2026 Cambridge IAS-Level Biology (9700) Practice Paper with Answers

1. Determine the actual distance represented by the stage micrometer divisions:
\(12\text{ stage divisions} = 12 \times 0.01\text{ mm} = 0.12\text{ mm} = 120\ \mu\text{m}\).

2. Calculate the value of one eyepiece graticule unit (epu):
\(40\text{ epu} = 120\ \mu\text{m} \implies 1\text{ epu} = \frac{120\ \mu\text{m}}{40} = 3.0\ \mu\text{m}\).

3. Calculate the actual length of the guard cell:
\(\text{Actual length} = 18\text{ epu} \times 3.0\ \mu\text{m} = 54\ \mu\text{m}\).

Therefore, the correct option is C.

[1] C — \(54\ \mu\text{m}\)

- Amylose: Polymer of \(\alpha\)-glucose containing only \(\alpha\)-1,4-glycosidic bonds, which coil into an unbranched helical structure.
- Amylopectin: Polymer of \(\alpha\)-glucose containing \(\alpha\)-1,4-glycosidic bonds and \(\alpha\)-1,6-glycosidic bonds at branch points, forming a branched chain.
- Cellulose: Polymer of \(\beta\)-glucose containing only \(\beta\)-1,4-glycosidic bonds (with alternating inverted glucose residues), forming unbranched, straight chains bonded together into microfibrils by hydrogen bonds.

Option A correctly lists all features.

[1] A — Amylose: \(\alpha\)-glucose, 1,4-only, unbranched helical; Amylopectin: \(\alpha\)-glucose, 1,4 and 1,6, branched; Cellulose: \(\beta\)-glucose, 1,4-only, straight chain

- Inhibitor X increases \(K_{\text{m}}\) without changing \(V_{\text{max}}\), which is characteristic of a competitive inhibitor. Competitive inhibitors have a shape complementary to the active site, competing with the substrate for binding. High substrate concentrations overcome the inhibition, so \(V_{\text{max}}\) is reached, but higher substrate concentration is required to reach \(\frac{1}{2}V_{\text{max}}\), raising \(K_{\text{m}}\).
- Inhibitor Y decreases \(V_{\text{max}}\) without altering \(K_{\text{m}}\), which is characteristic of a non-competitive inhibitor. Non-competitive inhibitors bind to an allosteric site, altering the enzyme conformation and reducing the number of functional active sites, thus decreasing \(V_{\text{max}}\).

Therefore, B is correct.

[1] B — Inhibitor X competes with substrate for the active site; inhibitor Y binds to an allosteric site and reduces the number of functional active sites.

Let \(s\) be the side length of a cube in mm.
The total volume of the cube is \(V_{\text{total}} = s^3\).
Acid penetrates \(1.5\text{ mm}\) from each surface, leaving an inner unpenetrated cube of side length \(s - 2(1.5) = s - 3\text{ mm}\).
The volume remaining pink is \(V_{\text{pink}} = (s - 3)^3\).

The percentage of volume remaining pink is:
\[\text{Percentage pink} = \left(\frac{s - 3}{s}\right)^3 \times 100\% = \left(1 - \frac{3}{s}\right)^3 \times 100\%\]

- Cube \(10\text{ mm}\): \(\left(1 - \frac{3}{10}\right)^3 = (0.7)^3 = 0.343 = 34.3\%\)
- Cube \(20\text{ mm}\): \(\left(1 - \frac{3}{20}\right)^3 = (0.85)^3 \approx 0.614 = 61.4\%\)
- Cube \(30\text{ mm}\): \(\left(1 - \frac{3}{30}\right)^3 = (0.9)^3 = 0.729 = 72.9\%\)
- Cube \(40\text{ mm}\): \(\left(1 - \frac{3}{40}\right)^3 = (0.925)^3 \approx 0.791 = 79.1\%\)

As the size of the cube increases, the surface area to volume ratio decreases, meaning a smaller fraction of the total volume is penetrated by diffusion within a fixed time. Thus, the \(40\text{ mm}\) cube has the greatest percentage of volume remaining pink.

[1] D — \(40\text{ mm}\) cube

1. In \(\text{G}_1\) phase, the cell contains \(24\) unreplicated chromosomes, containing \(6.8\text{ pg}\) of DNA.
2. During the S phase of interphase, semi-conservative DNA replication occurs, doubling the DNA content from \(6.8\text{ pg}\) to \(13.6\text{ pg}\).
3. By metaphase of mitosis, each of the \(24\) chromosomes consists of two sister chromatids joined at the centromere. Thus, there are \(24 \times 2 = 48\) chromatids.

Therefore, at metaphase, there are 48 chromatids and \(13.6\text{ pg}\) of DNA (Option D).

[1] D — 48 chromatids and \(13.6\text{ pg}\) of DNA

1. Number of tRNA molecules: Each of the 240 amino acids in the polypeptide is delivered to the ribosome by a tRNA molecule. The stop codon does not code for an amino acid and does not bind a tRNA molecule (it binds release factors). Thus, a minimum of 240 tRNA molecules is required.
2. Number of nucleotides in coding region: The coding region consists of 240 sense codons (including the start codon coding for methionine) plus 1 stop codon = 241 codons in total.
Since each codon consists of 3 nucleotides: \(241 \times 3 = 723\text{ nucleotides}\).

Therefore, the correct row is B.

[1] B — 240 tRNA molecules and 723 nucleotides

The correct sequence for phloem loading and mass flow:
1. Proton pumps actively transport \(\text{H}^+\) ions out of companion cells into the cell wall (apoplast), creating an electrochemical/concentration gradient.
2. \(\text{H}^+\) ions diffuse back into the companion cells through cotransporter proteins, bringing sucrose molecules with them against the sucrose concentration gradient.
3. Sucrose diffuses down its concentration gradient from companion cells into sieve tube elements via plasmodesmata (symplast pathway).
4. High concentration of sucrose lowers the water potential in sieve tube elements, causing water to enter from adjacent xylem vessels by osmosis, generating high hydrostatic pressure that drives mass flow to sinks.

Option A correctly summarises this mechanism.

[1] A — \(\text{H}^+\) ions pumped out of companion cells \(\rightarrow\) \(\text{H}^+\) and sucrose cotransported into companion cells \(\rightarrow\) sucrose diffuses via plasmodesmata into sieve tubes \(\rightarrow\) water enters sieve tubes by osmosis

During ventricular systole:
- As the ventricles contract, intraventricular pressure rises rapidly. When ventricular pressure exceeds atrial pressure, the atrioventricular (AV) valves are pushed shut, preventing backflow of blood into the atria.
- As ventricular contraction continues, ventricular pressure rises further until it exceeds the pressure in the aorta/pulmonary artery. This pressure difference forces the semilunar valves open, allowing blood to be ejected into the arteries.

Therefore, Option B is correct.

[1] B — AV valves close when ventricular pressure exceeds atrial pressure; Semilunar valves open when ventricular pressure exceeds arterial pressure.

1. 10 stage micrometer divisions = \(10 \times 10\text{ μm} = 100\text{ μm}\).
2. 40 eyepiece graticule divisions = \(100\text{ μm}\), so 1 eyepiece graticule division = \(\frac{100}{40} = 2.5\text{ μm}\).
3. Actual length of the cell = \(18 \times 2.5\text{ μm} = 45\text{ μm}\).

B is correct [1 mark].
A represents incorrect division (4.5 μm).
C is an incorrect graticule value (72 μm).
D is a factor-of-10 error (450 μm).

