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

1. At ×100 magnification, 50 eyepiece units (epu) = 1.0 mm = 1000 μm.
Therefore, 1 epu at ×100 = 1000 / 50 = 20 μm.

2. When changing magnification from ×100 to ×400, the magnification increases by a factor of 4 (×400 / ×100 = 4).
Therefore, the actual distance represented by 1 epu decreases by a factor of 4:
1 epu at ×400 = 20 μm / 4 = 5.0 μm.

3. The cell measures 7 epu at ×400:
Actual diameter = 7 × 5.0 μm = 35 μm.

B [1]
1 epu at ×100 = 20 μm; 1 epu at ×400 = 5 μm; 7 × 5 μm = 35 μm.

A linear polypeptide of 42 amino acids contains 41 peptide bonds (number of bonds = n - 1 = 42 - 1 = 41). Hydrolysis requires the addition of one water molecule per peptide bond broken, so 41 water molecules are consumed. Each resulting free amino acid contains both a basic amino group (-NH2) and an acidic carboxyl group (-COOH).

B [1]
41 peptide bonds hydrolysed requires 41 water molecules; each monomer has an amino group and a carboxyl group.

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. At high substrate concentrations, the substrate outcompetes the inhibitor, allowing the reaction to reach the same maximum velocity (Vmax unchanged). However, a higher substrate concentration is required to reach half Vmax, meaning Km increases (representing a decreased apparent affinity of the enzyme for the substrate).

B [1]
Unchanged Vmax and increased Km is characteristic of reversible competitive inhibition.

At 0.3 mol dm-3 sucrose, the water potential of the external solution equals that of the potato cells (incipient equilibrium). In a 0.5 mol dm-3 sucrose solution, the solution has a lower (more negative) water potential than the potato cell sap. Water leaves the cells by osmosis down the water potential gradient, causing the protoplast to shrink away from the cell wall (plasmolysis).

C [1]
Water leaves cells in 0.5 mol dm-3 sucrose because solution water potential is lower than cell sap water potential, leading to plasmolysis.

Telomeres are non-coding repeating sequences of DNA located at the ends of eukaryotic chromosomes. During DNA replication, DNA polymerase cannot replicate the extreme ends of the lagging strand. Telomeres ensure that vital protein-coding genes are not lost with successive rounds of cell division and protect chromosome ends from degradation.

A [1]
Telomeres are non-coding terminal DNA regions preventing loss of genes during replication.

1. Total number of bases = 2400 base pairs × 2 = 4800 bases.
2. In double-stranded DNA, % cytosine (C) = % guanine (G) = 28%.
Total C + G = 28% + 28% = 56%.
3. % adenine (A) + % thymine (T) = 100% - 56% = 44%.
Since % A = % T, % thymine = 44% / 2 = 22%.
4. Total number of thymine bases = 0.22 × 4800 = 1056.

C [1]
Total bases = 4800; % T = (100 - (28 × 2)) / 2 = 22%; 0.22 × 4800 = 1056.

1. Proton pumps actively transport H+ ions out of the companion cell cytoplasm into the cell wall space, establishing an electrochemical proton gradient.
2. H+ ions diffuse back into the companion cell through cotransporter carrier proteins, carrying sucrose molecules with them against the sucrose concentration gradient.
3. Sucrose then diffuses from the companion cell into the sieve tube element via plasmodesmata.

A [1]
Active pumping of H+ out -> cotransport of H+ and sucrose in -> symplastic movement of sucrose into sieve tube element via plasmodesmata.

The flexible hinge region of an antibody molecule allows the two antigen-binding sites (variable regions) to move and bind simultaneously to two separate antigen epitopes that are situated variable distances apart on pathogen surfaces. Variable regions (not constant regions) form the antigen-binding sites. Heavy and light chains are held together by disulfide bonds as well as non-covalent interactions. Each antibody has two identical antigen-binding sites, not four.

B [1]
Hinge region confers flexibility to bind antigens at variable distances.

1. Find 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 division (epu):
\(40\text{ epu} = 200\ \mu\text{m} \implies 1\text{ epu} = \frac{200}{40} = 5\ \mu\text{m}\).

3. Calculate the actual length of the cell:
\(\text{Actual length} = 36\text{ epu} \times 5\ \mu\text{m} = 180\ \mu\text{m}\).

B is correct [1].

Calculation of 1 eyepiece unit \(= 5\ \mu\text{m}\); actual length \(= 36 \times 5 = 180\ \mu\text{m}\).

A is incorrect (corresponds to dividing by 2 instead of multiplying: \(36 \times 2.5 = 90\ \mu\text{m}\)).
C is incorrect (corresponds to miscalculating 1 epu as \(10\ \mu\text{m}\)).
D is incorrect (unit error or magnification multiplication).

- Amylose is an unbranched polymer of \(\alpha\)-glucose linked exclusively by \(\alpha\)-1,4-glycosidic bonds.
- Amylopectin is a branched polymer of \(\alpha\)-glucose linked by \(\alpha\)-1,4-glycosidic bonds with \(\alpha\)-1,6-glycosidic bonds at branch points.
- Cellulose is an unbranched polymer of \(\beta\)-glucose linked by \(\beta\)-1,4-glycosidic bonds, where consecutive glucose monomers are rotated by \(180^{\circ}\) relative to each other.

C is correct [1].

- Row C correctly identifies \(\alpha\)-1,4 for amylose, \(\alpha\)-1,4 and \(\alpha\)-1,6 for amylopectin, and \(\beta\)-1,4 with rotated monomers for cellulose.

The maximum rate (\(V_{\max}\)) remains unchanged, while the Michaelis-Menten constant (\(K_m\)) increases from \(0.4\) to \(1.2\text{ mmol dm}^{-3}\). An increased \(K_m\) indicates a lower affinity of the enzyme for its substrate because a higher concentration of substrate is required to achieve half-maximal velocity (\(0.5\ V_{\max}\)). This is characteristic of a competitive inhibitor, which binds reversibly to the active site and competes directly with substrate molecules.

A is correct [1].

Competitive inhibitor binds to active site, increasing \(K_m\) (lowering affinity) without altering \(V_{\max}\).

B is incorrect because non-competitive inhibitors decrease \(V_{\max}\).
C is incorrect because \(K_m\) increase means affinity decreases, not increases.
D is incorrect because competitive inhibitors do not bind allosteric sites.

At lower temperatures, phospholipids tend to pack more closely together, decreasing fluidity. Membrane fluidity is maintained or increased at low temperatures by:
1. An increased proportion of unsaturated fatty acid hydrocarbon tails (which have 'kinks' that prevent close packing).
2. Shorter fatty acid hydrocarbon chains (which have fewer intermolecular hydrophobic interactions).
3. Cholesterol preventing close packing of phospholipid tails.

C is correct [1].

Increasing the proportion of unsaturated fatty acid tails introduces kinks/bends, preventing tight packing of lipids at low temperatures.

A is incorrect (saturated tails pack tightly, reducing fluidity).
B is incorrect (longer chains increase hydrophobic interactions, reducing fluidity).
D is incorrect (glycolipid carbohydrate chains do not increase hydrocarbon tail mobility in this manner).

- In metaphase of mitosis, each of the 12 chromosomes has replicated during S phase and consists of 2 sister chromatids held by 1 centromere.
- Number of chromatids \(= 12 \times 2 = 24\).
- Number of centromeres \(= 12\).
- Mitosis maintains the diploid chromosome number in daughter cells.
- Immediately after cytokinesis, each daughter cell has 12 chromosomes (each consisting of a single chromatid).

D is correct [1].

- Metaphase chromatids \(= 24\)
- Metaphase centromeres \(= 12\)
- Daughter cell chromosomes \(= 12\)

Rows A, B, and C have incorrect counts for chromatids, centromeres, or daughter chromosomes.

1. Total number of bases in double-stranded DNA \(= 3000\text{ base pairs} \times 2 = 6000\text{ bases}\).
2. Adenine (A) \(= 22\%\). By complementary base pairing, Thymine (T) \(= 22\%\).
3. Combined percentage of \(A + T = 22\% + 22\% = 44\%\).
4. Combined percentage of \(G + C = 100\% - 44\% = 56\%\).
5. Percentage of Cytosine (C) \(= \frac{56\%}{2} = 28\%\).
6. Total number of cytosine bases \(= 0.28 \times 6000 = 1680\).

