2026 Cambridge IAL Biology (9700) 模擬試題連答案詳解

Protons (\(\text{H}^+\)) are actively transported out of the companion cell cytoplasm into the cell wall by proton pumps using ATP. This creates an electrochemical gradient with a higher concentration of \(\text{H}^+\) in the cell wall than inside the companion cell. Protons then move back down their concentration gradient into the companion cell via cotransporter proteins, bringing sucrose molecules along with them against the concentration gradient of sucrose.

B is correct [1].
A is incorrect because protons are pumped out using ATP, not diffused out.
C is incorrect because protons move back down their gradient, not against it.
D is incorrect because sucrose enters via cotransport, not facilitated diffusion alone.

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. Increasing substrate concentration overcomes the inhibition, so \(V_{\text{max}}\) remains unchanged, but a higher substrate concentration is required to reach half \(V_{\text{max}}\), meaning \(K_{\text{m}}\) increases (reflecting a lower affinity for the substrate in the presence of the inhibitor).

A is correct [1].
B is incorrect because non-competitive inhibitors decrease \(V_{\text{max}}\).
C is incorrect because competitive inhibitors do not change \(V_{\text{max}}\).
D is incorrect because competitive inhibitors do not bind allosteric sites.

Statement 1 is correct: both contain glycerol linked to fatty acid chains via ester bonds.
Statement 2 is correct: phospholipids have two fatty acid chains (two ester bonds) and one phosphate group, whereas triglycerides have three fatty acid chains (three ester bonds).
Statement 3 is incorrect: triglycerides are entirely hydrophobic (non-polar) and do not have a hydrophilic head.
Statement 4 is correct: the negatively charged phosphate head is hydrophilic and interacts with polar water molecules, allowing bilayer formation.
Therefore, 1, 2, and 4 are correct.

C is correct [1].
A, B, and D are incorrect because statement 3 is incorrect (triglycerides lack a hydrophilic head).

When there is no net movement of water into or out of the plant tissue by osmosis, the percentage change in length (or mass) is zero. This point corresponds to the x-intercept where the line crosses the horizontal axis (0% change in length), indicating that the water potential of the external solution equals the internal water potential of the tissue cells.

B is correct [1].
A is incorrect because maximum increase occurs when water enters cells (hypotonic solution).
C is incorrect because maximum decrease occurs when water leaves cells (hypertonic solution).
D is incorrect because the y-intercept is pure water (0 mol dm⁻³), where water potential is 0 kPa.

Veins are characterized by a wide lumen, a relatively thin wall with less smooth muscle and elastic tissue in the tunica media compared to arteries, and valves to prevent the backflow of blood under low pressure. Arteries have thick walls, narrow lumens, no valves (except at the base of major arteries leaving the heart), and carry blood under high pressure.

D is correct [1].
A is incorrect because arteries do not have valves in their length and have high pressure.
B is incorrect because veins have low blood pressure, not high.
C is incorrect because arteries have thick muscular walls and lack valves.

5 stage micrometer divisions = \(5 \times 10\,\mu\text{m} = 50\,\mu\text{m}\).
20 eyepiece graticule units = \(50\,\mu\text{m}\).
1 eyepiece graticule unit = \(\frac{50\,\mu\text{m}}{20} = 2.5\,\mu\text{m}\).
Actual length of chloroplast = \(3.2 \times 2.5\,\mu\text{m} = 8.0\,\mu\text{m}\).

A is correct [1].
Calculation: 1 epu = (5 × 10) / 20 = 2.5 µm; 3.2 × 2.5 = 8.0 µm.
B (16.0 µm), C (4.0 µm), and D (1.6 µm) arise from arithmetic scale errors.

In double-stranded DNA, the percentage of adenine (A) equals the percentage of thymine (T), and the percentage of cytosine (C) equals the percentage of guanine (G).
If \(\text{A} = 28\%\), then \(\text{T} = 28\%\).
Total \(\text{A} + \text{T} = 28\% + 28\% = 56\%\).
Therefore, \(\text{G} + \text{C} = 100\% - 56\% = 44\%\).
Since \(\text{G} = \text{C}\), the percentage of cytosine is \(\frac{44\%}{2} = 22\%\).

C is correct [1].
A (28%) is the percentage of thymine.
B (44%) is the combined percentage of G + C.
D (56%) is the combined percentage of A + T.

Cholera is caused by the bacterium Vibrio cholerae and is transmitted via water-borne ingestion of contaminated water/food.
Malaria is caused by the protoctist Plasmodium (e.g., P. falciparum) and is transmitted by the female Anopheles mosquito vector.
Tuberculosis is caused by the bacterium Mycobacterium tuberculosis (or M. bovis) and is transmitted via airborne droplets.
HIV/AIDS is caused by the virus Human Immunodeficiency Virus and is transmitted via exchange of body fluids / blood-to-blood contact / sexual contact.

D is correct [1].
A is incorrect because malaria is caused by a protoctist, not a bacterium.
B is incorrect because Vibrio cholerae is a bacterium, not a virus, and is water-borne.
C is incorrect because Mycobacterium tuberculosis is transmitted by airborne droplets, not an insect vector.

1. Determine the actual length represented by 15 divisions of the stage micrometer:
\(15 \times 0.01\text{ mm} = 0.15\text{ mm} = 150\ \mu\text{m}\).

2. Calculate the length represented by 1 eyepiece graticule unit (epu):
\(50\text{ epu} = 150\ \mu\text{m} \implies 1\text{ epu} = \frac{150\ \mu\text{m}}{50} = 3.0\ \mu\text{m}\).

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

B is correct [1].

A is incorrect: calculates \(15 / 50 = 0.3\text{ mm}\) without converting units correctly.
C is incorrect: uses \(18 \times (50/15)\).
D is incorrect: inverted scale calculation.

Amylopectin contains \(\alpha\)-glucose monomers linked by \(\alpha\)-1,4-glycosidic bonds and \(\alpha\)-1,6-glycosidic branches.

Glycogen also contains \(\alpha\)-glucose monomers with \(\alpha\)-1,4-glycosidic bonds and more frequent \(\alpha\)-1,6-glycosidic branches.

Cellulose consists of unbranched chains of \(\beta\)-glucose monomers joined by \(\beta\)-1,4-glycosidic bonds, where alternate glucose units are rotated \(180^{\circ}\).

C is correct [1].

A is incorrect: cellulose is formed from \(\beta\)-glucose, not \(\alpha\)-glucose.
B is incorrect: glycogen has \(\alpha\)-1,6 bonds, not \(\beta\)-1,6 bonds.
D is incorrect: amylopectin does not contain \(\beta\)-glycosidic bonds.

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. As the substrate concentration is increased, substrate molecules outcompete the inhibitor for the active sites. Therefore, at very high substrate concentrations, the maximum velocity (\(V_{\max}\)) remains unchanged, but the value of the Michaelis-Menten constant (\(K_{\mathrm{m}}\)) increases because a higher concentration of substrate is required to achieve \(\frac{1}{2}V_{\max}\).

D is correct [1].

A is incorrect: \(V_{\max}\) is unchanged in competitive inhibition.
B is incorrect: \(K_{\mathrm{m}}\) increases, not decreases.
C is incorrect: \(V_{\max}\) can still be reached at high substrate concentrations.

When there is \(0\%\) change in mass, there is no net movement of water into or out of the potato cells by osmosis. This means the water potential of the potato tissue is equal to the water potential of the \(0.35\text{ mol dm}^{-3}\) sucrose solution. Since a higher concentration of sucrose gives a lower (more negative) water potential, the potato cells have a higher water potential than a \(0.50\text{ mol dm}^{-3}\) sucrose solution, and a lower water potential than pure water.

A is correct [1].

B is incorrect: water molecules still move across the membrane in both directions at equal rates (dynamic equilibrium).
C is incorrect: in \(0.20\text{ mol dm}^{-3}\) sucrose, water potential of the solution is higher than the cells, so water enters and cells become turgid, not flaccid.
D is incorrect: at \(0.35\text{ mol dm}^{-3}\), the cells are at incipient plasmolysis on average or in osmotic equilibrium, but water potential is equal, not zero.