Statement 1 is correct: both cellulose (made of \(\beta\)-glucose) and amylopectin (made of \(\alpha\)-glucose) are polymers composed of hexose monomers.
Statement 2 is correct: cellulose has \(\beta\)-1,4-glycosidic bonds requiring alternate glucose molecules to be rotated 180°, whereas amylopectin has \(\alpha\)-1,4-bonds along chains and \(\alpha\)-1,6-bonds at branch points.
Statement 3 is correct: straight unbranched cellulose chains form extensive intermolecular hydrogen bonds yielding microfibrils with high tensile strength, whereas amylopectin is branched and compact, serving as an energy storage molecule without forming microfibrils.

A is correct [1 mark] - all 3 statements are biologically accurate.
B, C, D omit at least one correct statement.

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. This increases the Michaelis-Menten constant (\(K_m\)), reflecting a lower apparent affinity of the enzyme for its substrate. However, at sufficiently high substrate concentrations, substrate outcompetes the inhibitor, allowing the reaction to reach the original \(V_{\max}\).

B is correct [1 mark].
A describes non-competitive inhibition which decreases Vmax.
C refers to non-competitive inhibition altering turnover number.
D incorrectly states that affinity increases.

Water moves by osmosis down a water potential gradient from a region of higher water potential (less negative: \(-450\text{ kPa}\)) to a region of lower water potential (more negative: \(-750\text{ kPa}\)). Therefore, net movement of water is out of the cells into the sucrose solution. Loss of water from plant cells causes the protoplast to shrink away from the cell wall, resulting in plasmolysis.

B is correct [1 mark].
A and D have incorrect direction of net water movement.
C predicts turgid cells despite water loss.

Telomeres are non-coding repetitive DNA sequences capping the ends of eukaryotic linear chromosomes. During DNA replication, DNA polymerase cannot fully replicate the 3’ terminal end of lagging strands. Telomeres prevent the loss of essential vital coding sequences (genes) during successive cycles of replication.

C is correct [1 mark].
A is the role of centromeres.
B is the role of the nucleolus.
D is the role of cohesin/centromere.

1. Total number of nucleotides/bases = \(1200 \times 2 = 2400\).
2. Adenine = 22%, so Thymine = 22%. Total A + T = 44%.
3. Guanine + Cytosine = \(100\% - 44\% = 56\%\). Therefore, Cytosine = \(28\%\).
4. Number of Cytosine bases = \(0.28 \times 2400 = 672\).
5. Each strand contains 1200 nucleotides. A single linear strand of \(n\) nucleotides has \(n - 1\) phosphodiester bonds = \(1200 - 1 = 1199\) bonds.
6. For two linear strands, total phosphodiester bonds = \(2 \times 1199 = 2398\).

B is correct [1 mark].
A calculates base percentage incorrectly based on single-strand count.
C fails to subtract the ends of linear polynucleotide chains (gives 2400 bonds).
D gives values for only one strand.

In the root cortex, water travels via both the apoplast pathway (through cell walls and intercellular spaces) and symplast pathway (through cytoplasm via plasmodesmata). At the endodermis, the Casparian strip contains waterproof suberin in the radial and transverse cell walls, completely blocking the apoplast pathway. Water is therefore forced to cross the plasma membrane into the symplast pathway.

A is correct [1 mark].
B incorrectly states lignin blocks symplast.
C claims suberin actively transports water.
D incorrectly claims cellulose allows apoplastic passage through the Casparian strip.

During exercise, active respiration in muscle tissue releases high concentrations of carbon dioxide (\(\text{CO}_2\)). In red blood cells, carbonic anhydrase catalyses the hydration of \(\text{CO}_2\) to carbonic acid, which dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). The higher concentration of \(\text{H}^+\) ions binds to oxyhaemoglobin, forming haemoglobinic acid (\(\text{HHb}\)) and causing haemoglobin to change shape and release oxygen more readily at any given partial pressure of oxygen (a rightward shift).

C is correct [1 mark].
A would shift the curve to the left (higher pH / lower [H+]).
B describes decreased pCO2 which shifts the curve to the left.
D describes lower temperature which increases affinity and shifts curve left.

1. Determine the length of one stage micrometer division: \( 0.01\text{ mm} = 10\ \mu\text{m} \).
2. Find the total distance of 16 stage micrometer divisions: \( 16 \times 10\ \mu\text{m} = 160\ \mu\text{m} \).
3. Calculate the value of 1 eyepiece graticule unit (epu): \( 160\ \mu\text{m} / 40\text{ epu} = 4\ \mu\text{m}\text{ per epu} \).
4. Calculate the actual length of the guard cell: \( 25\text{ epu} \times 4\ \mu\text{m/epu} = 100\ \mu\text{m} \).

[1 mark] C is correct. (A = incorrect division 25/1; B = miscalculation; D = length of 40 epu).

Amylose is an unbranched polymer of \( \alpha \)-glucose with only \( \alpha \)-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 polymer of \( \beta \)-glucose with \( \beta \)-1,4-glycosidic bonds where adjacent monomers are rotated by \( 180^{\circ} \).

[1 mark] A is correct.

A competitive inhibitor binds reversibly to the active site, competing with the substrate. At high substrate concentrations, the substrate outcompetes the inhibitor, so the maximum rate of reaction (\( V_{\max} \)) remains unchanged. However, a higher substrate concentration is required to reach half of \( V_{\max} \), which means the affinity of the enzyme for the substrate decreases and \( K_{\text{m}} \) increases.

[1 mark] B is correct. (A = non-competitive inhibitor; C = incorrect effect on Km; D = both modified incorrectly).

Membrane fluidity at low temperatures is increased by having a higher proportion of unsaturated fatty acid tails with cis-double bonds. The kinks in unsaturated hydrocarbon chains prevent close packing of phospholipid molecules, maintaining membrane fluidity.

[1 mark] D is correct. (A = reduces fluidity; B = increases packing; C = longer chains increase hydrophobic interactions and decrease fluidity).

1. Total bases = \( 1200 \times 2 = 2400 \).
2. Adenine = \( 22\% \implies \text{Thymine} = 22\% \). Together \( \text{A} + \text{T} = 44\% \).
3. Guanine + Cytosine = \( 100\% - 44\% = 56\% \).
4. Guanine = \( 56\% / 2 = 28\% \). Number of guanine bases = \( 0.28 \times 2400 = 672 \).
5. Each linear single strand of \( 1200 \) nucleotides has \( 1200 - 1 = 1199 \) phosphodiester bonds.
6. Total phosphodiester bonds in both strands = \( 1199 \times 2 = 2398 \).

[1 mark] C is correct.

In phloem loading at a source, ATP is used by proton pumps to actively transport \( \text{H}^+ \) ions out of companion cells into cell walls. \( \text{H}^+ \) ions then diffuse back into companion cells down their concentration gradient through cotransporter carrier proteins, bringing sucrose along against its concentration gradient (cotransport / secondary active transport). Sucrose then diffuses into adjacent sieve tube elements via plasmodesmata (symplast pathway).

[1 mark] B is correct.

Statements 1, 2, and 3 are correct mechanisms contributing to the Bohr effect. Carbon dioxide diffuses into red blood cells where carbonic anhydrase catalyses the hydration of \( \text{CO}_2 \) to \( \text{H}_2\text{CO}_3 \), which dissociates into \( \text{H}^+ \) and \( \text{HCO}_3^- \). \( \text{H}^+ \) ions bind to haemoglobin forming haemoglobinic acid (HHb), causing an allosteric change that lowers haemoglobin's affinity for oxygen. Statement 4 is incorrect because hydrogencarbonate ions move out of the red blood cell into plasma (via the chloride shift) rather than binding to haem groups.

[1 mark] A is correct (1, 2 and 3 only).

