

(b) Sampling method [Max 3 marks]:
- mp1: Laying out two tape measures at right angles to create a grid / coordinate system [1].
- mp2: Using a random number generator / table to select coordinates [1].
- mp3: Placing a quadrat at the coordinates and sampling all snails of a standardised life stage / adult size (or setting a belt transect perpendicular to the shore) [1].

(c)(i) Justification for \(t\)-test [2 marks]:
- Any two from:
- Continuous data [1]
- Comparing the means of two distinct groups/populations [1]
- Data normally distributed (or standard deviations are similar) [1]

(c)(ii) Calculation of \(t\) [3 marks]:
- mp1: Calculation of difference in means (\(0.269\)) AND correct values of \(s_A^2/15\) (\(0.000123\)) and \(s_B^2/15\) (\(0.000160\)) [1].
- mp2: Correct denominator: \(\sqrt{0.0002833} = 0.01683\) [1].
- mp3: Correct final answer: \(16.0\) (allow \(15.96\) to \(16.0\)) [1]. (Allow ECF for minor arithmetic error in subtraction).

(c)(iii) Degrees of freedom [1 mark]:
- 28 (from \(15 + 15 - 2\)) [1].

(d) Conclusion [2 marks]:
- mp1: Calculated \(t\) (16.0) is greater than critical value (2.048) at \(p = 0.05\) (or \(p < 0.001\)) [1].
- mp2: Reject null hypothesis; the difference in shell aspect ratio between exposed and sheltered shores is statistically significant [1].

(e) Evaluation & modification [3 marks]:
- mp1: Identify two valid uncontrolled environmental factors (e.g. crab/predator density, food/algae availability, salinity/temperature fluctuations, desiccation) [1].
- mp2: Suggest measuring/quantifying the identified factor at both shore locations [1].
- mp3: Suggest a common-garden laboratory experiment / breeding both groups under identical controlled conditions to test whether the morphological difference is genetic or environmental [1].