Telomeres are non-coding, repetitive nucleotide sequences situated at the ends of eukaryotic linear chromosomes. During DNA replication, DNA polymerase cannot replicate the very end of the lagging strand. Telomeres ensure that vital protein-coding genes are not lost during successive rounds of DNA replication and prevent chromosome ends from being recognised as double-strand DNA breaks.

B is correct [1].

A is incorrect: centromeres attach to spindle microtubules, not telomeres.
C is incorrect: histone proteins are responsible for nucleosome condensation.
D is incorrect: DNA ligase joins Okazaki fragments.

In double-stranded DNA, the percentage of adenine (A) equals the percentage of thymine (T), and the percentage of guanine (G) equals the percentage of cytosine (C).

Given \(\text{A} = 32\%\):
\(\text{T} = 32\%\)
\(\text{Total (A + T)} = 32\% + 32\% = 64\%\)

Therefore, the remaining percentage for \(\text{G + C}\) is:
\(100\% - 64\% = 36\%\)

Since \(\text{G} = \text{C}\):
\(\text{Cytosine} = \frac{36\%}{2} = 18\%\).

C is correct [1].

A is incorrect: \(32\%\) is the proportion of thymine.
B is incorrect: \(36\%\) is the sum of guanine and cytosine.
D is incorrect: \(68\%\) is \(100 - 32\).

In phloem loading:
1. Proton pumps (\(\text{H}^+\)-ATPases) in companion cell surface membranes actively pump hydrogen ions (protons) out of the companion cells into the cell wall apoplast using ATP.
2. This sets up a high proton concentration gradient outside the companion cell.
3. Protons move back into the companion cells down their concentration gradient via cotransporter proteins, bringing sucrose molecules with them against the sucrose concentration gradient.
4. Sucrose then diffuses into the sieve tube element via plasmodesmata.

A is correct [1].

B is incorrect: \(\text{H}^+\) ions are pumped out of companion cells into cell walls, not into sieve tubes.
C is incorrect: sucrose enters via cotransport driven by the proton gradient, not direct active transport by ATP-driven pumps.
D is incorrect: proton movement back into companion cells is by facilitated diffusion down the proton gradient through cotransporters.

In actively respiring tissues:
1. \(\text{CO}_2\) diffuses into red blood cells where carbonic anhydrase catalyses its reaction with water 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. The increased \(\text{H}^+\) concentration causes \(\text{H}^+\) to bind to oxyhaemoglobin, forming haemoglobinic acid (\(\text{HHb}\)).
4. This causes haemoglobin to change shape and release its oxygen, shifting the oxygen dissociation curve to the right and reducing haemoglobin's affinity for oxygen at any given \(p\text{O}_2\).

D is correct [1].

A is incorrect: \(\text{CO}_2\) reacts to form \(\text{H}^+\), which binds to haemoglobin, shifting the curve to the right (not left).
B is incorrect: carbonic anhydrase forms carbonic acid, not carbaminohaemoglobin, and the curve shifts to the right.
C is incorrect: \(\text{H}^+\) binding decreases haemoglobin's affinity for oxygen, shifting the curve to the right.

During phloem loading:
1. Protons ($\text{H}^+$) are actively pumped out of the companion cell cytoplasm into the cell wall (apoplast) using ATP-powered proton pumps.
2. This sets up a proton concentration gradient; $\text{H}^+$ ions diffuse back into the companion cell through co-transporter proteins, bringing sucrose with them against its concentration gradient.
3. Sucrose then moves from the companion cell into the adjacent sieve tube element via symplastic connections called plasmodesmata down its concentration gradient.

B is correct [1].
- A is incorrect: $\text{H}^+$ ions are pumped out of the companion cell, not into it.
- C is incorrect: $\text{H}^+$ pumping is active, not facilitated diffusion, and movement into sieve tube elements is via plasmodesmata.
- D is incorrect: $\text{H}^+$ pumping requires active transport.

The potato tissue has a higher (less negative) water potential (\(-450\text{ kPa}\)) than the external sucrose solution (\(-1500\text{ kPa}\)). Water leaves the cells by osmosis down the water potential gradient, causing a loss in mass. Cylinder 2 has been cut into 8 pieces, giving it a greater total surface area to volume ratio ($SA:V$) than the intact Cylinder 1, which increases the rate of diffusion/osmosis. Therefore, Cylinder 2 loses mass at a faster rate and loses more mass over 45 minutes.

A is correct [1].
- B is incorrect: water moves out of the tissue, not into it.
- C is incorrect: potato cells lose water because the external solution has a lower water potential.
- D is incorrect: dynamic equilibrium is not reached immediately and surface area alters the rate of osmosis.

A competitive inhibitor binds reversibly to the active site of the enzyme, competing with the substrate. At sufficiently 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 the maximum velocity (\(\frac{1}{2}V_{\max}\)), so the Michaelis-Menten constant (\(K_m\)) increases (indicating lower affinity).

C is correct [1].
- A is incorrect: \(K_m\) increases, it does not remain unchanged.
- B is incorrect: non-competitive inhibitors decrease \(V_{\max}\).
- D is incorrect: competitive inhibitors do not change \(V_{\max}\) and they increase \(K_m\).

- Negative Biuret test indicates no protein/peptide bonds.
- Negative Iodine test indicates no starch (amylose/amylopectin).
- Negative initial Benedict's test indicates the absence of reducing sugars (such as glucose or maltose).
- Positive Benedict's test following acid hydrolysis and neutralisation indicates the presence of a non-reducing sugar, such as sucrose, which hydrolyses into reducing monosaccharides (glucose and fructose).

D is correct [1].
- A (amylose) would give a blue-black colour with iodine solution.
- B (maltose) is a reducing sugar and would give a positive Benedict's test before acid hydrolysis.
- C (globular protein) would give a purple/violet colour with Biuret reagent.

The organelle described has an outer membrane and an inner membrane folded into cristae, small circular DNA, and 70S ribosomes, which are characteristic features of a mitochondrion. Mitochondria are the site of aerobic respiration, where ATP is synthesised by chemiosmosis and substrate-linked phosphorylation.

B is correct [1].
- A is incorrect: chloroplasts have thylakoid membranes/grana, not cristae, and are not found in liver cells.
- C is incorrect: lysosomes are single-membrane bound and lack DNA/ribosomes.
- D is incorrect: the nucleolus is a non-membrane-bound subnuclear structure responsible for rRNA synthesis and ribosome subunit assembly.

In a diploid cell with \(2n = 24\):
- During metaphase of mitosis, there are 24 chromosomes aligned at the equator.
- Each chromosome consists of two sister chromatids joined at 1 centromere, giving a total of 24 centromeres.
- Each sister chromatid is a linear double-stranded DNA molecule having 2 telomeres (one at each end). Thus, each chromosome has \(2 \times 2 = 4\) telomeres.
- Total number of telomeres = \(24 \times 4 = 96\).

B is correct [1].
- A is incorrect: 48 telomeres would be present in an unreplicated chromosome state (\(G_1\)), not metaphase.
- C is incorrect: the number of centromeres is 24, not 48.
- D is incorrect: miscalculates both centromere and telomere numbers.

In arteries, the tunica media contains abundant elastic fibres. These stretch during ventricular systole to accommodate high-pressure surges of blood, and recoil during ventricular diastole, smoothing out pressure fluctuations and maintaining continuous flow.

Veins have a thin tunica media and do not actively pump blood; capillaries consist of a single layer of squamous endothelial cells (not ciliated cells); and arteries have a relatively narrow lumen to maintain high hydrostatic pressure.

A is correct [1].
- B is incorrect: veins have a thin muscle layer and do not pump blood.
- C is incorrect: capillaries are lined with squamous endothelium, not ciliated epithelium.
- D is incorrect: arteries have a relatively narrow lumen to maintain pressure.