Monoclonal antibody production involves fusing antibody-secreting plasma cells (isolated from the spleen of an immunized mammal) with myeloma cells (cancerous plasma cells that can divide indefinitely in culture). The resulting hybridoma cells divide indefinitely by mitosis while secreting a single specific monoclonal antibody.

[1 mark] D is correct. (A & B = incorrect cell types; C = myeloma cell does not secrete the target antibody).

1. Determine the length of 16 stage micrometer divisions:
\(16 \times 0.01\text{ mm} = 0.16\text{ mm} = 160\ \mu\text{m}\).

2. Determine the value of 1 eyepiece graticule division (epu):
\(40\text{ epu} = 160\ \mu\text{m}\)
\(1\text{ epu} = \frac{160}{40} = 4.0\ \mu\text{m}\).

3. Calculate the actual length of the palisade mesophyll cell:
\(\text{Actual length} = 25\text{ epu} \times 4.0\ \mu\text{m} = 100\ \mu\text{m}\).

C is correct (1 mark).
[1 mark for selecting C: \(100\ \mu\text{m}\)]
[A, B, and D are incorrect due to miscalculation of graticule division values or failure to convert mm to \(\mu\text{m}\)]

Amylopectin is a branched polysaccharide composed of \(\alpha\)-glucose residues linked by \(\alpha\)-1,4-glycosidic bonds along chains and \(\alpha\)-1,6-glycosidic bonds at branch points. Glucose monomers do not alternate in orientation.

Cellulose is an unbranched polymer of \(\beta\)-glucose residues linked by \(\beta\)-1,4-glycosidic bonds. To form the \(\beta\)-1,4-glycosidic bond, every alternate \(\beta\)-glucose residue is rotated \(180^\circ\) relative to its neighbour.

B is correct (1 mark).
Row B correctly identifies the monomers, glycosidic bond types, and the \(180^\circ\) monomer inversion present only in cellulose.

A non-competitive inhibitor binds to an allosteric site on the enzyme, altering the tertiary structure of the active site so that catalytic activity decreases, which lowers \(V_{\max}\).

Because the inhibitor does not compete with the substrate for the active site, the affinity of the unaffected active sites for the substrate remains unchanged. Since \(K_m\) is an inverse measure of enzyme affinity for its substrate, \(K_m\) remains unchanged.

D is correct (1 mark).
[1 mark for identifying non-competitive inhibition with \(K_m\) remaining unchanged]

Unsaturated fatty acid hydrocarbon tails contain carbon–carbon double bonds (\(\text{C}=\text{C}\)), which create kinks in the hydrocarbon chains. These kinks prevent the phospholipid molecules from packing closely together as temperature drops, reducing intermolecular hydrophobic interactions and lowering the temperature at which the membrane solidifies.

A is correct (1 mark).
[1 mark for identifying that kinks created by \(\text{C}=\text{C}\) double bonds prevent close packing of phospholipid tails]

1. Total number of nucleotides/bases = \(1200 \times 2 = 2400\).

2. Percentage of cytosine (C) = \(28\%\), so number of C bases = \(0.28 \times 2400 = 672\).
By complementary base pairing, number of guanine (G) bases = \(672\).
Total (C + G) = \(672 + 672 = 1344\).

3. Total (A + T) = \(2400 - 1344 = 1056\).
Since A = T, number of adenine (A) bases = \(\frac{1056}{2} = 528\).

4. For a linear double-stranded DNA molecule of \(1200\) base pairs, each of the two strands contains \(1200\) nucleotides.
Number of phosphodiester bonds per linear single strand = \(1200 - 1 = 1199\).
Total phosphodiester bonds in both strands = \(1199 \times 2 = 2398\).

A is correct (1 mark).
[1 mark for: Adenine = 528 and Phosphodiester bonds = 2398]

The correct sequence for sucrose loading into the phloem is:
1. Proton pumps actively transport \(\text{H}^+\) ions out of the companion cell cytoplasm into the cell wall using ATP (Stage 2).
2. This sets up a high electrochemical gradient of \(\text{H}^+\) outside the cell. Protons then diffuse back into the companion cell through cotransporter proteins, bringing sucrose with them against its concentration gradient (Stage 1).
3. Sucrose diffuses from the companion cell into the sieve tube element via plasmodesmata (Stage 3).
4. The accumulation of sucrose lowers the water potential inside the sieve tube, drawing in water by osmosis and generating high hydrostatic pressure (Stage 4).

Therefore, the correct sequence is 2 \(\rightarrow\) 1 \(\rightarrow\) 3 \(\rightarrow\) 4.

B is correct (1 mark).
[1 mark for selecting the correct biological sequence: 2 -> 1 -> 3 -> 4]

In actively respiring tissue:
- Carbon dioxide reacts with water inside the RBC, catalysed by carbonic anhydrase, forming carbonic acid which dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\).
- \(\text{HCO}_3^-\text{ ions}\) diffuse out of the red blood cell into the blood plasma.
- Chloride ions (\(\text{Cl}^-\)) move into the red blood cell down the electrochemical gradient to maintain electrical neutrality (the chloride shift).
- The \(\text{H}^+\) ions combine with oxyhaemoglobin to form haemoglobinic acid (\(\text{HHb}\)), causing haemoglobin to release oxygen (affinity of haemoglobin for oxygen decreases, shifting the dissociation curve to the right / Bohr effect).

C is correct (1 mark).
[1 mark for: Movement of \(\text{HCO}_3^-\): out of RBC; Movement of \(\text{Cl}^-\): into RBC; Affinity of haemoglobin for oxygen: decreases]

In the production of monoclonal antibodies, an animal (such as a mouse) is injected with the target antigen. Plasma cells (activated B-lymphocytes) producing the specific antibody are harvested from the spleen and fused with myeloma cells (cancerous/tumour plasma cells). The resulting hybridoma cells combine the ability to produce a specific antibody with the ability to divide indefinitely in cell culture.

A is correct (1 mark).
[1 mark for identifying fusion of plasma cells/B-lymphocytes with myeloma cells allowing continuous division and specific antibody production]

Step 1: Determine the length of one small division on the stage micrometer:
\(0.10\text{ mm} / 10 = 0.010\text{ mm} = 10\,\mu\text{m}\).

Step 2: Calculate the total length corresponding to \(20\) stage micrometer divisions:
\(20 \times 10\,\mu\text{m} = 200\,\mu\text{m}\).

Step 3: Calculate the value of \(1\) eyepiece graticule unit (epu):
\(200\,\mu\text{m} / 50\text{ epu} = 4.0\,\mu\text{m per epu}\).

Step 4: Calculate the actual diameter of the pollen grain:
\(15\text{ epu} \times 4.0\,\mu\text{m} = 60\,\mu\text{m}\).

B is correct [1 mark].
Calculation:
1 small division of stage micrometer = \(0.01\text{ mm} = 10\,\mu\text{m}\);
\(50\text{ epu} = 20 \times 10\,\mu\text{m} = 200\,\mu\text{m}\);
\(1\text{ epu} = 4.0\,\mu\text{m}\);
Actual diameter = \(15 \times 4.0 = 60\,\mu\text{m}\).

Glycogen is a polymer of \(\alpha\)-glucose linked by \(\alpha\)-1,4-glycosidic bonds with \(\alpha\)-1,6-glycosidic branch points. Cellulose is an unbranched polymer of \(\beta\)-glucose containing only \(\beta\)-1,4-glycosidic bonds, where alternate glucose residues are rotated through \(180^\circ\).

C is correct [1 mark].
Glycogen contains \(\alpha\)-glucose with 1,4 and 1,6 links (branched).
Cellulose contains \(\beta\)-glucose with 1,4 links only (unbranched straight chains).