C is correct [1].

\(28\%\) of 6000 bases \(= 1680\).

A is incorrect (calculated \(28\%\) of 3000 base pairs instead of total bases \(= 840\)).
B is incorrect (corresponds to the number of adenine bases \(= 0.22 \times 6000 = 1320\)).
D is incorrect (combined \(G + C\) count \(= 3360\)).

- Statement 1 is correct: Proton pumps actively transport \(\text{H}^+\) ions out of companion cells into cell walls using ATP.
- Statement 2 is incorrect: Pumping \(\text{H}^+\) out of the cell creates a higher concentration of \(\text{H}^+\) in the cell wall (apoplast), not the cytoplasm.
- Statement 3 is correct: \(\text{H}^+\) ions move down their concentration gradient through cotransporter membrane proteins, driving the uptake of sucrose against its concentration gradient.
- Statement 4 is correct: Sucrose diffuses into sieve tube elements through connecting plasmodesmata down its concentration gradient.

B is correct [1].

Statements 1, 3, and 4 are correct.
Statement 2 is false as the proton gradient has a higher \(\text{H}^+\) concentration outside the cell wall than inside the companion cell.

- B-lymphocytes differentiate into plasma cells that synthesise and secrete large quantities of specific antibodies.
- T-helper cells possess CD4 receptors and release signalling cytokines that stimulate B-lymphocyte clonal expansion and activate macrophages.
- T-killer (cytotoxic) cells destroy infected host cells displaying foreign antigens by secreting perforins/toxins.
- Neutrophils and macrophages act as phagocytes to engulf and digest pathogens.

A is correct [1].

- Row A correctly describes B-lymphocytes (plasma cells) secreting specific antibodies, T-helper cells releasing cytokines, and T-killer cells destroying virally infected cells.
- Rows B, C, and D confuse cytokine secretion, antibody secretion, and direct cytotoxicity.

1. Determine the actual length represented by the stage micrometer divisions: \(24\text{ divisions} \times 0.01\text{ mm} = 0.24\text{ mm} = 240\ \mu\text{m}\).
2. Calculate the value of 1 eyepiece graticule unit: \(\frac{240\ \mu\text{m}}{40\text{ units}} = 6.0\ \mu\text{m}\text{ per unit}\).
3. Calculate the actual length of the cell: \(35\text{ units} \times 6.0\ \mu\text{m} = 210\ \mu\text{m}\).

[1] C is correct (210 \(\mu\text{m}\)); A, B, and D result from incorrect arithmetic or inversion of the calibration ratio.

Cellulose is an unbranched polymer composed exclusively of \(\beta\)-glucose units joined by \(\beta\)-1,4-glycosidic bonds. To form these bonds, alternate \(\beta\)-glucose molecules are rotated \(180^\circ\). The straight chains lie parallel to one another, allowing extensive hydrogen bonding between hydroxyl groups on adjacent chains to produce high tensile strength microfibrils.

[1] C is correct (Cellulose); Amylose and glycogen are made of \(\alpha\)-glucose, and amylopectin/glycogen contain 1,6-glycosidic branch points.

A competitive inhibitor binds reversibly to the active site, competing directly with the substrate. At sufficiently high substrate concentrations, the substrate outcompetes the inhibitor, reaching the same \(V_{\text{max}}\). However, a higher substrate concentration is required to reach \(\frac{1}{2}V_{\text{max}}\), which increases the \(K_m\) value and indicates a decreased apparent affinity of the enzyme for its substrate.

[1] B is correct; A and D describe non-competitive inhibitors (which reduce \(V_{\text{max}}\)), and C incorrectly states that an increased \(K_m\) reflects increased affinity.

Because the external sucrose solution has a more negative (lower) water potential than the potato cell sap, water moves down the water potential gradient out of the cells by osmosis. As water leaves the vacuole and cytoplasm, the volume of the protoplast decreases, causing it to pull away from the rigid cell wall (plasmolysis).

[1] B is correct; A is incorrect because sucrose cannot readily cross the cell surface membrane by simple facilitated diffusion in uninjured cells, C describes tissue placed in hypotonic solution, and D contradicts osmosis thermodynamics.

During anaphase of mitosis, centromeres split/divide, allowing sister chromatids to separate into individual daughter chromosomes. Kinetochore microtubules of the spindle shorten, pulling the chromosomes towards opposite poles. The nuclear envelope broke down during prophase and does not reform until telophase, so it is not intact.

[1] A is correct; centromeres divide and chromatids separate as spindle fibres shorten in the absence of an intact nuclear envelope.

1. The non-transcribed coding strand is \(\text{5' - ATG CCT GAG TAC - 3'}\).
2. The mRNA transcribed from the template strand will have the same sequence as the coding strand (with U replacing T): \(\text{5' - AUG CCU GAG UAC - 3'}\).
3. The second codon of the mRNA is \(\text{5' - CCU - 3'}\).
4. The tRNA anticodon binds complementarily and antiparallel to the codon: \(\text{3' - GGA - 5'}\).

[1] B is correct (\(\text{3' - GGA - 5'}\)); A contains thymine (tRNA has uracil/no thymine), C is the mRNA codon, and D represents an incorrect complementary pairing.

All four statements accurately describe the mechanism of active phloem loading:
1. ATP-driven proton pumps pump \(\text{H}^+\) from companion cells into the cell wall.
2. This creates an electrochemical/proton concentration gradient.
3. Sucrose enters through sucrose-\(H^+\) cotransporters using the secondary active transport driven by the proton gradient.
4. The high concentration of dissolved sucrose in companion cells and sieve tube elements lowers the water potential, drawing water in from adjacent xylem by osmosis.

[1] A is correct; statements 1, 2, 3, and 4 are all correct descriptions of phloem loading mechanism.

A higher partial pressure of carbon dioxide in actively metabolising tissues leads to increased formation of carbonic acid and hydrogen ions (\(\text{H}^+\)). These \(\text{H}^+\) ions bind to oxyhaemoglobin to form haemoglobinic acid (HHb), reducing haemoglobin's affinity for oxygen and shifting the oxygen dissociation curve to the right. This facilitates the unloading of oxygen to respiring cells.

[1] C is correct; A, B, and D all cause a leftward shift of the oxygen dissociation curve (increased affinity for oxygen).

1. At \(\times 100\) magnification, 25 eyepiece graticule units = \(40 \times 10\text{ }\mu\text{m} = 400\text{ }\mu\text{m}\).
2. Therefore, at \(\times 100\), 1 eyepiece graticule unit = \(\frac{400}{25} = 16\text{ }\mu\text{m}\).
3. When magnification is increased by a factor of 4 (from \(\times 100\) to \(\times 400\)), each eyepiece graticule unit represents \(\frac{16\text{ }\mu\text{m}}{4} = 4\text{ }\mu\text{m}\).
4. Actual thickness = \(15 \times 4\text{ }\mu\text{m} = 60\text{ }\mu\text{m}\).

B is correct [1 mark].

A is incorrect: calculates using \(1\text{ }\mu\text{m}\) per unit.
C is incorrect: does not scale for the 4-fold increase in magnification (\(15 \times 16 = 240\text{ }\mu\text{m}\)).
D is incorrect: multiplies by 4 instead of dividing (\(240 \times 4 = 960\text{ }\mu\text{m}\)).

1 is correct: Amylose is an unbranched polymer of \(\alpha\)-glucose and cellulose is an unbranched polymer of \(\beta\)-glucose (both are hexoses).
2 is correct: Amylose contains \(\alpha\)-1,4-glycosidic bonds and cellulose contains \(\beta\)-1,4-glycosidic bonds.
3 is incorrect for amylose: Alternate monomer rotation by \(180^\circ\) occurs only in cellulose due to the orientation of \(-\text{OH}\) groups on \(\beta\)-glucose.
4 is correct: Hydrogen bonds form within amylose (stabilising the helix) and between parallel chains in cellulose (forming microfibrils).

B is correct [1 mark] (1, 2 and 4 only).
A, C, and D are incorrect because statement 3 applies exclusively to cellulose.