1. Total number of base pairs = \(1400\), so total number of bases in the original double helix = \(1400 \times 2 = 2800\).
2. Percentage of cytosine (\(\text{C}\)) = \(32\%\). By complementary base pairing, percentage of guanine (\(\text{G}\)) = \(32\%\).
3. Total \(\text{C} + \text{G} = 64\%\).
4. Total \(\text{A} + \text{T} = 100\% - 64\% = 36\%\).
5. Percentage of adenine (\(\text{A}\)) = \(36\% \div 2 = 18\%\).
6. Number of adenine bases in the original double helix = \(0.18 \times 2800 = 504\).
7. During semi-conservative replication, the two parent strands separate, and new complementary strands are synthesised. The set of newly synthesised strands contains the same number of adenine nucleotides as the original DNA molecule, which is \(504\).

C is correct [1].
- A (252) represents the number of adenine bases in a single strand, not the total required for both newly synthesised daughter strands.
- B (448) is \(16\%\) of 2800 (calculation error on base percentages).
- D (1008) doubles the required adenine nucleotides incorrectly.

When magnification is increased from \(\times 100\) to \(\times 400\), the linear magnification increases by a factor of \(4\).

Consequently, the actual distance represented by each eyepiece graticule unit (epu) decreases by a factor of \(4\):
\[\text{Value of } 1.0\text{ epu at } \times 400 = \frac{10.0\ \mu\text{m}}{4} = 2.5\ \mu\text{m}\]

The actual diameter of the nucleus is therefore:
\[6.0\text{ epu} \times 2.5\ \mu\text{m} = 15.0\ \mu\text{m}\]

B is correct [1].

A is incorrect (incorrectly divided by 40 instead of 4).
C is incorrect (incorrectly multiplied calibration value by 4 and divided by 10).
D is incorrect (used \(\times 100\) calibration value of \(10.0\ \mu\text{m}\) per epu without adjusting for the \(\times 400\) magnification).

Analysing each test outcome:
- Iodine test remains yellow-brown: starch is absent.
- Biuret test turns purple: peptide bonds are present, confirming the presence of protein.
- Benedict's test remains blue both before and after acid hydrolysis: reducing sugars and non-reducing sugars are absent.
- Ethanol emulsion test forms a cloudy white emulsion: lipids are present.

Therefore, the sample contains protein and lipid only.

B is correct [1].

A is incorrect as non-reducing sugars were ruled out by negative Benedict's following acid hydrolysis.
C is incorrect as reducing sugars gave a negative Benedict's test.
D is incorrect as starch gave a negative iodine test.

A competitive inhibitor binds reversibly to the active site of an enzyme. Because it competes directly with substrate molecules:
- \(V_{\text{max}}\) remains unchanged because sufficiently high substrate concentrations outcompete the inhibitor (Statement 1 and Statement 2 are correct).
- The \(K_{\text{m}}\) increases, which signifies a lower (decreased) affinity of the enzyme for its substrate in the presence of the inhibitor (Statement 3 is correct).
- Non-competitive inhibitors bind to allosteric sites and decrease \(V_{\text{max}}\), so Statement 4 is incorrect.

Hence, statements 1, 2 and 3 are correct.

C is correct [1].

A and D are incorrect as statement 4 describes non-competitive inhibition.
B is incorrect as it omits statement 3, which correctly defines the relationship between increased Km and decreased affinity.

The percentage mass changes are:
- \(0.10\ \text{mol}\ \text{dm}^{-3}\): \(+12.0\%\)
- \(0.30\ \text{mol}\ \text{dm}^{-3}\): \(+3.0\%\)
- \(0.50\ \text{mol}\ \text{dm}^{-3}\): \(-7.0\%\)
- \(0.70\ \text{mol}\ \text{dm}^{-3}\): \(-16.0\%\)

The point of zero percentage change in mass (isotonic point) lies between \(0.30\) and \(0.50\ \text{mol}\ \text{dm}^{-3}\) (approximately \(0.36\ \text{mol}\ \text{dm}^{-3}\)).

In the \(0.70\ \text{mol}\ \text{dm}^{-3}\) sucrose solution, the solution has a lower (more negative) water potential than the cell sap of the potato cells. Water leaves the cells by osmosis down a water potential gradient, causing protoplasts to pull away from cell walls, making the cells plasmolysed.

B is correct [1].

A is incorrect as cells lose water in 0.70 mol dm^-3 and become plasmolysed, not turgid.
C and D are incorrect because 0.50 mol dm^-3 resulted in net mass loss (-7%), so the isotonic point must be below 0.50 mol dm^-3.

During S phase of interphase, semi-conservative DNA replication doubles the mass of DNA from \(4.0\ \text{pg}\) to \(8.0\ \text{pg}\).

In metaphase of mitosis:
- Each of the \(16\) chromosomes consists of two genetically identical sister chromatids: \(16 \times 2 = 32\) chromatids.
- Each chromosome has \(1\) centromere holding its two sister chromatids together: \(16\) centromeres.
- The total mass of DNA remains \(8.0\ \text{pg}\) until cytokinesis divides the cell.

B is correct [1].

A is incorrect as chromatid number doubles and DNA mass doubles before metaphase.
C is incorrect because sister chromatids share a single centromere per chromosome prior to anaphase.
D is incorrect as centromere count equals the chromosome number (16), not haploid (8).

In double-stranded DNA:
- \(\text{G} = 340\), so \(\text{C} = 340\). The total \(\text{G} + \text{C} = 680\) nucleotides.
- The total \(\text{A} + \text{T} = 1200 - 680 = 520\) nucleotides (with \(\text{A} = 260\) and \(\text{T} = 260\)).

During transcription, mRNA is synthesised from one template DNA strand, so the mRNA has \(\frac{1200}{2} = 600\) nucleotides.

Because base pairing dictates that DNA \(\text{A}\) pairs with RNA \(\text{U}\) and DNA \(\text{T}\) pairs with RNA \(\text{A}\), the sum of \(\text{A} + \text{U}\) in the single-stranded mRNA must equal the number of \(\text{A} + \text{T}\) on the template strand, which is \(\frac{520}{2} = 260\).

Therefore, statement B is correct.

B is correct [1].

A is incorrect as mRNA is single-stranded and contains 600 nucleotides.
C is incorrect as the double-stranded DNA contains 260 adenine nucleotides (not 340).
D is incorrect as mRNA is a nucleic acid and contains phosphodiester bonds, not peptide bonds.

All four statements correctly outline the active loading mechanism of sucrose into phloem:
1 \(\text{H}^+\) ions are actively pumped out of companion cells into the apoplast/cell wall by ATP-driven proton pumps.
2 A high \(\text{H}^+\) electrochemical gradient drives the secondary active transport of sucrose into companion cells via \(\text{H}^+\)-sucrose cotransporter proteins.
3 Sucrose diffuses down its concentration gradient from companion cells into adjacent sieve tube elements via plasmodesmata (symplastic route).
4 High solute (sucrose) concentration in the sieve tube element lowers its water potential (makes it more negative), drawing water in from nearby xylem by osmosis, generating hydrostatic pressure.

A is correct [1].

B, C and D are incorrect because all four statements (1, 2, 3 and 4) are accurate biological facts describing phloem loading in the syllabus.

During vigorous exercise, respiring muscle tissues produce large amounts of carbon dioxide (\(\text{CO}_2\)) and lactic acid, causing an increase in \(p\text{CO}_2\) and a decrease in blood pH.

This causes the oxygen dissociation curve of haemoglobin to shift to the right (the Bohr shift).

A rightward shift means that at any given partial pressure of oxygen (\(p\text{O}_2\)), haemoglobin has a lower percentage saturation with oxygen, facilitating increased dissociation (release) of oxygen to the respiring tissues.

B is correct [1].

A is incorrect as a decrease in pCO2 shifts the curve to the left.
C is incorrect as an increase in pH shifts the curve to the left.
D is incorrect as lower temperature shifts the curve to the left and haemoglobin affinity increases.

1. Determine the length of the stage micrometer divisions:
\(5\text{ divisions} = 5 \times 0.01\text{ mm} = 0.05\text{ mm} = 50\,\mu\text{m}\).