A competitive inhibitor binds reversibly to the active site, competing directly with the substrate. At very high substrate concentrations, the substrate outcompetes the inhibitor, so the maximum velocity (\(V_{\max}\)) remains unchanged. However, a higher substrate concentration is required to achieve half of the maximum velocity (\(0.5\,V_{\max}\)), which means the Michaelis-Menten constant (\(K_{\text{m}}\)) increases.

A is correct [1 mark].
Competitive inhibitor:
- \(V_{\max}\) remains unchanged because excess substrate overcomes inhibition.
- \(K_{\text{m}}\) increases because apparent affinity for substrate decreases.

Cholesterol regulates membrane fluidity: at high temperatures, it restricts phospholipid movement, preventing the membrane from becoming too fluid; at low temperatures, it prevents close packing of hydrocarbon fatty acid tails, maintaining membrane fluidity and preventing crystallisation.

D is correct [1 mark].
At lower temperatures, cholesterol prevents close packing of fatty acid tails, preventing the membrane from becoming too rigid.

Event 2 (DNA replication / S phase of interphase) occurs first. Event 3 (prophase) occurs next, during which centrioles move and the nuclear envelope breaks down. Event 4 (metaphase) is when chromosomes align on the equator. Event 1 (anaphase) occurs when sister chromatids separate to opposite poles. The correct chronological sequence is 2 \(\rightarrow\) 3 \(\rightarrow\) 4 \(\rightarrow\) 1.

B is correct [1 mark].
Order:
- 2: Interphase (S phase)
- 3: Prophase
- 4: Metaphase
- 1: Anaphase

DNA template strand: \(3'\text{—TAC\,GCA\,TGG\,ACT—}5'\).
mRNA strand transcribed (complementary and antiparallel): \(5'\text{—AUG\,CGU\,ACC\,UGA—}3'\).
tRNA anticodons (complementary to mRNA codons): \(3'\text{—UAC—}5'\), \(3'\text{—GCA—}5'\), \(3'\text{—UGG—}5'\), \(3'\text{—ACU—}5'\), which written \(5'\rightarrow 3'\) or simply base-by-base correspond to UAC, GCA, UGG, ACU (identical to the DNA template base sequence with U replacing T).

C is correct [1 mark].
DNA template: TAC GCA TGG ACT
mRNA codons: AUG CGU ACC UGA
tRNA anticodons: UAC GCA UGG ACU

Sucrose loading involves:
1. Proton pumps actively transport \(\text{H}^+\) ions out of companion cells into the cell wall (apoplast) using ATP (Statement 1 is true).
2. \(\text{H}^+\) ions diffuse back into the companion cell down their concentration gradient via cotransporter proteins, bringing sucrose with them against its concentration gradient (Statement 3 is true; Statement 2 is false).
3. Sucrose then diffuses into the sieve tube element from the companion cell through connecting plasmodesmata (Statement 4 is true).
Therefore, 1, 3 and 4 are correct.

D is correct [1 mark].
Statements 1, 3, and 4 are correct mechanisms in phloem loading.
Statement 2 is incorrect because sucrose enters via cotransport, not simple diffusion.

In actively respiring tissues, carbon dioxide concentration is high. Carbon dioxide reacts with water inside red blood cells to form carbonic acid, catalysed by carbonic anhydrase. The carbonic acid dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). \(\text{H}^+\) ions combine with haemoglobin to form haemoglobinic acid (HHb), causing haemoglobin to release oxygen more readily. This shifts the oxygen dissociation curve to the right, increasing the unloading of oxygen to respiring tissues.

A is correct [1 mark].
High \(p\text{CO}_2\) causes a shift to the right, releasing more oxygen at the same \(p\text{O}_2\).

The Michaelis-Menten constant (\(K_\text{m}\)) is defined as the substrate concentration at which the enzyme-catalysed reaction proceeds at half of its maximum rate (\(\frac{1}{2}V_\text{max}\)). The relationship between \(K_\text{m}\) and enzyme affinity is inverse: an enzyme with a low \(K_\text{m}\) requires a lower concentration of substrate to reach half-maximal velocity, demonstrating a high affinity of its active site for the substrate.

1 mark for: substrate concentration at half \(V_\text{max}\) / \(0.5\,V_\text{max}\) / half maximum velocity; [1]
0.5 mark for: (low \(K_\text{m}\) means) high affinity of enzyme for substrate; [0.5]

Proton pumps in the companion cell membrane use ATP to actively pump \(\text{H}^+\) ions out of the cytoplasm into the cell wall apoplast. This builds up a steep electrochemical and concentration gradient of \(\text{H}^+\) ions. Protons return into the companion cell by facilitated diffusion via cotransporter (symporter) proteins, driving the cotransport of sucrose against its concentration gradient into the companion cell before it diffuses into the sieve tube element via plasmodesmata.

0.5 mark for: \(\text{H}^+\) / protons actively pumped out of companion cell (into cell wall) using ATP; [0.5]
0.5 mark for: establishes a \(\text{H}^+\) concentration / electrochemical gradient; [0.5]
0.5 mark for: \(\text{H}^+\) diffuse back in through cotransporter protein bringing sucrose (into companion cell) against its concentration gradient; [0.5]

Amylose is a plant storage polysaccharide consisting of \(\alpha\)-glucose residues linked by \(\alpha\)-1,4-glycosidic bonds, resulting in a compact helical conformation. Cellulose is a structural polysaccharide consisting of \(\beta\)-glucose monomers linked by \(\beta\)-1,4-glycosidic bonds, requiring alternate monomers to be rotated by 180° along the chain, resulting in straight, unbranched linear chains that form extensive intermolecular hydrogen bonds.

0.5 mark each for any three valid differences [max 1.5]:
• amylose is composed of \(\alpha\)-glucose whereas cellulose is composed of \(\beta\)-glucose;
• amylose has \(\alpha\)-1,4-glycosidic bonds whereas cellulose has \(\beta\)-1,4-glycosidic bonds / in cellulose alternate glucose units are rotated by 180°;
• amylose forms a helical / coiled chain whereas cellulose forms straight / linear / unbranched chains / sheets.

Ribosomes located in the eukaryotic cytoplasm are 80S (composed of 60S and 40S subunits), unlike the 70S ribosomes of prokaryotes and organelles. Eukaryotic nuclear DNA is organised into linear chromosomes associated with basic histone proteins, whereas prokaryotic chromosomal DNA exists as a single, circular, naked (non-histone bound) DNA molecule.

0.5 mark for: 80S (ribosomes); [0.5]
0.5 mark for: eukaryotic DNA is linear AND prokaryotic DNA is circular; [0.5]
0.5 mark for: eukaryotic DNA is associated with histones / proteins AND prokaryotic DNA is naked / not associated with histones; [0.5] (Accept: eukaryotic DNA has telomeres / prokaryotic DNA lacks telomeres)

During DNA replication, DNA polymerase synthesises new DNA strands in a 5' to 3' direction by catalysing phosphodiester bond formation between adjacent deoxyribonucleotides complementary to the template strand. Because replication on the lagging strand is discontinuous, DNA ligase is required to join adjacent Okazaki fragments by catalysing the formation of phosphodiester bonds in the sugar-phosphate backbone.