An inhibitor that increases \(K_\text{m}\) while leaving \(V_\text{max}\) unchanged is a competitive inhibitor.
- Competitive inhibitors bind directly to the active site (statement 1 is correct).
- A higher \(K_\text{m}\) indicates a lower affinity of the enzyme for its substrate (statement 2 is correct).
- At high substrate concentrations, substrate outcompetes the inhibitor for active sites, reaching \(V_\text{max}\) (statement 3 is correct).
- Statement 4 describes non-competitive inhibition, which lowers \(V_\text{max}\).

A is correct [1 mark] (1, 2 and 3).
B, C, and D are incorrect as statement 4 is characteristic of non-competitive inhibition.

A is incorrect: The mass increased in \(\mathbf{W}\), meaning water entered cells by osmosis down a water potential gradient; \(\mathbf{W}\) has a higher water potential than the tissue.
B is incorrect: In solution \(\mathbf{X}\), there is dynamic equilibrium (water enters and leaves at equal rates), not zero movement.
C is correct: In solution \(\mathbf{Y}\), mass decreased from \(5.00\text{ g}\) to \(4.62\text{ g}\) because water moved out of the cells down a water potential gradient.
D is incorrect: In solution \(\mathbf{Z}\), the tissue lost the greatest amount of mass, which indicates \(\mathbf{Z}\) has the lowest water potential.

C is correct [1 mark].
A, B, and D describe incorrect interpretations of water potential, dynamic equilibrium, and osmosis trends.

1. At metaphase: There are 8 chromosomes, each consisting of 2 sister chromatids (16 chromatids in total). Each chromatid is a double-stranded DNA molecule with 2 telomeres (one at each end). Thus, total telomeres = \(16 \times 2 = 32\).
2. At anaphase: The centromeres divide and sister chromatids separate to become individual daughter chromosomes. There are now 16 chromosomes in the cell. The total amount of DNA has not changed, so there are still 32 telomeres.

A is correct [1 mark].
B, C, and D miscount chromatid separation as maintaining 8 chromosomes or halve telomere count incorrectly.

1. 3600 base pairs = \(3600 \times 2 = 7200\) total bases.
2. In double-stranded DNA, percentage of \(\text{G} = \text{C} = 22\%\).
3. Percentage of \(\text{G} + \text{C} = 44\%\).
4. Therefore, percentage of \(\text{A} + \text{T} = 100\% - 44\% = 56\%\).
5. Percentage of \(\text{A} = \frac{56\%}{2} = 28\%\).
6. Number of adenine bases = \(0.28 \times 7200 = 2016\).

D is correct [1 mark].
A calculates 22% of 3600.
B calculates 28% of 3600 base pairs instead of total bases.
C calculates 22% of 7200 (the number of guanine bases).

The apoplast pathway moves water exclusively through non-living cell walls and intercellular spaces (not through plasmodesmata or living protoplasts). The movement of water through the apoplast is driven by mass flow due to transpiration pull and cohesive/adhesive forces. At the endodermis, the waterproof suberin in the Casparian strip blocks the apoplast pathway, forcing water into the symplast.

A is correct [1 mark].
B, C, and D contain incorrect descriptions of the apoplast pathway regarding cellular structures or Casparian strip blockage.

A rightward shift of the oxygen dissociation curve (Bohr effect) means haemoglobin has a lower affinity for oxygen, promoting oxygen unloading:
- 1 is correct: High \(\text{pCO}_2\) promotes formation of carbonic acid and hydrogen ions, shifting the curve to the right.
- 2 and 4 are correct: Increased \([\text{H}^+]\) / decreased pH promotes binding of \(\text{H}^+\) to oxyhaemoglobin, causing it to release oxygen and shift the curve to the right.
- 3 is incorrect: A decrease in temperature increases haemoglobin's affinity for oxygen, shifting the curve to the left.

B is correct [1 mark] (1, 2 and 4 only).
A, C, and D are incorrect as statement 3 shifts the curve to the left.

1. Find the total distance represented by 40 stage micrometer divisions:
\(40 \times 0.01\text{ mm} = 0.40\text{ mm} = 400\ \mu\text{m}\).
2. Calculate the length represented by 1 epu:
\(1\text{ epu} = \frac{400\ \mu\text{m}}{25} = 16\ \mu\text{m}\).
3. Calculate the actual length of the cell:
\(\text{Actual length} = 15\text{ epu} \times 16\ \mu\text{m} = 240\ \mu\text{m}\).

[1 mark] C is correct (240 \(\mu\text{m}\)).
A is incorrect: corresponds to arithmetic error dividing by 10 (24 \(\mu\text{m}\)).
B is incorrect: calculated from \(15 \times 0.40 \times 16\) incorrectly.
D is incorrect: calculated using inverse ratio \(25 / 40 \times 15 \times 40\).

- Statement 1 is correct: amylose consists of \(\alpha\)-glucose; cellulose consists of \(\beta\)-glucose.
- Statement 2 is correct: to form \(\beta\)-1,4-glycosidic bonds in cellulose, alternate \(\beta\)-glucose units are inverted (rotated \(180^\circ\)). In amylose, all \(\alpha\)-glucose units have the same orientation.
- Statement 3 is incorrect: amylose is unbranched and contains only \(\alpha\)-1,4-glycosidic bonds (amylopectin contains \(\alpha\)-1,6 bonds).
- Statement 4 is correct: unbranched cellulose chains lie parallel and form extensive hydrogen bonds giving microfibrils with high tensile strength, while amylose coils into a helix via internal hydrogen bonds.

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

[1 mark] D is correct (1, 2 and 4 only).
A, B, and C include statement 3 (incorrect because amylose has only 1,4-glycosidic bonds; 1,6-glycosidic bonds are found in amylopectin).

- Inhibitor P increases \(K_{\text{m}}\) (from 2.5 to 7.2) without altering \(V_{\text{max}}\) (remains 120), which is the characteristic behaviour of a competitive inhibitor.
- Inhibitor Q decreases \(V_{\text{max}}\) (from 120 to 65) without altering \(K_{\text{m}}\) (remains 2.5), which is the characteristic behaviour of a non-competitive inhibitor.
- In competitive inhibition (inhibitor P), adding excess substrate increases the probability that substrate binds to the active site instead of inhibitor, so the rate increases until \(V_{\text{max}}\) is restored.

[1 mark] A is correct (P: competitive; Q: non-competitive; effect on P: reaches uninhibited \(V_{\text{max}}\)).
B, C, and D are incorrect due to misidentifying competitive vs non-competitive kinetics or the overcome-ability of competitive inhibition by substrate.

- The sucrose concentration at which no net gain or loss of water occurs (percentage change in mass = 0%) represents the point where the water potential of the solution equals that of the beetroot tissue.
- Between \(0.2\text{ mol dm}^{-3}\) (+6.4%) and \(0.4\text{ mol dm}^{-3}\) (-3.2%), a linear interpolation gives \(0.2 + \left(\frac{6.4}{6.4 - (-3.2)}\right) \times 0.2 = 0.2 + \left(\frac{6.4}{9.6}\right) \times 0.2 \approx 0.33\text{ mol dm}^{-3}\).
- Therefore, the water potential of the sucrose solution equals the internal water potential of the tissue cells at approximately \(0.33\text{ mol dm}^{-3}\).

[1 mark] B is correct (approx. 0.33 mol dm\(^{-3}\) is the isotonic point where mass change is zero).
A is incorrect: net uptake of water occurs up to 0.33 mol dm\(^{-3}\), not 0.4 mol dm\(^{-3}\).
C is incorrect: cells in pure water do not burst due to rigid cellulose cell walls providing wall pressure.
D is incorrect: at 0.8 mol dm\(^{-3}\), cells lose water and become plasmolysed, rather than turgid.

1. In \(\text{G}_1\), the cell has 36 chromosomes (each composed of a single chromatid) and \(6.0\text{ pg}\) of DNA.
2. During S phase, DNA undergoes semi-conservative replication. By metaphase, each chromosome consists of two sister chromatids joined at the centromere. The chromosome number remains 36, while the DNA mass doubles to \(12.0\text{ pg}\).
3. During anaphase, sister chromatids separate. Cytokinesis divides the cytoplasm into two identical daughter cells, each having 36 chromosomes and \(6.0\text{ pg}\) of nuclear DNA.

[1 mark] C is correct (Metaphase: 36 chromosomes, 12.0 pg; Daughter cell: 36 chromosomes, 6.0 pg).
A and B are incorrect: confused chromatid number with chromosome number in metaphase (72 is chromatid number).
D is incorrect: failed to account for DNA doubling in S phase.