2. Calculate the value of \(1\) eyepiece graticule unit (epu):
\(25\text{ epu} = 50\,\mu\text{m}\)
\(1\text{ epu} = \frac{50\,\mu\text{m}}{25} = 2.0\,\mu\text{m}\).

3. Calculate the actual length of the palisade mesophyll cell:
\(\text{Actual length} = 18\text{ epu} \times 2.0\,\mu\text{m} = 36\,\mu\text{m}\).

[1] for correct option B (36 \(\mu\text{m}\)).
Incorrect options correspond to common calibration and unit conversion errors.

Maltose is a reducing disaccharide and gives a positive Benedict's test in Sample 1 without prior hydrolysis. Sucrose is a non-reducing disaccharide and does not react initially. When Sample 2 is boiled with dilute acid, the glycosidic bonds in sucrose (and maltose) are hydrolysed. Hydrolysis of one sucrose molecule produces one molecule of glucose and one molecule of fructose, both of which are reducing sugars. This substantially increases the concentration of reducing sugar monomers available to reduce copper(II) ions, producing a heavier precipitate.

[1] for correct option C;
Reject A, B, and D as scientifically incorrect regarding the reducing properties and hydrolysis products of sucrose and maltose.

A competitive inhibitor binds reversibly to the active site, competing with the substrate. This increases the Michaelis constant (\(K_{\text{m}}\)) because a higher substrate concentration is required to achieve half the maximum rate, but \(V_{\max}\) remains unchanged because high substrate concentrations displace the inhibitor. Thus, X is competitive and overcome by increasing substrate concentration. A non-competitive inhibitor binds to an allosteric site, decreasing \(V_{\max}\) without affecting \(K_{\text{m}}\), so Y is non-competitive.

[1] for correct option A.

The incipient plasmolysis / isotonic point is at \(0.32\,\text{mol}\,\text{dm}^{-3}\), where the water potential of the tissue matches the solution. In a \(0.50\,\text{mol}\,\text{dm}^{-3}\) sucrose solution, the solution has a higher solute concentration and therefore a more negative water potential than the potato cells. Consequently, water moves out of the cells down the water potential gradient by osmosis, leading to a loss in mass and plasmolysis.

[1] for correct option B.

In \(\text{G}_1\), there are 8 chromosomes, each consisting of 1 double-stranded DNA molecule (8 centromeres, 16 telomeres, 0 replicated sister chromatid pairs).
During S phase, DNA replicates. By metaphase:
- Each chromosome consists of 2 identical sister chromatids: \(8 \times 2 = 16\) chromatids.
- Sister chromatids are joined at \(1\) centromere per chromosome: \(8\) centromeres.
- Each chromatid has \(2\) telomeres (one at each end): \(16 \text{ chromatids} \times 2 = 32\) telomeres.

[1] for correct option B.

According to the semi-conservative model:
- Generation 0: 1 heavy molecule (\(^{15}\text{N}/^{15}\text{N}\)).
- Generation 1: 2 hybrid molecules (\(^{15}\text{N}/^{14}\text{N}\)).
- Generation 2: 2 hybrid molecules and 2 light molecules (\(^{14}\text{N}/^{14}\text{N}\)), total 4 molecules.
- Generation 3: 2 hybrid molecules and 6 light molecules (\(^{14}\text{N}/^{14}\text{N}\)), total 8 molecules.

Proportion of hybrid DNA = \(\frac{2}{8} = \frac{1}{4}\) (25%).
Proportion of light DNA = \(\frac{6}{8} = \frac{3}{4}\) (75%).

[1] for correct option B.

Sucrose loading involves:
1. Proton pumps (\(\text{H}^+\)-ATPases) in the companion cell surface membrane actively transport \(\text{H}^+\) ions out of the companion cell into the surrounding cell wall apoplast using ATP.
2. This creates a high concentration of protons (electrochemical gradient) in the apoplast.
3. Protons diffuse back into the companion cell down their electrochemical gradient through cotransporter (symporter) proteins, carrying sucrose molecules into the cell against the sucrose concentration gradient.

[1] for correct option A.

An increase in \(p\text{CO}_2\) and hydrogen ion concentration (lower pH) causes the oxygen dissociation curve of haemoglobin to shift to the right (the Bohr effect). This decreases haemoglobin's affinity for oxygen at any given partial pressure of oxygen, promoting the dissociation and release (unloading) of oxygen to the vigorously respiring muscle tissues where it is needed for aerobic respiration.

[1] for correct option A.

(a)
1. Sieve tube elements lack a nucleus, vacuoles, and major organelles / have a thin peripheral layer of cytoplasm: this provides a wide, open lumen that reduces resistance to the mass flow of sap.
2. Sieve plates with open sieve pores at the end walls: these allow continuous flow of hydrostatic pressure-driven sap between adjacent sieve tube elements.

(b)
1. Active transport of protons ( ext{H}^+): Proton pumps ( ext{H}^+-ATPase) in the companion cell surface membrane actively pump ext{H}^+ ions out of the cytoplasm into the cell wall, using ATP.
2. Concentration / electrochemical gradient: This establishes a higher concentration of ext{H}^+ outside the companion cell compared to the inside.
3. Cotransport of sucrose: ext{H}^+ ions diffuse back down their electrochemical gradient into the companion cell through cotransporter (symporter) proteins.
4. Sucrose entry: Sucrose is transported simultaneously against its concentration gradient alongside ext{H}^+ ions through the cotransporter.

(c)
Any two from:
- Actively growing root tips / roots
- Storage organs (e.g. tubers / taproots / bulbs / seeds)
- Developing flowers / fruits
- Growing shoot tips / young developing leaves

(a) 1 mark for structural feature, 1 mark for corresponding explanation, max 4 marks:
- lack of nucleus / lack of tonoplast / thin peripheral layer of cytoplasm / reduced organelles; [1]
- reduces resistance to flow / provides large / unobstructed lumen for mass flow; [1]
- presence of sieve plates / perforated end walls / sieve pores; [1]
- allows continuous longitudinal flow of sap / connects lumen of adjacent cells; [1]
- elongated cells joined end-to-end; [1]
- forms a continuous tube / column; [1]
(Note: feature and explanation must match)

(b) max 4 marks:
- ext{H}^+ / protons actively pumped / transported out of companion cell (into cell wall); [1]
- using ATP / hydrolysis of ATP / via ext{H}^+ pump; [1]
- creates a concentration / electrochemical / ext{H}^+ gradient (higher ext{H}^+ outside); [1]
- ext{H}^+ moves back into companion cell by facilitated diffusion / down its concentration gradient; [1]
- via cotransporter / symporter protein; [1]
- sucrose is carried together with ext{H}^+ (against sucrose concentration gradient); [1]

(c) 1 mark each, max 2 marks:
- roots / root tips / root apical meristems; [1]
- developing fruit / seeds; [1]
- storage organs / tubers / bulbs / rhizomes; [1]
- shoot tips / developing flower / young leaves; [1]
(A: any named valid plant storage/growth organ; R: mature leaves / photosynthetic leaves)

(a)(i) At the start of the reaction, substrate concentration is at its highest and is known / not limiting. Over time, substrate is used up and its concentration decreases, leading to fewer enzyme-substrate collisions, so the rate decreases. Measuring the initial rate ensures the measured rate reflects the set independent variable without substrate depletion.

(a)(ii) As hydrogen peroxide concentration increases from 10 to $160\,\text{mmol}\,\text{dm}^{-3}$, the rate increases from 12.0 to $54.0\,\text{cm}^3\,\text{min}^{-1}$. Above $160\,\text{mmol}\,\text{dm}^{-3}$ (from 160 to $320\,\text{mmol}\,\text{dm}^{-3}$), the rate plateaus / levels off around $54.0\text{--}55.0\,\text{cm}^3\,\text{min}^{-1}$ as $V_{\max}$ is reached and enzyme active sites become saturated.