1 mark for DNA polymerase: adds (free/complementary) DNA nucleotides / forms phosphodiester bonds between adjacent nucleotides (in 5' to 3' direction / to 3' end); [1]
0.5 mark for DNA ligase: joins / links Okazaki fragments (on the lagging strand) / forms phosphodiester bonds between fragments / seals nicks in sugar-phosphate backbone; [0.5]

Active immunity results from an individual's own immune response to an antigen, where B-lymphocytes differentiate into antibody-secreting plasma cells and long-lived memory cells are formed. Passive immunity involves the direct acquisition of pre-formed antibodies from another individual or organism (e.g., across the placenta, via breast milk, or by injection of antitoxins); because host lymphocytes are not stimulated by antigens, no memory cells are produced and immunity is short-lived.

0.5 mark for: active immunity involves antibodies produced by own body / plasma cells / lymphocytes, whereas passive immunity involves antibodies received from external source / another individual / donor; [0.5]
0.5 mark for: active immunity results in the production of memory cells; [0.5]
0.5 mark for: passive immunity does not produce memory cells / provides temporary / short-term protection only; [0.5]

(a) The alginate gel matrix acts as a physical barrier to diffusion. The substrate molecules must diffuse through the pores of the alginate bead to reach the enzyme's active site, resulting in a lower concentration of substrate immediately surrounding the active sites compared to the bulk solution. This decreases the frequency of successful collisions between substrate and active site.

(b) 1. The enzyme can be easily recovered, separated from the product stream, and reused multiple times, reducing operational costs.
2. The product is not contaminated with the enzyme, eliminating the need for complex downstream purification.
3. The immobilization matrix often stabilizes the tertiary structure of the enzyme, making it more resistant to denaturation by changes in temperature or pH.

(a) max 2 marks:
1. alginate matrix / gel acts as a barrier to diffusion / slows movement of substrate;
2. substrate concentration at active site is lower than in bulk solution;
3. fewer (effective / successful) collisions / fewer enzyme-substrate complexes (ESCs) formed per unit time;

(b) max 2 marks (any two from):
4. enzyme can be easily recovered / separated / reused;
5. product is not contaminated with enzyme / no downstream separation needed;
6. enzyme has increased thermal / chemical stability / less easily denatured by heat / pH change;
7. allows continuous production process / enzyme withstands higher flow rates;

1. Proton pumps (\(\text{H}^+\) ATPases) in the companion cell surface membrane actively pump hydrogen ions (\(\text{H}^+\)) out of the cytoplasm into the cell wall / apoplast, using energy from ATP hydrolysis.
2. This creates an electrochemical / concentration gradient of \(\text{H}^+\) across the cell surface membrane (higher \(\text{H}^+\) concentration in cell wall).
3. Hydrogen ions diffuse back into the companion cell through co-transporter / symporter membrane proteins down their concentration gradient.
4. Sucrose molecules are co-transported simultaneously against their concentration gradient alongside the hydrogen ions.
5. Sucrose then diffuses from the companion cells into the sieve tube elements through plasmodesmata via the symplast pathway.

max 4 marks:
1. \(\text{H}^+\) / protons pumped out of companion cell (into cell wall / apoplast);
2. using ATP / active transport / against concentration gradient;
3. builds up a high \(\text{H}^+\) concentration / electrochemical gradient in cell wall / apoplast;
4. \(\text{H}^+\) moves back in by facilitated diffusion through a co-transporter (carrier) protein;
5. sucrose is transported (into companion cell) with \(\text{H}^+\) / against its concentration gradient;
6. sucrose diffuses into sieve tube element via plasmodesmata;

1. Collagen is a fibrous structural protein, whereas haemoglobin is a globular functional / transport protein.
2. Collagen consists of three polypeptide chains wound into a triple helix (tropocollagen), whereas haemoglobin consists of four polypeptide chains (two \(\alpha\)-globin and two \(\beta\)-globin chains) folded into a spherical quaternary conformation.
3. Collagen has a repetitive primary structure where every third amino acid is glycine, whereas haemoglobin has a non-repeating, diverse amino acid sequence.
4. Haemoglobin contains a non-protein prosthetic group (a haem group containing \(\text{Fe}^{2+}\)) in each subunit, whereas collagen does not have a prosthetic group.
5. Collagen is insoluble in water due to exposed hydrophobic R-groups, whereas haemoglobin is soluble in water with hydrophilic R-groups on the exterior.

max 4 marks (must give paired comparative points for full marks):
1. collagen is fibrous / insoluble whereas haemoglobin is globular / soluble;
2. collagen has 3 polypeptide chains whereas haemoglobin has 4 (chains / 2 \(\alpha\) and 2 \(\beta\));
3. collagen forms a (tight) triple helix / rope-like structure whereas haemoglobin is folded into a spherical / compact shape;
4. collagen has no prosthetic group whereas haemoglobin contains 4 haem groups / \(\text{Fe}^{2+}\);
5. collagen has a repeating amino acid sequence / glycine at every third position whereas haemoglobin has a variable / non-repeating sequence;
6. collagen has hydrophobic R-groups on exterior whereas haemoglobin has hydrophilic R-groups on exterior;

1. Rough endoplasmic reticulum (RER): Ribosomes attached to the cytosolic surface translate mRNA and synthesize the polypeptide chain, which enters the RER lumen/cisternae where it folds into its tertiary structure and undergoes initial processing.
2. Transport vesicles bud off from the RER membrane containing the folded protein and travel to the cis face of the Golgi body.
3. Golgi body (apparatus): Modifies the protein by adding carbohydrate/oligosaccharide chains (glycosylation) to form glycoproteins, and sorts/packages them into secretory vesicles.
4. Secretory vesicles bud off from the trans face of the Golgi body, move along microtubules toward the cell surface membrane, and fuse with it to release the glycoprotein outside the cell via exocytosis.

max 4 marks:
1. RER: ribosomes translate mRNA / synthesize polypeptide / folds protein (in lumen / cisternae);
2. transport vesicles transport protein from RER to (cis face of) Golgi body;
3. Golgi body: modifies protein by adding carbohydrate / oligosaccharide / glycosylation;
4. Golgi body: packages glycoprotein into secretory vesicles / sorts protein;
5. secretory vesicles fuse with cell surface membrane to release contents by exocytosis (ATP-requiring);

1. A phagocyte (such as a macrophage) engulfs the pathogen by endocytosis, digests it, and displays the non-self antigen on its cell surface membrane (antigen presentation).
2. A specific T-helper lymphocyte with complementary T-cell receptors binds to the presented antigen.
3. The activated T-helper cell secretes cytokines to stimulate the activation and proliferation of specific B-lymphocytes (clonal selection).
4. Selected B-lymphocytes undergo repeated mitotic divisions (clonal expansion).
5. These B-lymphocytes differentiate into antibody-secreting plasma cells (and memory cells).
6. Plasma cells synthesize and secrete large quantities of specific monoclonal antibodies into the blood and lymph.

max 4 marks:
1. macrophage / antigen-presenting cell (APC) engulfs pathogen and displays antigen on membrane;
2. T-helper cell with complementary receptor binds to antigen on APC;
3. cytokines released by T-helper cell stimulate specific B-lymphocyte / clonal selection;
4. B-lymphocyte undergoes mitosis / clonal expansion;
5. B-lymphocytes differentiate into plasma cells (and memory B cells);
6. plasma cells synthesize and secrete specific antibodies (into blood / lymph);

1. Carbon dioxide produced by respiring tissues diffuses into red blood cells down its concentration gradient.
2. Carbonic anhydrase catalyses the reversible reaction between \(\text{CO}_2\) and \(\text{H}_2\text{O}\) to form carbonic acid (\(\text{H}_2\text{CO}_3\)).
3. Carbonic acid dissociates spontaneously into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)).
4. Hydrogencarbonate ions diffuse down their concentration gradient out of the red blood cell into the blood plasma.
5. Chloride ions (\(\text{Cl}^-\)) move from the plasma into the red blood cell via an exchange protein to maintain electrochemical neutrality / balance electrical charges; this is known as the chloride shift.
6. The \(\text{H}^+\) ions bind to oxyhaemoglobin to form haemoglobinic acid (HHb), releasing oxygen (Bohr effect) and buffering intracellular pH.