1. Total bases in double-stranded DNA \(= 450 \times 2 = 900\text{ bases}\).
2. Percentage of cytosine (C) \(= 32\%\), so percentage of guanine (G) \(= 32\%\).
\(\text{Total } (\text{C} + \text{G}) = 64\%\).
3. Percentage of \((\text{A} + \text{T}) = 100\% - 64\% = 36\%\).
Percentage of adenine (A) \(= 18\%\).
Number of adenine bases \(= 0.18 \times 900 = 162\).
4. Length of template strand \(= 450\text{ nucleotides}\).
Total codons in mRNA \(= \frac{450}{3} = 150\text{ codons}\).
5. Since the last codon is a non-coding stop codon, maximum amino acids \(= 150 - 1 = 149\).

[1 mark] B is correct (162 adenine bases, 149 amino acids).
A is incorrect: calculated codons without subtracting stop codon (150).
C is incorrect: calculated adenine count from 450 instead of 900 total bases (81).
D is incorrect: calculated A + T total (324) instead of adenine alone.

- \(\text{H}^+\) ions (protons) are actively pumped out of companion cells into cell wall spaces using ATP.
- This establishes a steep proton electrochemical gradient across the companion cell membrane.
- \(\text{H}^+\) ions diffuse back into companion cells down this gradient through cotransporter membrane proteins, co-transporting sucrose against its concentration gradient.
- High sucrose concentration in companion cells and sieve tube elements lowers (makes more negative) the solute potential, causing water to enter by osmosis from the xylem, which raises hydrostatic pressure and drives mass flow.

[1 mark] A is correct (Protons pumped into cell wall; cotransport into companion cell; water potential decreases; water enters by osmosis).
B is incorrect: sucrose is cotransported into, not out of, companion cells.
C is incorrect: protons are pumped out using ATP, not diffused out actively, and water leaves xylem, not enters xylem.
D is incorrect: water potential decreases (becomes more negative), rather than increases.

- A rightward shift of the oxygen dissociation curve is known as the Bohr shift.
- It is caused by an increased concentration of carbon dioxide (\(p\text{CO}_2\)), which reacts with water (catalysed by carbonic anhydrase in RBCs) to form carbonic acid, dissociating into \(\text{H}^+\) and \(\text{HCO}_3^-\). The resulting lower pH / higher \([\text{H}^+]\) alters haemoglobin's tertiary structure, reducing its affinity for oxygen.
- Advantage: At any given \(pO_2\) in actively respiring tissues, haemoglobin has a lower percentage saturation, allowing more oxygen to dissociate and be released for aerobic cellular respiration.

[1 mark] D is correct (Increase in \([\text{H}^+]\) / decrease in pH causes right shift; haemoglobin releases more oxygen to respiring tissue).
A and B describe a leftward shift (increase in affinity).
C is incorrect: higher \(p\text{CO}_2\) decreases haemoglobin affinity, not increases it.

In the ring form of \(\alpha\)-glucose, the hydroxyl (-OH) group on carbon-1 (C1) projects downwards (below the ring plane), whereas in \(\beta\)-glucose, the -OH group on C1 projects upwards (above the ring plane). When two \(\alpha\)-glucose molecules condense, a 1,4-glycosidic bond is formed with the loss of a water molecule.

1 mark for: on carbon 1 / C1, the hydroxyl / -OH group is below the ring in \(\alpha\)-glucose and above the ring in \(\beta\)-glucose [A: -OH on C1 points in opposite directions / AW];
0.5 marks for: \(\alpha\)-1,4-glycosidic bond / 1,4-glycosidic bond [R: if peptide or ester bond].

Convert image length to micrometres: \(48\text{ mm} = 48\,000\,\mu\text{m}\).
Use the formula: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\).
\(\text{Actual size} = \frac{48\,000\,\mu\text{m}}{30\,000} = 1.6\,\mu\text{m}\).

0.5 marks for correct conversion of units (e.g. \(48\text{ mm} = 48\,000\,\mu\text{m}\) or \(0.048\text{ m}\));
1 mark for correct calculation showing \(1.6\,\mu\text{m}\) [A: \(1.6\) if units are already provided in the answer prompt].

The Casparian strip contains waterproof suberin in the radial and transverse walls of endodermal cells. This blocks the apoplast pathway (movement through cell walls), directing water and dissolved mineral ions into the living protoplast (symplast pathway) where uptake can be regulated by transport proteins in the plasma membrane.

0.5 marks for: suberin [A: lignin];
1 mark for: blocks / waterproofs the apoplast pathway / cell wall pathway AND forces water into the symplast pathway / through the cytoplasm / across the cell surface membrane.

Monoclonal antibody production requires fusing an antibody-secreting plasma cell (or spleen B-lymphocyte isolated from an immunised animal) with an immortal myeloma (cancer) cell. This forms a hybridoma cell capable of both continuous cell division and secreting a single specific antibody clone.

0.5 marks for: (splenic) B-lymphocyte / plasma cell AND myeloma / cancer / tumour cell [both required];
1 mark for: hybridoma cells can divide indefinitely / are immortal AND produce / secrete the specific antibody [AW].

Purine bases (adenine and guanine) consist of two fused rings (a 6-membered ring and a 5-membered ring), while pyrimidine bases (thymine, cytosine, uracil) consist of a single 6-membered ring. A cytosine-guanine (C-G) complementary base pair forms three hydrogen bonds.

1 mark for: purines have two rings / double-ring structure AND pyrimidines have one ring / single-ring structure;
0.5 marks for: 3 / three (hydrogen bonds).

Carbonic anhydrase catalyses the conversion of \(\text{CO}_2\) and \(\text{H}_2\text{O}\) into \(\text{H}_2\text{CO}_3\), which dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). An increase in \(\text{CO}_2\) produces more \(\text{H}^+\), which binds to haemoglobin forming haemoglobinic acid (HHb), lowering haemoglobin's affinity for oxygen and promoting oxygen unloading.

0.5 marks for: catalyses the reaction between carbon dioxide and water to form carbonic acid / \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\);
1 mark for: decreases the affinity of haemoglobin for oxygen / promotes oxygen release / shifts oxygen dissociation curve to the right / Bohr shift.

Cholera is caused by the Gram-negative bacterium Vibrio cholerae (a prokaryote). It is transmitted via the faecal-oral route through ingestion of water or food contaminated with human faeces containing the pathogen.

1 mark for: Vibrio cholerae [must be spelt correctly] AND bacterium / prokaryote;
0.5 marks for: water-borne / food-borne / ingestion of water or food contaminated with faeces / faecal-oral route.

\(K_m\) is defined as the substrate concentration required to reach half the maximum velocity (\(0.5\,V_{\text{max}}\)). An enzyme with a low \(K_m\) requires a lower concentration of substrate to reach \(\frac{1}{2}V_{\text{max}}\), which reflects a high binding affinity between the enzyme active site and the substrate.

1 mark for: the substrate concentration at which the enzyme achieves half its maximum rate / \(\frac{1}{2}V_{\text{max}}\) / half the maximum velocity [A: formula or graph reference showing \([S]\) at \(0.5 V_{\text{max}}\)];
0.5 marks for: (a low \(K_m\) indicates) high affinity of the enzyme for the substrate.

(a) Adjacent amino acid monomers are joined by condensation reactions to form covalent peptide bonds between the amine group of one amino acid and the carboxyl group of the next.

(b) The specific three-dimensional tertiary conformation of a globular protein is stabilised by non-covalent interactions such as hydrogen bonds, ionic bonds (electrostatic attractions), or hydrophobic interactions (disulfide bonds are covalent and excluded by the question stem).

(a) peptide (bond) [0.75]

(b) any one from:
- hydrogen bond(s) ;
- ionic bond(s) / electrovalent bond(s) / electrostatic attraction(s) ;
- hydrophobic interaction(s) / hydrophobic bond(s) ; [0.75]

[Total: 1.5]

(a) Prokaryotic cells contain 70S ribosomes free in their cytoplasm, unlike eukaryotic cytoplasmic ribosomes which are 80S.

(b) The bacterial cell wall is composed of a rigid meshwork polymer known as peptidoglycan (also accepted: murein).