(b) $V_{\max}$ for both free and immobilized enzyme is approximately $55\,\text{cm}^3\,\text{min}^{-1}$. Half $V_{\max}$ is $27.5\,\text{cm}^3\,\text{min}^{-1}$. For free catalase, half $V_{\max}$ is reached at a lower substrate concentration (approx. $26\text{--}28\,\text{mmol}\,\text{dm}^{-3}$) than for immobilized catalase ($80\,\text{mmol}\,\text{dm}^{-3}$). Therefore, free catalase has a lower $K_m$ and a higher affinity for hydrogen peroxide than immobilized catalase, because the alginate matrix creates a diffusion barrier slowing substrate access to the active site.

(c)
1. The enzyme can be easily recovered and reused.
2. The product is not contaminated with enzyme, reducing downstream purification costs.
3. Immobilized enzymes generally have greater thermal and pH stability.

(a)(i) max 2 marks:
- substrate concentration is not limiting / at a known / maximum concentration at $t = 0$; [1]
- rate decreases over time as substrate is depleted / product accumulates / reaction slows down; [1]
- allows a valid comparison of enzyme activity at specified substrate concentrations; [1]

(a)(ii) max 2 marks:
- as substrate concentration increases, initial rate increases and then levels off / plateaus / approaches $V_{\max}$; [1]
- quoted data with units from Table 2.1 (e.g. increases from $12.0\,\text{cm}^3\,\text{min}^{-1}$ at $10\,\text{mmol}\,\text{dm}^{-3}$ to $54.0\,\text{cm}^3\,\text{min}^{-1}$ at $160\,\text{mmol}\,\text{dm}^{-3}$ and levels off at $55.0\,\text{cm}^3\,\text{min}^{-1}$ at $320\,\text{mmol}\,\text{dm}^{-3}$); [1]
- active sites become saturated / enzyme concentration becomes the limiting factor at high substrate concentrations; [1]

(b) max 3 marks:
- $K_m$ is the substrate concentration at half $V_{\max}$; [1]
- free catalase has a lower $K_m$ / requires lower substrate concentration to reach half $V_{\max}$ than immobilized catalase; [1]
- data quote: half $V_{\max}$ ($\,\approx 27.5\,\text{cm}^3\,\text{min}^{-1}$) reached at $80\,\text{mmol}\,\text{dm}^{-3}$ for immobilized vs approx. $26\text{--}28\,\text{mmol}\,\text{dm}^{-3}$ for free; [1]
- lower $K_m$ means free catalase has a higher affinity for substrate / alginate matrix acts as a diffusion barrier; [1]

(c) max 3 marks:
- enzyme can be recovered / separated and reused; [1]
- product is uncontaminated / less downstream processing / purification required; [1]
- enzyme is more stable at higher temperatures / wider range of pH; [1]
- continuous production / flow-through bioreactor is possible; [1]

(a)(i) 'Fluid' refers to the ability of phospholipids and proteins to move laterally within their monolayer. 'Mosaic' refers to the scattered, irregular arrangement of various proteins embedded in or associated with the phospholipid bilayer.

(a)(ii)
1. Regulates/maintains membrane fluidity at different temperatures (prevents membrane becoming too fluid at high temperatures and prevents closely packed hydrocarbon chains from becoming too rigid at low temperatures).
2. Increases mechanical stability of the membrane / reduces permeability to small water-soluble molecules/ions.

(b)
- $\text{K}^+$ ions are moved against their concentration gradient by active transport.
- This requires specific carrier / transport proteins / $\text{K}^+$ pumps in the cell surface membrane.
- Active transport requires ATP / metabolic energy.
- In the presence of oxygen, aerobic respiration occurs in mitochondria, generating ATP to power the carrier proteins.
- In the absence of oxygen, aerobic respiration stops / ATP production is insufficient, so active transport cannot occur.

(c)
- Simple diffusion occurs directly through the phospholipid bilayer (between phospholipid molecules), whereas facilitated diffusion requires specific channel or carrier transport proteins.
- Simple diffusion typically transports small, non-polar / hydrophobic molecules, whereas facilitated diffusion transports polar, charged, or larger molecules (e.g. ions, glucose).

(a)(i) 1 mark each, max 2 marks:
- fluid: phospholipids (and proteins) can move laterally / diffuse within the layer; [1]
- mosaic: proteins are scattered / randomly distributed / embedded within the phospholipid bilayer; [1]

(a)(ii) 1 mark each, max 2 marks:
- maintains / regulates membrane fluidity / prevents membrane becoming too fluid at high temperatures / too rigid at low temperatures; [1]
- increases mechanical stability / strengthens membrane; [1]
- decreases permeability to hydrophilic substances / water-soluble ions / polar molecules; [1]

(b) max 4 marks:
- uptake is by active transport; [1]
- movement is against a concentration gradient / from lower to higher concentration; [1]
- involves carrier proteins / intrinsic transport proteins / ion pumps; [1]
- requires ATP / energy from respiration; [1]
- oxygen is required for aerobic respiration / oxidative phosphorylation (to produce ATP); [1]
- lack of oxygen prevents / reduces ATP production, so carrier cannot change shape / transport stops; [1]

(c) max 2 marks (must be comparative points):
- simple diffusion is directly through the phospholipid bilayer, whereas facilitated diffusion requires transport / channel / carrier proteins; [1]
- simple diffusion transports non-polar / small lipid-soluble molecules, whereas facilitated diffusion transports charged / polar / large water-soluble molecules; [1]
(R: statements suggesting either requires ATP)

(a) Adult haemoglobin consists of four polypeptide chains (two $\alpha$-globin and two $\beta$-globin subunits). Each polypeptide subunit is associated with a non-protein prosthetic group called a haem group, containing an iron(II) ion ($\text{Fe}^{2+}$). The subunits are held together by hydrophobic interactions, hydrogen bonds, and ionic bonds.

(b)(i) Bohr effect / Bohr shift.

(b)(ii) Actively respiring tissues produce high amounts of $\text{CO}_2$, lowering pH. At higher $p\text{CO}_2$, haemoglobin's affinity for oxygen decreases (dissociation curve shifts to the right). Therefore, at any given $p\text{O}_2$ in respiring tissues, haemoglobin releases / unloads more oxygen, making more $\text{O}_2$ available for aerobic cellular respiration.

(c) Carbonic anhydrase catalyses the reversible reaction between $\text{CO}_2$ and $\text{H}_2\text{O}$ inside red blood cells to form carbonic acid ($\text{H}_2\text{CO}_3$), which dissociates into $\text{H}^+$ and hydrogencarbonate ions ($\text{HCO}_3^-$). $\text{HCO}_3^-$ ions diffuse out of erythrocytes into the plasma down a concentration gradient. To maintain electrical neutrality, chloride ions ($\text{Cl}^-$) diffuse from plasma into red blood cells (the chloride shift).

(a) max 3 marks:
- four polypeptide subunits / chains; [1]
- two alpha ($\alpha$) and two beta ($\beta$) chains / globins; [1]
- each polypeptide contains a haem prosthetic group / four haem groups total; [1]
- each haem contains an iron ion / $\text{Fe}^{2+}$; [1]
- subunits held together by hydrophobic interactions / ionic bonds / hydrogen bonds; [1]

(b)(i) 1 mark:
- Bohr effect / Bohr shift; [1]

(b)(ii) max 3 marks:
- respiring tissues have high $p\text{CO}_2$ / lower pH / acidic conditions; [1]
- affinity of haemoglobin for oxygen decreases / curve shifts to the right; [1]
- more oxygen is released / dissociated / unloaded from oxyhaemoglobin at the same $p\text{O}_2$; [1]
- delivers more oxygen to tissues for aerobic respiration / ATP production; [1]

(c) max 3 marks:
- carbonic anhydrase catalyses the reaction: $\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3$ / formation of carbonic acid; [1]
- $\text{H}_2\text{CO}_3$ dissociates into $\text{H}^+$ and $\text{HCO}_3^-$ / hydrogencarbonate; [1]
- $\text{HCO}_3^-$ diffuses out of erythrocyte / into plasma; [1]
- chloride shift: chloride ions ($\text{Cl}^-$) move into erythrocyte to maintain electrical neutrality / balance charge; [1]

(a)
- Cholera: Type of pathogen = Bacterium; Name = Vibrio cholerae
- Tuberculosis: Type of pathogen = Bacterium; Name = Mycobacterium tuberculosis (or Mycobacterium bovis)
- Malaria: Name of pathogen = Plasmodium falciparum / P. vivax / P. malariae / P. ovale

(b)(i)
- Penicillin inhibits the enzyme transpeptidase (glycopeptide transpeptidase).
- This prevents the formation of peptide cross-links between peptidoglycan chains during bacterial cell wall synthesis.
- When the bacterium grows or takes up water by osmosis, the weakened wall cannot withstand internal turgor pressure, leading to cell lysis / bursting.