max 4 marks:
1. \(\text{CO}_2\) reacts with \(\text{H}_2\text{O}\) catalysed by carbonic anhydrase to form carbonic acid (\(\text{H}_2\text{CO}_3\));
2. \(\text{H}_2\text{CO}_3\) dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\);
3. \(\text{HCO}_3^-\)(hydrogencarbonate) diffuses out of the red blood cell into the plasma;
4. \(\text{Cl}^-\) (chloride ions) diffuse / move into the red blood cell;
5. (chloride shift) maintains electrical neutrality / balances negative charge;
6. \(\text{H}^+\) binds to haemoglobin to form haemoglobinic acid / HHb (buffering pH / promoting oxygen release);

1. The two parent strands of the DNA double helix are antiparallel (one runs \(5'\) to \(3'\) and the other runs \(3'\) to \(5'\)).
2. DNA polymerase can only add free deoxyribonucleotides to the exposed \(3'\)-hydroxyl (\(-\text{OH}\)) end of an existing strand / elongates only in the \(5'\) to \(3'\) direction.
3. The leading strand is synthesized continuously in the \(5'\) to \(3'\) direction towards the advancing replication fork.
4. The lagging strand is oriented \(5'\) to \(3'\) towards the fork, so DNA polymerase must synthesize it away from the replication fork in short segments known as Okazaki fragments.
5. DNA ligase forms phosphodiester bonds between adjacent Okazaki fragments to create a continuous sugar-phosphate backbone on the lagging strand.

max 4 marks:
1. DNA strands are antiparallel / run in opposite directions (\(5'\) to \(3'\) and \(3'\) to \(5'\));
2. DNA polymerase can only synthesize / add nucleotides in the \(5'\) to \(3'\) direction / to the \(3'\) \(-\text{OH}\) end;
3. leading strand is synthesized continuously towards the replication fork / unwinding site;
4. lagging strand is synthesized discontinuously / in fragments / away from replication fork;
5. producing Okazaki fragments;
6. DNA ligase forms phosphodiester bonds / joins Okazaki fragments together (into a continuous strand);

1. Anaphase:
- The centromere of each chromosome splits / divides.
- Spindle fibres (kinetochore microtubules) shorten / contract.
- Sister chromatids are pulled apart and move to opposite poles of the cell, centromere leading, becoming individual daughter chromosomes.

2. Telophase:
- The separated sets of chromosomes reach the opposite poles of the cell.
- Chromosomes uncoil / decondense / elongate into thin chromatin threads.
- A new nuclear envelope reforms around each group of chromosomes from fragments of endomembrane system.
- Nucleoli reappear inside each newly formed nucleus and spindle fibres disintegrate.

max 4 marks:
Anaphase (max 2 marks):
1. centromeres divide / split;
2. sister chromatids pulled apart / separated (towards opposite poles);
3. spindle fibres / microtubules shorten / contract;

Telophase (max 2 marks):
4. chromatids / chromosomes reach opposite poles;
5. chromosomes uncoil / decondense / become diffuse / chromatin;
6. nuclear envelopes reform (around each group of chromosomes);
7. nucleoli reappear / spindle breaks down;

(a) Protons (\(\text{H}^+\)) that were actively pumped out of the companion cell move back into the cell down their electrochemical / concentration gradient via facilitated diffusion through a cotransporter (symporter) protein in the cell surface membrane. The movement of protons provides the energy to transport sucrose molecules simultaneously into the companion cell against their concentration gradient.

(b) The loading of sucrose into the sieve tube element at the source lowers the water potential inside the phloem sap. Water enters the sieve tube element from adjacent xylem vessels by osmosis, generating a high hydrostatic pressure. At the sink, sucrose is actively unloaded and utilised or stored, raising the water potential so water exits by osmosis, resulting in a low hydrostatic pressure. Phloem sap moves down this hydrostatic pressure gradient from source to sink by mass flow.

(a)
1. \(\text{H}^+\) / protons move down their concentration / electrochemical gradient through a cotransporter / symporter (protein) (into the companion cell); [1]
2. sucrose moves (simultaneously) with \(\text{H}^+\) / protons against its concentration gradient (into the companion cell); [1]
[Total: 2]

(b)
1. (entry of sucrose) lowers water potential in sieve tube so water enters by osmosis, generating high hydrostatic pressure (at source); [1]
2. (unloading of sucrose at sink) raises water potential so water leaves by osmosis / lowers hydrostatic pressure AND sap moves down hydrostatic pressure gradient (from source to sink); [1]
[Total: 2]

(a)(i) Compound Z is a competitive inhibitor because \(V_{\max}\) is unchanged while \(K_m\) increases.

(a)(ii) Compound Z has a complementary shape similar to that of the substrate and binds reversibly to the enzyme's active site, competing directly with substrate molecules. Because inhibitor molecules occupy some active sites, a higher substrate concentration is needed to achieve half the maximum velocity (\(\frac{1}{2}V_{\max}\)), which increases the \(K_m\) value. At very high (saturating) substrate concentrations, substrate molecules outcompete the inhibitor, allowing all active sites to be occupied, so the original \(V_{\max}\) can still be reached.

(b) A non-competitive inhibitor binds to an allosteric site, altering the shape of the active site. This decreases the turnover rate of the enzyme, so \(V_{\max}\) decreases, while the affinity of the remaining unaffected active sites for the substrate remains the same, so \(K_m\) remains unchanged.

(a)(i)
competitive (inhibition); [1]

(a)(ii)
1. inhibitor / compound Z has a similar shape to substrate / binds to active site / competes with substrate (for active site); [1]
2. higher substrate concentration needed to reach half \(V_{\max}\) / outcompete inhibitor (hence \(K_m\) increases) AND at high / saturating [substrate], substrate outcompetes inhibitor so \(V_{\max}\) is reached; [1]
[Total: 2]

(b)
\(V_{\max}\) decreases / lowers AND \(K_m\) remains unchanged / stays the same; [1]
[Total: 1]

(i) Percentage increase = \(\frac{\text{final value} - \text{initial value}}{\text{initial value}} \times 100\)
Increase in rate for free urease \(= 41.4 - 18.4 = 23.0\ \mu\text{g min}^{-1}\)
Percentage increase \(= \frac{23.0}{18.4} \times 100 = 125\%\)

(ii) Ratio \(= \frac{\text{rate of free urease}}{\text{rate of immobilised urease}}\)
For \(30\text{ }^{\circ}\text{C}\) ([A]): \(\frac{18.4}{12.0} = 1.5333... \approx 1.53\)
For \(50\text{ }^{\circ}\text{C}\) ([B]): \(\frac{41.4}{33.6} = 1.2321... \approx 1.23\)

(i) [1.5 marks total]
- Correct calculation of change in rate: \(41.4 - 18.4 = 23.0\) [0.5 mark];
- Correct formula application and final answer given as \(125\) or \(125\%\) (must be 3 significant figures) [1 mark];
(Allow full marks if correct answer given with no working shown)

(ii) [1 mark total]
- [A] = 1.53 [0.5 mark];
- [B] = 1.23 [0.5 mark];
(Reject if not rounded to 2 decimal places)

For [A] (Percentage change at \(0.2\text{ mol dm}^{-3}\)):
\(\text{Percentage change} = \frac{\text{Change in mass}}{\text{Initial mass}} \times 100 = \frac{+0.29}{4.15} \times 100 = +6.9879...\% \approx +7.0\%\)