(a) 70S (ribosomes) [0.75]
[R] 80S

(b) peptidoglycan / murein [0.75]
[R] cellulose / chitin

[Total: 1.5]

1. Protons (\(\text{H}^+\) ions) are actively pumped out of the companion cell into the surrounding cell wall space using ATP via a proton pump / \(\text{H}^+\)-ATPase.
2. This creates a steep proton concentration / electrochemical gradient (higher \(\text{H}^+\) concentration in cell wall than inside the companion cell cytoplasm).
3. Protons move back into the companion cell down their concentration gradient through a cotransporter / symporter protein, which co-transports sucrose into the cell against its concentration gradient (secondary active transport).

1. protons / \(\text{H}^+\) (ions) actively transported / pumped out of companion cell (into apoplast / cell wall) using ATP / via \(\text{H}^+\)-ATPase ;
2. creates a concentration gradient / electrochemical gradient of \(\text{H}^+\) (ions) (between apoplast and cytoplasm) ;
3. \(\text{H}^+\) (ions) enter companion cell (down their gradient) through a cotransporter (protein) / symporter ;
4. sucrose moves into companion cell against its concentration gradient (together with \(\text{H}^+\)) ; [max 3]

1. High rate of respiration produces a high \(p\text{CO}_2\) in tissues.
2. \(\text{CO}_2\) diffuses into erythrocytes and reacts with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)), catalysed by carbonic anhydrase.
3. Carbonic acid dissociates to yield \(\text{H}^+\) ions and \(\text{HCO}_3^-\).
4. \(\text{H}^+\) ions bind to oxyhaemoglobin to form haemoglobinic acid (\(\text{HHb}\)), causing an allosteric conformational change in haemoglobin that lowers its affinity for oxygen, shifting the dissociation curve to the right and promoting oxygen release.

1. (higher rate of aerobic respiration releases) high \(p\text{CO}_2\) / \(\text{CO}_2\) diffuses into red blood cells ;
2. \(\text{CO}_2\) reacts with \(\text{H}_2\text{O}\) to form \(\text{H}_2\text{CO}_3\) / carbonic acid, catalysed by carbonic anhydrase, which dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\);
3. \(\text{H}^+\) (ions) bind to haemoglobin / oxyhaemoglobin (forming haemoglobinic acid / \(\text{HHb}\)) ;
4. causes conformational change in haemoglobin / lowers affinity of haemoglobin for oxygen (so oxygen is released) ; [max 3]

1. Inhibitor binds to an allosteric site (a site other than the active site) of the enzyme.
2. This changes the tertiary structure and active site conformation, preventing catalysis or substrate binding in bound enzymes, so \(V_{\max}\) decreases even at saturating substrate concentrations.
3. The affinity of unaffected active sites for substrate remains constant (or the inhibitor binds equally well to free enzyme and ES complex), so \(K_{\text{m}}\) remains unchanged.

1. inhibitor binds to allosteric site / site other than active site ;
2. alters tertiary structure / 3D shape of active site so active site is no longer complementary / fewer ES complexes formed per unit time ;
3. \(V_{\max}\) decreases (and cannot be restored by increasing substrate concentration) ;
4. \(K_{\text{m}}\) remains unchanged / constant because affinity of remaining active sites for substrate is not altered ; [max 3]

1. The two parental DNA strands are antiparallel (one runs 5' to 3' and the other 3' to 5').
2. DNA polymerase can only extend a growing strand in the 5' to 3' direction (adding nucleotides to the 3' -OH group).
3. Because the replication fork unzips continuously in one direction, the lagging strand must be synthesised in short segments (Okazaki fragments) moving away from the replication fork, which are later joined together by DNA ligase.

1. DNA strands are antiparallel / lagging strand template runs 5' to 3' into the replication fork ;
2. DNA polymerase can only add (free) nucleotides to the 3' (-OH) end (of the new strand) / synthesises DNA only in the 5' to 3' direction ;
3. synthesis on the lagging strand occurs away from the replication fork (as DNA unzips) ;
4. produces Okazaki fragments / short DNA fragments (which are joined by DNA ligase) ; [max 3]

1. Penicillin inhibits the enzyme transpeptidase, which forms peptide cross-links between peptidoglycan polymers during cell wall synthesis.
2. The newly formed bacterial cell wall becomes weak and fragile; as water enters the cell down a water potential gradient by osmosis, the cell cannot withstand internal hydrostatic pressure and undergoes osmotic lysis.
3. Viruses do not have a cell wall, peptidoglycan, or their own metabolic enzymes (transpeptidase) for penicillin to inhibit.

1. penicillin inhibits transpeptidase / prevents formation of peptide cross-links between peptidoglycan (chains / molecules) ;
2. cell wall is weakened / cannot resist osmotic pressure / internal turgor pressure ;
3. water enters cell by osmosis / down a water potential gradient causing cell to burst / lyse ;
4. viruses have no cell wall / no peptidoglycan / no transpeptidase / are non-cellular ; [max 3]

1. Primary structure has a repeating tripeptide sequence where glycine is every third amino acid; because glycine has only a hydrogen atom as its R-group, the three polypeptide chains can pack tightly together.
2. Three helical polypeptide chains are held together by hydrogen bonds to form a rigid triple helix (tropocollagen).
3. Tropocollagen molecules are arranged parallel and staggered, held together by covalent cross-links between lysine residues, forming fibrils and fibres that resist pulling (tensile) forces.

1. glycine as every third amino acid / small R-group allows close / tight packing of (three) polypeptide chains ;
2. three polypeptide chains form a triple helix held together by hydrogen bonds ;
3. (tropocollagen) molecules arranged in parallel / staggered layout (to prevent points of weakness) ;
4. covalent bonds / cross-links between adjacent molecules / form fibrils (and fibres) which resist stretching / tension ; [max 3]

1. Phagocytes / macrophages engulf and hydrolyse the bacterial pathogen, presenting non-self antigens on their cell surface membrane.
2. Only specific B-lymphocytes (and T-helper lymphocytes) possessing surface receptors / antibodies that are complementary to the specific antigen will bind to it (clonal selection).
3. Binding activates the selected lymphocytes, stimulating them to divide repeatedly by mitosis (clonal expansion) to form antibody-secreting plasma cells and memory cells.

1. macrophage / phagocyte / dendritic cell engulfs pathogen and presents (foreign / non-self) antigens on its cell surface membrane ;
2. B-lymphocyte / T-helper cell with complementary (surface) receptor / antibody binds to the specific antigen (clonal selection) ;
3. T-helper cells release cytokines / activate selected B-cells ;
4. selected B-cells divide by mitosis / undergo clonal expansion to differentiate into plasma cells (which secrete antibodies) and memory cells ; [max 3]

1. Alveolar wall (squamous epithelium) and capillary wall (endothelium) are each one cell thick, creating an extremely short diffusion distance (thin barrier).
2. Large number of alveoli (and extensive capillary network) provides a very large surface area to volume ratio for diffusion.
3. Continuous blood flow through capillaries and ventilation of the lungs maintain steep concentration / partial pressure gradients for both \(\text{O}_2\) and \(\text{CO}_2\).

1. thin wall / single layer of squamous epithelial cells (alveolus) AND endothelial cells (capillary) / short diffusion pathway ;
2. large number of alveoli / extensive capillary network gives large surface area (for diffusion) ;
3. blood flow / continuous circulation removes \(\text{O}_2\) and brings \(\text{CO}_2\) (maintaining steep concentration / partial pressure gradient) ;
4. ventilation / breathing constantly refreshes alveolar air / brings \(\text{O}_2\) and removes \(\text{CO}_2\) (maintaining steep gradient) ; [max 3]

Sucrose loading into companion cells involves secondary active transport:
1. Proton pumps (H+-ATPases) in the cell surface membrane of companion cells actively pump hydrogen ions (H+ / protons) out of the companion cell cytoplasm into the cell wall / apoplast pathway using energy from ATP hydrolysis.
2. This creates a steep proton / electrochemical concentration gradient, with a higher concentration of H+ in the cell wall / apoplast than inside the companion cell cytoplasm.
3. Protons move back into the companion cell by facilitated diffusion down their concentration gradient through sucrose-H+ cotransporter proteins (symporters). The kinetic energy released by proton movement is used to transport sucrose molecules into the companion cell simultaneously against their concentration gradient.