(b)(ii) Viruses do not have a cellular structure, cell wall, or peptidoglycan / do not have their own metabolic enzymes or transpeptidase for penicillin to target.

(c)
Any two from:
- Vaccination using the BCG vaccine in vulnerable populations.
- Contact tracing and screening (e.g. chest X-rays / sputum tests) of individuals in close contact with infected patients.
- Treatment with a full multi-drug antibiotic regimen (DOTS / directly observed therapy) to ensure compliance and avoid resistance.
- Isolation of active TB patients in well-ventilated rooms to prevent aerosol droplet transmission.
- Pasteurization of milk to eliminate Mycobacterium bovis.

(a) 1 mark each for correct row / pathogen name, max 4 marks:
- Cholera type: bacterium / bacteria AND name: Vibrio cholerae; [1]
- TB type: bacterium / bacteria; [1]
- TB name: Mycobacterium tuberculosis / Mycobacterium bovis; [1]
- Malaria name: Plasmodium falciparum / Plasmodium vivax / Plasmodium malariae / Plasmodium ovale; [1]
(Note: genus names must have capital letter, species lowercase)

(b)(i) max 3 marks:
- inhibits transpeptidase / glycoprotein peptidase enzyme; [1]
- prevents cross-link formation between peptidoglycan chains / prevents cell wall synthesis; [1]
- cell wall is weakened / autolysins continue to break down wall; [1]
- cell bursts / lyses due to osmotic uptake of water / hydrostatic pressure; [1]
(A: only acts on growing/dividing bacteria)

(b)(ii) 1 mark:
- viruses do not have a cell wall / peptidoglycan / cell structure / metabolic enzymes; [1]

(c) max 2 marks (1 mark per distinct measure):
- BCG vaccination (of infants / high risk groups); [1]
- isolating infected individuals with active pulmonary TB; [1]
- contact tracing of people exposed to infected patients; [1]
- ensuring patients complete full course of multiple antibiotics / DOTS programme; [1]
- pasteurisation of milk (to prevent M. bovis); [1]
- improving housing / reducing overcrowding / improving ventilation; [1]

(a)(i) DNA polymerase catalyses the addition of free deoxyribonucleotides to the 3' end of a growing DNA strand via complementary base pairing ($A-T$ and $C-G$). It catalyses the formation of phosphodiester bonds between adjacent nucleotides in the 5' to 3' direction.

(a)(ii) DNA ligase joins Okazaki fragments on the lagging strand by catalysing the formation of phosphodiester bonds between the sugar-phosphate backbones of adjacent DNA fragments.

(b)(i) $5'\text{ - A U G G C A U U G C U A A G U - }3'$

(b)(ii) 5 amino acids (15 bases / 3 bases per codon = 5 codons).

(c)
- Primary structure: The substitution changes one mRNA codon. Due to the degenerate nature of the genetic code, the new codon may code for the same amino acid (silent mutation, no change in primary structure), or it may code for a different amino acid (missense mutation, one amino acid substituted in primary structure), or code for a stop codon (nonsense mutation, truncated polypeptide).
- Tertiary structure: If an amino acid with different R-group properties (e.g. polar vs non-polar, charged vs uncharged) is inserted, it may disrupt the formation of hydrogen bonds, ionic bonds, disulfide bridges, or hydrophobic interactions.
- This alters the specific 3D folding / conformation of the tertiary structure, which can lead to a loss of function (e.g. altered active site shape in an enzyme).

(a)(i) max 2 marks:
- adds free (deoxyribo)nucleotides to the 3' end / extends growing strand; [1]
- by complementary base pairing (A to T, C to G); [1]
- catalyses the formation of phosphodiester bonds (between phosphate and deoxyribose); [1]

(a)(ii) max 2 marks:
- joins Okazaki fragments / discontinuous fragments on lagging strand; [1]
- forms phosphodiester bonds (between adjacent sugar-phosphate backbones); [1]

(b)(i) 1 mark:
- $5'\text{ - A U G G C A U U G C U A A G U - }3'$ (must have correct $5'$ and $3'$ orientation and U instead of T); [1]

(b)(ii) 1 mark:
- 5 / five; [1]

(c) max 4 marks:
- changes one base in a codon / triplet on mRNA; [1]
- silent mutation: code is degenerate, so may code for the same amino acid (no effect on primary structure); [1]
- missense / substitution: codes for a different amino acid, altering primary structure / sequence of amino acids; [1]
- nonsense mutation: may create a premature stop codon, producing a shorter / truncated polypeptide; [1]
- change in R-group alters ionic bonds / hydrogen bonds / disulfide bridges / hydrophobic interactions; [1]
- alters 3D folding / conformation / tertiary structure (or active site shape / loss of function); [1]

(a) (i) Using \(C_1 V_1 = C_2 V_2\) to prepare \(20.0\text{ cm}^3\) total volume:
- For \(2.4\%\): \(V_1 = \frac{2.4 \times 20.0}{3.0} = 16.0\text{ cm}^3\) of H, and \(20.0 - 16.0 = 4.0\text{ cm}^3\) of W.
- For \(1.8\%\): \(V_1 = \frac{1.8 \times 20.0}{3.0} = 12.0\text{ cm}^3\) of H, and \(20.0 - 12.0 = 8.0\text{ cm}^3\) of W.
- For \(1.2\%\): \(V_1 = \frac{1.2 \times 20.0}{3.0} = 8.0\text{ cm}^3\) of H, and \(20.0 - 8.0 = 12.0\text{ cm}^3\) of W.
- For \(0.6\%\): \(V_1 = \frac{0.6 \times 20.0}{3.0} = 4.0\text{ cm}^3\) of H, and \(20.0 - 4.0 = 16.0\text{ cm}^3\) of W.

(a) (ii)
- Independent variable: concentration of hydrogen peroxide / substrate (\(\%\)).
- Dependent variable: volume of oxygen gas produced in \(60\text{ s}\) (\(\text{cm}^3\)).

(a) (iii)
Table of Results:
| Concentration of hydrogen peroxide / \(\%\) | Volume of oxygen gas produced in \(60\text{ s}\) / \(\text{cm}^3\) — Trial 1 | Volume of oxygen gas produced in \(60\text{ s}\) / \(\text{cm}^3\) — Trial 2 | Volume of oxygen gas produced in \(60\text{ s}\) / \(\text{cm}^3\) — Trial 3 | Mean volume of oxygen gas produced in \(60\text{ s}\) / \(\text{cm}^3\) |
|---|---|---|---|---|
| \(0.6\) | \(7.5\) | \(8.0\) | \(7.0\) | \(7.5\) |
| \(1.2\) | \(14.0\) | \(15.0\) | \(14.5\) | \(14.5\) |
| \(1.8\) | \(20.5\) | \(21.0\) | \(19.5\) | \(20.3\) |
| \(2.4\) | \(24.0\) | \(25.5\) | \(24.5\) | \(24.7\) |
| \(3.0\) | \(26.0\) | \(27.0\) | \(26.5\) | \(26.5\) |

(b) (i)
- \(x\)-axis: Concentration of hydrogen peroxide / \(\%\)
- \(y\)-axis: Mean volume of oxygen gas produced in \(60\text{ s}\) / \(\text{cm}^3\)
- Scales: Linear, occupying more than half the grid in both dimensions.
- Points plotted accurately to within half a small square.
- Points joined point-to-point with straight ruled lines or a smooth curve of best fit.