For [B] (Change in mass at \(0.4\text{ mol dm}^{-3}\)):
\(\text{Change in mass} = \text{Final mass} - \text{Initial mass} = 4.08 - 4.25 = -0.17\text{ g}\)

For [C] (Percentage change at \(0.6\text{ mol dm}^{-3}\)):
\(\text{Percentage change} = \frac{-0.41}{4.10} \times 100 = -10.000...\% \approx -10.0\%\)

[2.5 marks total]
- Value for [A]: \(+7.0\) or \(+6.99\) with positive sign indicated (or clearly implied as a gain) [1 mark];
- Value for [B]: \(-0.17\) with negative sign indicated [0.5 mark];
- Value for [C]: \(-10.0\) with negative sign indicated to 1 decimal place (accept \(-10\)) [1 mark];
(Deduct 0.5 mark overall if signs (+ / -) are completely omitted)

1. Prepare a fully enclosed, ruled results table.
2. Label the first column with the independent variable: \'concentration of starch / %\'.
3. Label the dependent variable columns: \'time taken to reach end-point / s\' subdivided into \'Trial 1\', \'Trial 2\', and \'Mean\'.
4. List all concentrations in descending or ascending order (1.0, 0.8, 0.6, 0.4, 0.2, and 0.0).
5. Record raw times as whole numbers (or to a consistent degree of precision).
6. Calculate the mean time accurately for each concentration; for 0.0% starch, state \'0\' or \'no reaction / no colour change\'.
7. Ensure no units are written within the body cells of the table.

[1] Ruled table with clearly drawn borders and all internal cell lines present, with the independent variable in the first column.
[1] Correct column headings with units using solidus: \'concentration of starch / %\' AND \'time taken to reach end-point / s\' (or \'time / s\') with sub-headings for trials and mean.
[1] All raw timing data recorded to the same degree of precision (whole seconds / no fractions of a second) and no units written inside the body of the table.
[1] Correctly calculated mean times corresponding to raw data, showing that time taken increases as starch concentration increases (or time taken decreases if end-point reaction rate is measured appropriately).

1. Draw a ruled table with complete outer borders and grid lines.
2. First column heading: \'concentration of sucrose solution / mol dm^{-3}\'.
3. Subsequent column headings: \'initial mass / g\', \'final mass / g\', \'change in mass / g\', and \'percentage change in mass / %\'.
4. Enter concentrations: 0.0, 0.2, 0.4, 0.6, 0.8.
5. Record all raw masses to a consistent precision matching the balance (e.g., 2 decimal places).
6. Calculate change in mass = final mass - initial mass (with correct + or - signs).
7. Calculate percentage change in mass = \(\frac{\text{change in mass}}{\text{initial mass}} \times 100\) to 1 or 2 decimal places consistently.
8. Ensure no units appear in the data cells.

[1] Fully ruled table with all cell lines drawn, independent variable in the first column.
[1] All column headings correct with units separated by a solidus: \'concentration of sucrose solution / mol dm^{-3}\', \'initial mass / g\', \'final mass / g\', \'change in mass / g\', and \'percentage change in mass / %\'.
[1] All raw mass measurements recorded to a consistent degree of precision (e.g., all to 2 decimal places) with no units in individual data cells.
[1] Correctly calculated change in mass (with appropriate + or - signs or indication of gain/loss) and percentage change in mass to a consistent number of significant figures / decimal places.

1. Draw a completely bordered, ruled table.
2. Headings for dilution preparation: \'concentration of glucose / %\', \'volume of 1.0% glucose solution / cm^3\', and \'volume of distilled water / cm^3\'.
3. Heading for dependent variable: \'time taken for first colour change / s\'.
4. Include rows for all serial dilution concentrations (1.0, 0.5, 0.25, 0.125, 0.0625), control (0.0), and unknown sample U.
5. Record dilution volumes correctly to make a constant total volume (e.g., 10 cm^3 or 5 cm^3).
6. Record timing data in whole seconds with no units in the body of the table.
7. Results show a clear trend: higher glucose concentrations change colour in a shorter time.

[1] Completely ruled table with outer border and internal grid lines; independent variable in the first column.
[1] All column headings correct with units: \'concentration of glucose / %\', \'volume of (1.0%) glucose (solution) / cm^{3}\', \'volume of (distilled) water / cm^{3}\', and \'time (taken for first colour change) / s\'.
[1] Complete row entries showing correct proportional/serial dilution volumes for 5 concentrations, the 0.0% control, and sample U.
[1] All raw times recorded to consistent precision (whole seconds) with no units in data cells, showing time decreasing as glucose concentration increases.

To plot the graph correctly according to Cambridge practical examination conventions:

1. Axes and Labels:
- Place the independent variable (concentration of zinc sulfate / $\text{mmol dm}^{-3}$) on the horizontal ($x$) axis.
- Place the dependent variable (time taken for starch hydrolysis / $\text{s}$) on the vertical ($y$) axis.
- Ensure both axis titles include the appropriate solidus and units exactly as given in Table 1.1.

2. Scale:
- Choose a sensible linear scale for both axes (e.g., $x$-axis: $0.1\text{ mmol dm}^{-3}$ per $2\text{ cm}$; $y$-axis: $50\text{ s}$ per $2\text{ cm}$).
- Ensure that the plotted points occupy more than $50\%$ of the grid in both horizontal and vertical directions.
- Avoid awkward scales (such as multiples of $3$ or $7$).

3. Plotting:
- Plot all 5 coordinate points accurately within $\pm 0.5\text{ mm}$ (half a small grid square):
- $(0.0, 45)$
- $(0.2, 68)$
- $(0.4, 102)$
- $(0.6, 155)$
- $(0.8, 230)$
- Mark each point using a small, neat cross ($\times$) or a small dot enclosed within a circle ($\odot$).

4. Line:
- Join points point-to-point using a sharp pencil and a ruler, or draw a smooth, continuous curve of best fit.
- Ensure lines are thin, sharp, and uninterrupted (not feathery).
- Do not extrapolate the line beyond $0.0\text{ mmol dm}^{-3}$ or $0.8\text{ mmol dm}^{-3}$.

1 mark per marking point (max 4 marks):

- A (Axes): Correct labels with units on both axes: horizontal ($x$) axis labelled 'concentration of zinc sulfate / $\text{mmol dm}^{-3}$' AND vertical ($y$) axis labelled 'time taken for starch hydrolysis / $\text{s}$' [1]
(A: 'time / s' only if qualified by hydrolysis / starch breakdown; R: missing units or incorrect abbreviations)

- S (Scale): Linear, evenly spaced intervals on both axes such that data covers more than half the grid along both the $x$-axis and $y$-axis [1]
(R: awkward scales such as increments of 3, 7, etc.)

- P (Plotting): All 5 points plotted accurately to within $\pm 0.5$ small square using small, neat crosses ($\times$) or encircled dots [1]
(R: large blobs / points larger than 1 mm diameter)

- L (Line): Sharp, clear, ruled straight lines connecting adjacent points from point to point (or a smooth curve through all points) with no extrapolation beyond the first ($0.0\text{ mmol dm}^{-3}$) and last ($0.8\text{ mmol dm}^{-3}$) data points [1]
(R: thick, sketchy/feathery lines > 0.5 mm width; R: extrapolation to origin if not in data)

To produce a high-scoring low-power tissue plan diagram:
1. Use of Space and Linework: Use a sharp HB pencil to draw clear, unbroken, single lines. Avoid any sketchy, overlapping, or feathery lines. Do not add any shading or cross-hatching. The diagram should occupy at least 50% of the allotted space.
2. Tissue Layers: Accurately represent all visible tissue boundaries without drawing individual cells. For a dicotyledonous stem sector, include the outer epidermal boundary, cortex, a discrete vascular bundle, and the border to the central pith.
3. Tissue Proportions: Ensure relative layer thicknesses reflect the specimen (e.g., cortex depth relative to the radial length of the vascular bundle and pith).
4. Vascular Bundle Detail: Clearly demarcate the distinct tissue regions within the vascular bundle (sclerenchyma cap / bundle cap, phloem, cambium band, and xylem).
5. Labelling: Draw a straight, ruled label line (without an arrowhead) ending precisely on the xylem tissue, and clearly write the label 'xylem' horizontally.