1. hydrogen ions / protons / \(\text{H}^+\) pumped out of companion cell (into cell wall / apoplast) by active transport / using ATP / via proton pump ;
2. creates a concentration / electrochemical gradient of \(\text{H}^+\) (higher in apoplast / cell wall than inside companion cell) ;
3. \(\text{H}^+\) move back into companion cell (down concentration gradient) through / via a cotransporter (protein) / symport ;
4. sucrose moves / transported into companion cell against its concentration gradient (along with \(\text{H}^+\)) ;

[Max 3 marks]

To calculate the magnification of the micrograph:

1. State the formula for magnification:
\[\text{Magnification} = \frac{\text{size of image (I)}}{\text{actual size of specimen (A)}}\]

2. Convert both measurements into the same unit (e.g., micrometres, \(\mu\text{m}\)):
\[\text{Image length } (I) = 68\text{ mm} = 68 \times 1000\ \mu\text{m} = 68\,000\ \mu\text{m}\]
\[\text{Actual length } (A) = 1.6\ \mu\text{m}\]

3. Calculate magnification:
\[\text{Magnification} = \frac{68\,000\ \mu\text{m}}{1.6\ \mu\text{m}} = 42\,500\]

Magnification = \(\times 42\,500\)

1. Correct conversion of measurement into consistent units AND division of image size by actual size (e.g. \(\frac{68\,000}{1.6}\) or \(\frac{68}{0.0016}\)) ; [1]
2. Correct final answer of \(42\,500\) / \(\times 42\,500\) / \(4.25 \times 10^4\) ; [1]

Notes:
- A \(\times\) before or after the number, but do not penalise if missing.
- R if units (e.g. mm or \(\mu\text{m}\)) are given in the final magnification answer.

(a)(i) From Table 1.1, the initial rate without inhibitor plateaus at a constant value of 50.0 $\mu$mol dm$^{-3}$ s$^{-1}$ from 8.0 mmol dm$^{-3}$ onwards. Therefore, $V_{\text{max}} = 50.0\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}$.

(a)(ii) $K_{\text{m}}$ is the substrate concentration at which the reaction rate is half of $V_{\text{max}}$. Half of $V_{\text{max}} = \frac{50.0}{2} = 25.0\text{ }\mu\text{mol dm}^{-3}\text{ s}^{-1}$. Looking at Table 1.1 without inhibitor, an initial rate of 25.0 $\mu$mol dm$^{-3}$ s$^{-1}$ occurs at a lactose concentration of 2.0 mmol dm$^{-3}$. Therefore, $K_{\text{m}} = 2.0\text{ mmol dm}^{-3}$.

(b) Competitive inhibitors bind reversibly to the active site. The data shows that at a high substrate concentration (20.0 mmol dm$^{-3}$), the rate with inhibitor reaches the same $V_{\text{max}}$ (50.0 $\mu$mol dm$^{-3}$ s$^{-1}$) as without inhibitor because substrate molecules outcompete the inhibitor for active sites. However, at lower substrate concentrations, the rate is lower in the presence of the inhibitor, indicating an increased apparent $K_{\text{m}}$ (requires a higher substrate concentration to reach $\frac{1}{2}V_{\text{max}}$).

(a)(i) 50.0 / 50 ($\mu$mol dm$^{-3}$ s$^{-1}$); [1]
(a)(ii) 2.0 / 2 (mmol dm$^{-3}$); [1]
(b) (same $V_{\text{max}}$) $V_{\text{max}}$ remains unchanged / reaches 50.0 $\mu$mol dm$^{-3}$ s$^{-1}$ at high substrate concentration / at 20.0 mmol dm$^{-3}$; [1]
(higher $K_{\text{m}}$ / outcompeting) higher substrate concentration is required to achieve the same rate / substrate outcompetes inhibitor at high substrate concentrations / $K_{\text{m}}$ is increased (from 2.0 to 4.0 mmol dm$^{-3}$); [1]

(a)
Initial transmission at 40 $^{\circ}$C = 80.0%
Final transmission at 60 $^{\circ}$C = 16.0%
Decrease in transmission = $80.0 - 16.0 = 64.0$
Percentage decrease = $\frac{64.0}{80.0} \times 100 = 80.0\%$.

(b) As temperature rises above 40 $^{\circ}$C:
1. Phospholipids gain kinetic energy, increasing membrane fluidity and creating gaps in the bilayer.
2. Membrane proteins (integral/transmembrane proteins in cell surface membrane and tonoplast) denature due to disruption of hydrogen and ionic bonds, destroying tertiary structure.
3. This loss of membrane integrity allows the water-soluble betalain pigment stored in the vacuole to diffuse out into the solution, increasing pigment concentration and decreasing light transmission.

(a) $(80.0 - 16.0) / 80.0 \times 100$ OR $64.0 / 80.0 \times 100$; [1]
80 / 80.0 (%); [1]
(b) Any two from:
1. (phospholipids) increased kinetic energy of phospholipids / phospholipids move apart / membrane becomes more fluid; [1]
2. (proteins) denaturation / change in tertiary structure / loss of 3D shape of membrane / intrinsic / carrier / channel proteins; [1]
3. (pigment leakage) tonoplast / cell surface membrane loses integrity / becomes more permeable so pigment / betalain diffuses / leaks out (into bathing solution); [1]

(a) Increasing the partial pressure of carbon dioxide decreases the affinity of haemoglobin for oxygen at any given partial pressure of oxygen (Bohr effect / shift).

(b)
1. Actively respiring muscle produces high amounts of carbon dioxide (elevating $p\text{CO}_2$ to 8.0 kPa) and hydrogen ions (lowering pH), which bind to haemoglobin forming haemoglobinic acid and altering its allosteric conformation.
2. In respiring tissue where $p\text{O}_2$ is low (e.g. at 4.0 kPa), haemoglobin saturation decreases from 56.0% to 34.0%, which releases an additional 22.0% of oxygen.
3. This ensures that more oxygen is unloaded/delivered directly to working muscle cells to sustain aerobic respiration and ATP synthesis, thereby preventing or delaying anaerobic respiration and fatigue.

(a) decreases affinity (of haemoglobin) for oxygen / curve shifts to the right / Bohr shift / oxygen dissociates more readily; [1]
(b) Any three from:
1. (conditions in muscle) actively respiring tissue produces more $\text{CO}_2$ / higher $p\text{CO}_2$ / forms $\text{H}^+$ / lowers pH / carbonic anhydrase produces $\text{H}^+$ / forms haemoglobinic acid (HHb); [1]
2. (data quote / comparison) at lower $p\text{O}_2$ in tissues (e.g. at 4.0 kPa / 2.0 kPa), saturation is lower at 8.0 kPa than at 5.3 kPa / 22.0% more oxygen is released at 4.0 kPa / 10.0% more oxygen released at 2.0 kPa; [1]
3. (oxygen unloading) more oxygen dissociates / is unloaded / released from oxyhaemoglobin to the tissue cells; [1]
4. (metabolic advantage) supplies oxygen for (aerobic) respiration / ATP production / delays onset of anaerobic respiration / lactate accumulation; [1]

(a) Similarities: In both species X and Y, increasing wind speed increases the rate of transpiration up to a plateau / maximum.
Differences: Species X has a consistently higher rate of transpiration than species Y at all wind speeds. Species X increases by $21.8 - 4.2 = 17.6\text{ mg dm}^{-2}\text{ min}^{-1}$, whereas species Y only increases by $4.0 - 1.8 = 2.2\text{ mg dm}^{-2}\text{ min}^{-1}$. Species Y plateaus earlier (at 5.0 m s$^{-1}$) than species X (which continues to rise to 6.0 m s$^{-1}$).

(b) Xerophytic structural adaptations:
1. Sunken stomata / stomatal crypts / pits: traps moist/humid air outside the stoma, which reduces the water potential gradient between inside the leaf and the atmosphere, and shields this boundary layer from being blown away by moving air.
2. Thick waxy cuticle: provides an impermeable barrier to water, preventing cuticular transpiration.
3. Hairs / trichomes on leaf surface: traps still air layer / reduces air movement near the stomata.
4. Rolled leaves: creates an enclosed microenvironment with high humidity around stomata.