(b) (ii)
From the graph at \(1.5\%\) hydrogen peroxide concentration, the mean volume of oxygen is approximately \(17.5\text{ cm}^3\) (acceptable range \(17.0 - 18.0\text{ cm}^3\)).
Rate = \(\frac{\text{Volume}}{\text{Time}} = \frac{17.5\text{ cm}^3}{60\text{ s}} = 0.292\text{ cm}^3\text{ s}^{-1}\) (or \(0.29\text{ cm}^3\text{ s}^{-1}\)).

(c) (i)
Source of error: Gas escapes while inserting the bung/stopper into the reaction flask after adding enzyme E.
Effect: The recorded volume of oxygen gas is lower than the actual volume produced, underestimating the rate.

(c) (ii)
1. Use a two-necked flask or a divided reaction vessel (or deliver enzyme via a syringe through a rubber septum) so no gas is lost when starting the reaction.
2. Use a thermostatically controlled water bath to keep the reaction temperature constant across all trials.

(c) (iii)
At high substrate concentrations, all enzyme active sites are occupied / catalase is saturated with substrate, so enzyme concentration becomes the limiting factor.

(a)(i) [3 marks total]
- 1 mark for correct volumes of H and W for \(2.4\%\) (\(16.0\) and \(4.0\)) and \(1.8\%\) (\(12.0\) and \(8.0\));
- 1 mark for correct volumes of H and W for \(1.2\%\) (\(8.0\) and \(12.0\)) and \(0.6\%\) (\(4.0\) and \(16.0\));
- 1 mark for all volumes given to one decimal place (e.g. \(16.0\), not \(16\));

(a)(ii) [2 marks total]
- 1 mark for Independent variable: concentration of hydrogen peroxide / substrate (concentration);
- 1 mark for Dependent variable: volume of oxygen / gas produced (in \(60\text{ s}\)) / rate of gas production;

(a)(iii) [4 marks total]
- 1 mark for fully enclosed table with fully ruled borders, correct headings with units separated by slash or brackets (`Concentration of hydrogen peroxide / %`, `Volume of oxygen / cm³`);
- 1 mark for correctly recording all raw data in trials 1, 2, and 3;
- 1 mark for calculating all means correctly (\(7.5\), \(14.5\), \(20.3\) or \(20.33\), \(24.7\) or \(24.67\), \(26.5\));
- 1 mark for consistent precision across raw data (all to 1 d.p.) and means to same or one more d.p.;

(b)(i) [4 marks total]
- 1 mark for axes correctly labelled with units (`Concentration of hydrogen peroxide / %` on x-axis and `Mean volume of oxygen gas / cm³` on y-axis);
- 1 mark for linear scale with values increasing from 0, using more than \(50\%\) of grid in both directions;
- 1 mark for all 5 points plotted accurately to within half a small square;
- 1 mark for points joined with thin, sharp ruled straight lines point-to-point OR drawn as a smooth curve of best fit without extrapolation;

(b)(ii) [2 marks total]
- 1 mark for correct reading from candidate's graph at \(1.5\%\) (e.g. \(17.0 - 18.0\text{ cm}^3\));
- 1 mark for dividing volume by \(60\) with correct units (\(0.28 - 0.31\text{ cm}^3\text{ s}^{-1}\));

(c)(i) [2 marks total]
- 1 mark for identifying gas loss during bung insertion / delay between adding enzyme and closing flask / temperature fluctuation;
- 1 mark for correct effect explained (e.g. gas lost leads to lower measured volume / lower calculated rate);

(c)(ii) [2 marks total]
- 1 mark each for any two valid improvements (max 2):
- Use a syringe to inject enzyme through a rubber septum to prevent gas loss;
- Use a water bath to maintain constant temperature;
- Use a magnetic stirrer to ensure continuous mixing;
- Use a data logger with gas pressure sensor to eliminate human timing error;

(c)(iii) [1 mark]
- Catalase active sites are saturated / enzyme concentration becomes limiting factor / all active sites are occupied.

(a) (i) Plan diagram requirements:
- Clear, continuous, unshaded lines drawn with a sharp pencil.
- Proper proportions: large cortex surrounding a central, compact stele/vascular cylinder.
- Distinct tissue boundaries: epidermis on the outside, wide cortex, distinct endodermis layer, central star-shaped / radial xylem with alternating phloem patches.
- No individual cells drawn.
- Correct label lines with ruler to cortex and xylem.

(a) (ii) High-power cell drawing requirements:
- Group of four touching xylem vessel elements drawn.
- Double lines for cell walls showing thickness and middle lamella.
- Angular / polygonal shapes of vessel lumina correctly depicted.
- Label lines with ruler to lumen and cell wall.

(b) (i)
- Total length of stage micrometer = \(1.0\text{ mm} = 1000\,\mu\text{m}\).
- Value of 1 stage micrometer division = \(\frac{1000\,\mu\text{m}}{100} = 10\,\mu\text{m}\).
- \(40\) stage micrometer divisions = \(40 \times 10\,\mu\text{m} = 400\,\mu\text{m}\).
- Since \(64\) eyepiece graticule units (epu) = \(400\,\mu\text{m}\):
\(1\text{ epu} = \frac{400\,\mu\text{m}}{64} = 6.25\,\mu\text{m}\).

(b) (ii)
- Actual diameter = \(18\text{ epu} \times 6.25\,\mu\text{m/epu} = 112.5\,\mu\text{m}\).

(b) (iii)
- Image size \(I = 33.8\text{ mm} = 33800\,\mu\text{m}\).
- Actual size \(A = 112.5\,\mu\text{m}\).
- Magnification \(M = \frac{I}{A} = \frac{33800}{112.5} = 300.44\approx \times 300\) (or \(\times 300.4\)).

(c) Table of Comparison:
| Feature / Tissue | Root (Figure 2.1) | Stem (Figure 2.2) |
|---|---|---|
| Arrangement of vascular tissue | Located in a central cylinder / stele | Arranged in a ring of separate vascular bundles near periphery |
| Arrangement of xylem | Star-shaped / cruciform in centre with radiating arms | Located on the inner side of each vascular bundle |
| Arrangement of phloem | Located in clusters between the xylem arms | Located on the outer side of each vascular bundle |
| Pith / central tissue | Absent (centre occupied by xylem) | Present (large central parenchyma / pith) |
| Endodermis / Casparian strip | Prominent single layer bounding the vascular cylinder | Inconspicuous / starch sheath / absent |

(a)(i) [5 marks total]
- 1 mark for clear, continuous outline, no feathering, no shading, filling at least half the available space;
- 1 mark for drawing correct proportion: wide cortex and central vascular cylinder (stele);
- 1 mark for correctly showing distinct tissue layers: epidermis, cortex boundary, endodermis ring;
- 1 mark for showing star-shaped/radial xylem core with phloem groups between xylem arms;
- 1 mark for correct labels with straight uncrossed label lines to `cortex` and `xylem`;

(a)(ii) [4 marks total]
- 1 mark for drawing exactly four adjacent/touching xylem vessel elements;
- 1 mark for cell walls represented by clean double lines with appropriate wall thickness;
- 1 mark for realistic polygonal/rounded lumen shapes with no gaps between adjacent cells except intercellular spaces if appropriate;
- 1 mark for correct labels with straight lines to `cell wall` and `lumen`;

(b)(i) [2 marks total]
- 1 mark for calculating distance for 40 stage divisions = \(400\,\mu\text{m}\) (or \(1\text{ div} = 10\,\mu\text{m}\));
- 1 mark for dividing \(400\) by \(64\) to give \(6.25\,\mu\text{m}\);

(b)(ii) [1 mark]
- 1 mark for \(18 \times 6.25 = 112.5\,\mu\text{m}\) (allow \(113\,\mu\text{m}\));

(b)(iii) [2 marks total]
- 1 mark for converting \(33.8\text{ mm}\) to \(33800\,\mu\text{m}\) and applying \(M = \frac{I}{A}\);
- 1 mark for correct magnification \(\times 300\) or \(\times 300.4\) (ecf from (b)(ii));

(c) [6 marks total]
- 1 mark for table format with clear column/row headings (`Feature`, `Root (Figure 2.1)`, `Stem (Figure 2.2)`);
- 1 mark for each valid direct comparative point between root and stem (max 3 differences, 1 mark per difference described for both organs, plus 1 mark for each accurate structural detail up to 5 marks for content):
- Difference 1: Vascular tissue centrally located in root vs arranged in a peripheral ring / separate bundles in stem;
- Difference 2: Xylem in central star/cross in root vs inner part of vascular bundles in stem;
- Difference 3: Phloem in patches between xylem arms in root vs outer region of bundles in stem;
- Difference 4: Pith absent in root vs pith present/prominent in centre of stem;
- Difference 5: Endodermis well-defined single layer in root vs endodermis absent / indistinct in stem.