Award 1 mark for each of the following points (up to 5 marks total):

1. Drawing Quality: Sharp, clear, continuous single lines without shading or sketching, drawing occupies at least 50% of the provided space.
2. No individual cells: Correct plan diagram convention showing tissue boundaries only, with no individual cells drawn anywhere.
3. Tissue layers present: Sector correctly depicts outer epidermis, cortex layer, at least one complete discrete vascular bundle, and the edge of the pith.
4. Correct proportions: Vascular bundle occupies an appropriate radial proportion of the stem sector (vascular bundle length is at least one-third of the radius from epidermis to pith center).
5. Label: Ruled label line touching the xylem region with the label 'xylem' written unambiguously.

To produce an accurate high-power cellular drawing:
1. Cell Count & Quality: Draw exactly four touching cells as specified. All lines must be sharp, smooth, single, and free of shading. The drawing should occupy more than 50% of the space provided.
2. Double Lines for Cell Walls: Plant cell walls must be drawn with double lines to show wall thickness accurately, with clean corners/junctions where cells touch.
3. Differential Wall Thickness: The xylem vessel elements should be drawn with visibly thicker secondary cell walls than the adjacent xylem parenchyma cells.
4. Cell Size and Morphology: The two xylem vessel elements must be drawn significantly larger than the two parenchyma cells, with realistic angular/circular outlines.
5. Labelling: Use a ruler to draw a straight label line ending inside the clear central space (lumen) of one xylem vessel element, labelled 'lumen'.

Award 1 mark for each of the following points (up to 5 marks total):

1. Drawing Quality & Scale: Clean, sharp, continuous single lines with no shading, overall drawing occupies at least 50% of the available space.
2. Cell Selection: Exactly four touching cells drawn (two larger vessels and two smaller parenchyma cells).
3. Cell Walls: Double lines drawn for all cell walls, showing distinct cell wall thickness (not single lines).
4. Structural Proportion & Detail: Cell walls of xylem vessel elements drawn visibly thicker than those of the adjacent parenchyma cells, with realistic lumen shapes.
5. Label: One ruled, straight label line terminating inside the cavity of a xylem vessel element, correctly labelled 'lumen'.

(i)
$$\text{Rate} = \frac{\text{Volume of oxygen produced}}{\text{Time taken in minutes}}$$
$$\text{Time in minutes} = \frac{48}{60} = 0.80\text{ min}$$
$$\text{Rate} = \frac{2.0\text{ cm}^3}{0.80\text{ min}} = 2.5\text{ cm}^3\text{ min}^{-1}$$

(ii)
Source of error: Alginate beads are made manually and may vary in diameter/surface area or enzyme content.
Improvement: Dispense sodium alginate-catalase mixture from a fixed-height micro-pipette or burette to produce uniform bead sizes.

(i) [1.5 marks]
- Correct conversion of seconds to minutes (48 / 60 = 0.80 min) OR working shown as \((2.0 / 48) \times 60\); [0.5]
- Final answer given as 2.5 with correct units \(\text{cm}^3\text{ min}^{-1}\); [1.0]

(ii) [1.0 mark]
- One valid error stated with matching improvement; [1.0]
e.g.,
- Error: bead diameter / volume / mass varies [A surface area variation];
Improvement: use a graduated syringe / fine capillary tube at constant drop height [A weigh beads to select equal mass];
- Error: gas escapes when inserting stopper/bung into boiling tube;
Improvement: use a partitioned tube / add substrate via syringe through a self-sealing suba-seal septum.

(i)
1 small stage micrometer division \(= \frac{0.1\text{ mm}}{10} = 0.01\text{ mm} = 10\,\mu\text{m}\).
20 small divisions \(= 20 \times 10\,\mu\text{m} = 200\,\mu\text{m}\).
$$1\text{ epu} = \frac{200\,\mu\text{m}}{40\text{ epu}} = 5.0\,\mu\text{m}$$

(ii)
$$\text{Actual diameter} = 13.5\text{ epu} \times 5.0\,\mu\text{m/epu} = 67.5\,\mu\text{m}$$

(iii)
Xylem vessels are often oval/elliptical or irregular due to lateral compression and secondary thickening, so measuring at several different angles provides an accurate representative mean diameter.

(i) [1.0 mark]
- Correct calculation: \(200 / 40 = 5.0\,\mu\text{m}\); [1.0]

(ii) [1.0 mark]
- Correct multiplication of candidate's (i) \(\times 13.5\) yielding \(67.5\,\mu\text{m}\) [A 68 \(\mu\text{m}\)]; [1.0]

(iii) [0.5 mark]
- Recognises vessel lumen is non-circular / oval / irregular / asymmetrical in transverse section; [0.5]

(i)
$$\text{Change in mass} = \text{Final mass} - \text{Initial mass} = 2.72\text{ g} - 3.20\text{ g} = -0.48\text{ g}$$
$$\text{Percentage change} = \left(\frac{-0.48}{3.20}\right) \times 100\,\% = -15.0\,\%$$

(ii)
Variable: Excess external solution adhering to the surface of the cylinder adds mass.
Standardization: Gently roll the potato cylinder a fixed number of times on a paper towel using identical light pressure before placing on the balance.

(i) [1.5 marks]
- Correct method: \(((2.72 - 3.20) / 3.20) \times 100\); [0.5]
- Correct value of \(-15.0\,\%\) or \(-15\,\%\) with negative sign explicitly stated; [1.0]
[R if sign missing or incorrect]

(ii) [1.0 mark]
- Identification of excess fluid / moisture on surface of cylinders AND standardisation by blotting in a specified, repeatable way (e.g., dab twice on paper towel / roll with constant pressure); [1.0]
[A evaporation of water during weighing AND standardisation by weighing in a closed weighing boat / immediately]

(i)
Using \(C_1 V_1 = C_2 V_2\):
$$V_1 = \frac{C_2 V_2}{C_1} = \frac{0.5\text{ g dm}^{-3} \times 10.0\text{ cm}^3}{2.0\text{ g dm}^{-3}} = 2.5\text{ cm}^3\text{ of stock }\mathbf{S}$$
$$\text{Volume of distilled water} = 10.0 - 2.5 = 7.5\text{ cm}^3$$

(ii)
Explanation: Color change is gradual and perceived differently by different observers or in varying room lighting, making the exact endpoint subjective.
Instrument: A colorimeter / light absorbance meter measuring absorbance or transmission at fixed intervals.

(i) [1.0 mark]
- \(2.5\text{ cm}^3\) of stock \(\mathbf{S}\) AND \(7.5\text{ cm}^3\) of distilled water; [1.0]

(ii) [1.5 marks]
- Explanation that the colour transition is continuous / judgment of green tint is subjective / depends on ambient lighting / human reaction time; [0.5]
- Instrument: colorimeter / spectrophotometer / light sensor; [1.0]
[R pH meter / thermometer]