(a) Two from:
1. (general trend) for both species, rate of transpiration increases with wind speed / both plateau / level off; [1]
2. (magnitude / difference) rate is higher in species X than species Y at all wind speeds / species X has a larger overall increase ($17.6\text{ mg dm}^{-2}\text{ min}^{-1}$) than species Y ($2.2\text{ mg dm}^{-2}\text{ min}^{-1}$); [1]
3. (plateau point) species Y plateaus at lower wind speed / at 5.0 m s$^{-1}$ whereas species X continues to increase up to 6.0 m s$^{-1}$; [1]
(b) Any two adaptations with linked explanations [max 2]:
- sunken stomata / stomata in pits / crypts; traps moist / humid air / shields boundary layer from wind / reduces water potential gradient; [1]
- trichomes / epidermal hairs; traps a layer of moist air / reduces air currents over stomata; [1]
- rolled / curled leaves; traps humid air / reduces exposed surface area to wind; [1]
- thick (waxy) cuticle; increases diffusion distance / impermeable barrier to water loss; [1]

To carry out a two-fold (halving) serial dilution where at least \(10\text{ cm}^3\) of each concentration is retained:

1. For \(2.0\text{ g dm}^{-3}\): Take \(10\text{ cm}^3\) of the \(4.0\text{ g dm}^{-3}\) stock solution (S) and add \(10\text{ cm}^3\) of distilled water (total volume = \(20\text{ cm}^3\)).
2. For \(1.0\text{ g dm}^{-3}\): Take \(10\text{ cm}^3\) of the \(2.0\text{ g dm}^{-3}\) solution and add \(10\text{ cm}^3\) of distilled water (leaving \(10\text{ cm}^3\) of \(2.0\text{ g dm}^{-3}\) solution).
3. For \(0.5\text{ g dm}^{-3}\): Take \(10\text{ cm}^3\) of the \(1.0\text{ g dm}^{-3}\) solution and add \(10\text{ cm}^3\) of distilled water (leaving \(10\text{ cm}^3\) of \(1.0\text{ g dm}^{-3}\) solution).
4. For \(0.25\text{ g dm}^{-3}\): Take \(10\text{ cm}^3\) of the \(0.5\text{ g dm}^{-3}\) solution and add \(10\text{ cm}^3\) of distilled water (leaving \(10\text{ cm}^3\) of \(0.5\text{ g dm}^{-3}\) solution and producing \(20\text{ cm}^3\) of \(0.25\text{ g dm}^{-3}\) solution).

Completed table:

| concentration of sucrose solution / \(\text{g dm}^{-3}\) | volume of sucrose solution / \(\text{cm}^3\) | source of sucrose solution | volume of distilled water / \(\text{cm}^3\) |
| :---: | :---: | :---: | :---: |
| 2.0 | 10 | \(4.0\text{ g dm}^{-3}\) / S | 10 |
| 1.0 | 10 | \(2.0\text{ g dm}^{-3}\) | 10 |
| 0.5 | 10 | \(1.0\text{ g dm}^{-3}\) | 10 |
| 0.25 | 10 | \(0.5\text{ g dm}^{-3}\) | 10 |

1. Shows halving serial dilution concentrations: 2.0, 1.0, 0.5, 0.25 (\(\text{g dm}^{-3}\)) in the first column ;
2. Shows correct source of sucrose solution for each subsequent dilution step (i.e. 2.0 from S / 4.0, 1.0 from 2.0, 0.5 from 1.0, 0.25 from 0.5) ;
3. Shows correct volumes of sucrose solution and distilled water in a \(1:1\) ratio with total volume at least \(20\text{ cm}^3\) at each step to allow \(10\text{ cm}^3\) transfer and leave \(\ge 10\text{ cm}^3\) (e.g. \(10\text{ cm}^3\) sucrose solution + \(10\text{ cm}^3\) distilled water) ;
[Total: 3 marks]

### Example Results Table

$$
\begin{array}{|c|c|c|c|}
\hline
\textbf{Ascorbic acid solution} & \textbf{Titre 1 / cm}^3 & \textbf{Titre 2 / cm}^3 & \textbf{Mean volume added / cm}^3 \\
\hline
\mathbf{A1}\text{ (1.00\%)} & 0.45 & 0.45 & 0.45 \\
\hline
\mathbf{A2}\text{ (0.50\%)} & 0.90 & 0.95 & 0.93 \\
\hline
\mathbf{A3}\text{ (0.25\%)} & 1.80 & 1.85 & 1.83 \\
\hline
\mathbf{A4}\text{ (0.125\%)} & 3.65 & 3.60 & 3.63 \\
\hline
\mathbf{U}\text{ (unknown)} & 1.20 & 1.25 & 1.23 \\
\hline
\end{array}
$$

### Marking Criteria Summary
1. Headings with units: Column/row headers clearly state the identity or concentration of the solution and the volumes with solidus unit (e.g. `/ cm^3`).
2. Replication & Mean: Includes repeated trials (Titre 1 and Titre 2 / replicate 1 and 2) and a calculated mean volume column.
3. Consistency of decimal places: All raw volume measurements recorded to the same precision (e.g. to \(0.05\text{ cm}^3\) or \(0.1\text{ cm}^3\)).
4. Trend: Correct biological trend observed where increasing ascorbic acid concentration requires a smaller volume to decolourise the fixed volume of DCPIP.
5. Unknown U included: Raw data and mean calculated for the unknown solution U within the expected range.

1. Table layout & headings: Table fully ruled with all column headings containing quantity and unit separated by a solidus (e.g., `concentration of ascorbic acid / %` or `solution` AND `volume / cm3` / `titre / cm3`); [1]
2. Replicates & Mean: Columns/rows provided for at least two trials/replicates and a separate column/row for calculated mean volume; [1]
3. Precision: All raw volume data recorded to a consistent degree of precision (all to 1 decimal place or all to 2 decimal places, e.g., to nearest \(0.05\text{ cm}^3\) or \(0.1\text{ cm}^3\)); [1]
4. Biological trend: Recorded volumes show that as the concentration of ascorbic acid increases from A4 to A1, the mean volume required to decolourise DCPIP decreases; [1]
5. Unknown recorded: Volume data for unknown U recorded in the table and mean calculated correctly without units in the body cells. [1]

(i) At 0% change in mass, the concentration is \(0.24\text{ mol dm}^{-3}\).

(ii) When the external sucrose solution has a concentration of \(0.24\text{ mol dm}^{-3}\), its water potential (\(\Psi\)) is exactly equal to the internal water potential of the potato tuber cells. Because the water potential gradient is zero, the rate of water entering the cells by osmosis equals the rate of water leaving the cells (dynamic equilibrium), resulting in no net movement of water and zero percentage change in mass.

(i) \(0.24\text{ mol dm}^{-3}\) (unit required); [0.5]

(ii) 1 mark for: water potential of external sucrose solution equals water potential of the potato cells / tissue / sap / \(\Psi_{\text{external}} = \Psi_{\text{cell}}\); [1.0]
1 mark for: no net movement of water / rate of water entering equals rate of water leaving (into or out of the cells) by osmosis / dynamic equilibrium; [1.0]

(a) \(\frac{1}{2}V_{\text{max}} = \frac{48.0}{2} = 24.0\text{ \mu mol dm}^{-3}\text{ s}^{-1}\). The substrate concentration at half of \(V_{\text{max}}\) is defined as the Michaelis-Menten constant (\(K_m\)).

(b) The value \(0.16\text{ mol dm}^{-3}\) is the \(K_m\). \(K_m\) provides an inverse measure of enzyme affinity for substrate. A lower \(K_m\) value indicates higher affinity between enzyme and substrate because fewer substrate molecules are required to saturate 50% of the enzyme active sites.

(a) \(24.0\text{ \mu mol dm}^{-3}\text{ s}^{-1}\) (allow 24) AND Michaelis-Menten constant / \(K_m\); [1.0]

(b) 1 mark for: \(K_m\) is an inverse measure of affinity / the lower the \(K_m\), the higher the enzyme affinity for the substrate (ora); [1.0]
0.5 marks for: indicates that at \(0.16\text{ mol dm}^{-3}\), 50% of the active sites are occupied / enzyme reaches half-maximal velocity; [0.5]

(a) When describing graphical trends in Paper 3: provide the overall direction of change with data quotes (both coordinates with units). From \(0\) to \(3\text{ m s}^{-1}\), the rate of transpiration increases rapidly from \(1.2\) to \(5.1\text{ mm}^3\text{ min}^{-1}\). Above \(3\text{ m s}^{-1}\), the increase slows down / reaches a plateau at \(5.3\text{ mm}^3\text{ min}^{-1}\).

(b) At high wind speeds, the boundary layer of moist air is completely blown away so diffusion gradient cannot increase further, or stomatal closure occurs to reduce dehydration, or another environmental parameter (light/temperature) becomes the limiting factor.

(a) 1 mark for: as wind speed increases, rate of transpiration increases up to \(3\text{ m s}^{-1}\) and then levels off / plateaus / increases at a much slower rate; [1.0]
0.5 marks for: data quote including both variables with correct units at two points (e.g. \(1.2\text{ mm}^3\text{ min}^{-1}\) at \(0\text{ m s}^{-1}\) to \(5.1\text{ mm}^3\text{ min}^{-1}\) at \(3\text{ m s}^{-1}\) or \(5.3\text{ mm}^3\text{ min}^{-1}\) at \(4\text{ m s}^{-1}\)); [0.5]

(b) 1 mark for any one from:
- stomata close / reduce aperture (in response to rapid water loss / wilt prevention);
- humid boundary layer is already completely removed / diffusion gradient at maximum;
- another factor is limiting (e.g. light intensity / temperature / rate of water absorption by roots); [1.0]

Between \(40\,^\circ\text{C}\) and \(70\,^\circ\text{C}\), thermal energy increases kinetic energy of phospholipids, increasing membrane fluidity and causing gaps in the bilayer. Concurrently, membrane transport proteins in the cell surface membrane and tonoplast denature as hydrogen bonds and ionic interactions are broken, altering their tertiary structure. This causes loss of membrane integrity and selective permeability, allowing betalain pigments to rapidly diffuse out of the vacuole and cytoplasm into the surrounding bathing water, resulting in higher colorimeter absorbance.

Marking points (max 2.5 marks):
- (increased temperature / kinetic energy) increases fluidity / permeability of phospholipid bilayer / disrupts hydrophobic interactions between phospholipids; [1.0]
- membrane / tonoplast proteins denature / tertiary structure altered / hydrogen/ionic bonds break; [1.0]
- loss of selective permeability / membrane integrity lost / creates gaps or pores allowing betalain pigment to diffuse / leak out; [0.5]

To gain full marks for a low-power plan tissue drawing:
1. Quality of execution and size: Use a sharp pencil (HB) to produce clean, unbroken, single lines with no shading, feathering, or overlapping lines. The drawing must occupy more than 50% of the allocated space.
2. Sector selection: Clearly delineate a wedge/sector from the outer surface into the central pith containing exactly two complete adjacent vascular bundles.
3. Tissue plan convention: Delineate tissue boundaries only; do not draw any individual cells or cell walls.
4. Proportional representation: Ensure the thickness of the cortex and the size/orientation of the vascular bundles relative to the overall radius are anatomically proportional (e.g. phloem situated external to xylem within each bundle).
5. Labelling: Use a ruler to draw a straight, uncrossed label line that stops exactly on the xylem region of one of the vascular bundles, accompanied by the clear label 'xylem'.

1. Line quality & size: Clear, sharp, continuous lines with no sketchy/overlapping lines or shading AND drawing occupies at least 50% of available space [1];
2. Sector representation: Correct sector drawn showing epidermis, cortex, two complete adjacent vascular bundles, and part of the central pith [1];
3. Plan drawing convention: No individual cells drawn anywhere in the diagram [1];
4. Proportion and organisation: Vascular bundles correctly oriented with phloem external to xylem AND vascular bundles shown separated by interfascial parenchyma / pith ray tissue [1];
5. Label: One ruled label line with a ruler touching the xylem tissue and correctly labelled 'xylem' / 'xylem tissue' / 'xylem vessel(s)' [1].

To gain full marks for a high-power cellular drawing in Paper 3:
1. Quality of drawing: Use a sharp HB pencil to draw clear, thin, continuous unbroken lines without sketchy overlaps or hatching/shading. The overall drawing must occupy at least 50% of the provided drawing frame/space.
2. Cell selection & arrangement: Accurately draw exactly four touching cells arranged as observed in the cortex.
3. Cell wall detail: Represent the cell walls of adjacent cells with double lines showing the distinct wall thickness/middle lamella rather than single lines.
4. Cell shape and proportions: Draw rounded or polygonal cells showing realistic relative sizes and small triangular/intercellular air spaces at junctions between cells.
5. Labelling: Use a ruler to draw an uncrossed, horizontal label line ending precisely on the feature (e.g., intercellular space or cell wall), with an unambiguous label written horizontally.

1. D1 (Clarity & Size): Sharp, continuous, thin unbroken outlines with no shading/cross-hatching AND drawing occupies more than 50% of the space provided;
2. D2 (Number & Configuration): Exactly four touching cells drawn forming a connected group;
3. D3 (Cell Wall Representation): Double lines drawn to represent the thickness of the cell walls between adjacent touching cells;
4. D4 (Proportion & Features): Cells drawn with rounded/polygonal shapes showing intercellular air spaces at cell junctions AND walls shown relatively thin and uniform;
5. L1 (Labelling): One ruled label line touching the intercellular space (or cell wall) with the correct label (intercellular space / air space / cell wall / primary cell wall).

To compare the microscopic anatomy of a dicotyledonous stem (T1) and a dicotyledonous root (T2):

1. Table Structure: Draw a neat, ruled table with explicit column headers: Feature / Characteristic, Stem (T1), and Root (T2).
2. Differences to Record:
* Arrangement of vascular tissue: In T1 (stem), vascular bundles are arranged in a distinct ring near the periphery; in T2 (root), vascular tissue forms a single central core / cylinder (stele).
* Central tissue (pith): In T1 (stem), a large central pith (parenchyma) is present; in T2 (root), there is no central pith (xylem occupies the central axis).
* Distribution of xylem and phloem: In T1 (stem), xylem is located on the inside and phloem on the outside of each vascular bundle; in T2 (root), phloem groups alternate with the radiating arms of xylem around the central cylinder.

| Feature | Stem (T1) | Root (T2) |
| :--- | :--- | :--- |
| Arrangement of vascular tissue | Vascular bundles arranged in a ring towards the outer edge / periphery | Vascular tissue concentrated in a single central core / cylinder |
| Pith / central tissue | Central pith present | Central pith absent (xylem fills center) |
| Organisation of xylem and phloem | Xylem on the inner side and phloem on the outer side within the same bundle | Xylem forms a central star/cross with phloem in separate patches alternating between the arms |

1 mark for table structure:
* Table drawn with ruled lines AND clear headings including Feature, Stem / T1, and Root / T2 ;

2 marks for observable differences (any 2 from the following paired statements [1 mark each]):
* Vascular bundles in a ring / at periphery in T1 AND vascular tissue in a single central cylinder / star-shaped core in T2 ;
* Pith / parenchyma present in center of T1 AND pith absent / xylem in center of T2 ;
* Xylem and phloem together in individual discrete bundles in T1 AND xylem and phloem in separate alternating groups/patches in T2 ;
* Endodermis not clearly defined / absent in T1 AND distinct endodermis / Casparian strip layer surrounding stele in T2 ;

Note: Reject cellular features not observable at plan power (e.g. companion cell organelles). Only credit paired, comparative statements.

1. Determine the length of one stage micrometer division:
\[\text{Length of } 1 \text{ division} = \frac{1.0\text{ mm}}{100} = 0.01\text{ mm} = 10\ \mu\text{m}\]

2. Calculate the distance represented by \(12\) stage micrometer divisions:
\[12 \times 10\ \mu\text{m} = 120\ \mu\text{m}\]

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

4. Calculate the actual lumen diameter of the xylem vessel:
\[\text{Actual diameter} = 23\text{ epu} \times 3.0\ \mu\text{m} = 69.0\ \mu\text{m} \text{ (or } 69\ \mu\text{m}\)\]

[1 mark] (i) Correct calculation of the calibration showing working: \(12 \times 10\ \mu\text{m} / 40\) (or \(0.12\text{ mm} / 40\)) leading to \(3.0\ \mu\text{m}\) (or \(0.003\text{ mm}\));
[1 mark] (ii) Correct multiplication of measured units by calibrated unit value: \(23 \times 3.0 = 69\ \mu\text{m}\) [ecf if calibration in (i) is incorrect];
[0.5 mark] Correct units (\(\mu\text{m}\)) and value given to appropriate significant figures (2 or 3 s.f.).