(a)
(i) The link reaction occurs inside the mitochondrial matrix.
(ii) Pyruvate is decarboxylated and dehydrogenated to form an acetyl group that binds with coenzyme A to form acetyl coenzyme A (acetyl CoA).

(b)
Rice adaptations to submergence:
1. Presence of aerenchyma (specialised tissue with large air spaces) in stems and roots, which facilitates the diffusion of oxygen from aerial parts down to submerged roots.
2. Rapid stem / internode elongation stimulated by ethylene, enabling leaves to stay above the surface of the floodwater for gas exchange and photosynthesis.
3. Roots can carry out ethanol fermentation under hypoxic/anaerobic conditions.
4. Root cells have higher tolerance to toxic ethanol due to elevated concentrations of the enzyme alcohol dehydrogenase.

(c)
(i) \(\text{RQ} = \frac{\text{volume of } \text{CO}_2 \text{ produced}}{\text{volume of } \text{O}_2 \text{ consumed}} = \frac{29.4}{42.0} = 0.70\).
(ii) The primary substrate is lipids / fatty acids / triglycerides, because lipids have a higher proportion of hydrogen-to-carbon/oxygen atoms, requiring more oxygen molecules per molecule of carbon dioxide released (giving an RQ of approximately \(0.7\)).
(iii) Under anaerobic conditions, \(\text{O}_2\) consumption becomes zero while \(\text{CO}_2\) is still released during fermentation, so the RQ becomes theoretically infinite / extremely high.

(a)(i)
- mitochondrial matrix; [1]

(a)(ii)
- acetyl coenzyme A / acetyl CoA; [1]

(b) (max 4 marks)
- aerenchyma / tissue with air spaces; [1]
- allows diffusion / transport of oxygen / gases to submerged roots; [1]
- increased / rapid stem / internode / petiole elongation / upward growth (stimulated by ethylene); [1]
- keeps leaves / photosynthetic parts above water level; [1]
- roots carry out anaerobic respiration / alcoholic / ethanol fermentation; [1]
- tolerant to ethanol / high levels of alcohol dehydrogenase; [1]

(c)(i)
- \(0.70\) / \(0.7\); [1]

(c)(ii)
- lipid / fatty acid / oil / triglyceride; [1]
- lipids have more C-H bonds / higher hydrogen:oxygen ratio / require more oxygen for oxidation per \(\text{CO}_2\) produced / carbohydrate has RQ of \(1.0\); [1]

(c)(iii)
- RQ increases / becomes very large / approaches infinity / undefined (no \(\text{O}_2\) taken up); [1]

(a)
Differences between non-cyclic and cyclic photophosphorylation:
1. Non-cyclic involves both Photosystem II (PSII) and Photosystem I (PSI), whereas cyclic involves Photosystem I (PSI) only.
2. Non-cyclic produces ATP, reduced NADP, and oxygen (via photolysis of water), whereas cyclic produces ATP only (no reduced NADP and no oxygen).
3. Non-cyclic involves the photolysis of water to replace lost electrons, whereas cyclic involves electrons returning back to the reaction centre chlorophyll of PSI.

(b)
- In the dark, the light-dependent stage stops immediately, so the supply of ATP and reduced NADP ceases.
- Carbon dioxide fixation by rubisco can temporarily continue, combining **RuBP with \(\text{CO}_2\)** to form GP, causing RuBP to be used up.
- Without ATP and reduced NADP, GP cannot be reduced to triose phosphate (TP), so GP accumulates and RuBP cannot be regenerated.

(c)
(i) At low light intensity, light intensity is the limiting factor; the light-dependent reactions are photochemical and independent of temperature.
At high light intensity, light is no longer limiting; the light-independent reactions (Calvin cycle) are enzyme-controlled (e.g. rubisco). Increasing the temperature increases the kinetic energy of enzyme and substrate molecules, leading to more frequent successful collisions and faster enzyme-substrate complex formation.
(ii) Rubisco (ribulose-1,5-bisphosphate carboxylase-oxygenase).

(a) (any 3 marks, must be comparative)
- non-cyclic involves PSI and PSII whereas cyclic involves PSI only; [1]
- non-cyclic photophosphorylation produces ATP and reduced NADP whereas cyclic produces ATP only / no reduced NADP; [1]
- non-cyclic photophosphorylation involves photolysis of water / oxygen evolution whereas cyclic does not; [1]
- in non-cyclic electrons do not return to reaction centre / flow to NADP whereas in cyclic electrons cycle back to PSI; [1]

(b) (max 3 marks)
- light-dependent reaction stops so no ATP / reduced NADP produced; [1]
- RuBP continues to combine with \(\text{CO}_2\) to form GP (so RuBP falls); [1]
- conversion of GP to triose phosphate (TP) requires ATP and reduced NADP; [1]
- without ATP / reduced NADP, GP cannot be reduced / converted / accumulates and RuBP cannot be regenerated; [1]

(c)(i) (max 3 marks)
- at low light intensity, light is the limiting factor / rate is independent of temperature; [1]
- at high light intensity, temperature / Calvin cycle is the limiting factor; [1]
- Calvin cycle / light-independent reaction is catalysed by enzymes / enzyme-controlled; [1]
- higher temperature increases kinetic energy of molecules / more frequent successful collisions between enzyme and substrate / more ES complexes formed; [1]

(c)(ii)
- rubisco / ribulose bisphosphate carboxylase / ribulose-1,5-bisphosphate carboxylase-oxygenase; [1]

(a)
- High hydrostatic pressure is established in the glomerulus because the afferent arteriole has a wider diameter than the efferent arteriole.
- This pressure forces water and small dissolved solutes out of the capillary blood.
- Solutes pass through three filtration barriers: the fenestrations (pores) in the capillary endothelium, the basement membrane (which acts as the main selective molecular sieve preventing large plasma proteins from passing), and the filtration slits between the foot processes of podocytes.
- The resulting fluid entering the lumen of Bowman's capsule is the glomerular filtrate.

(b)
- Sodium-potassium pumps (\(\text{Na}^+/\text{K}^+\)-ATPase) in the basal membrane actively transport \(\text{Na}^+\) ions out of the PCT epithelial cells into interstitial fluid/blood, lowering intracellular \(\text{Na}^+\) concentration.
- \(\text{Na}^+\) ions move down their electrochemical gradient from the lumen into the cell via co-transporter proteins, simultaneously bringing glucose into the cell (secondary active transport).
- Glucose then diffuses out of the PCT cell into tissue fluid and surrounding blood capillaries down its concentration gradient via facilitated diffusion.

(c)
- Osmoreceptors in the hypothalamus detect a decrease in blood water potential and stimulate the posterior pituitary gland to secrete ADH into the blood.
- ADH binds to complementary receptors on the surface of collecting duct (and DCT) epithelial cells, initiating a signalling cascade involving cyclic AMP (cAMP) and protein kinase A.
- Vesicles containing aquaporins fuse with the luminal cell surface membrane, increasing membrane permeability to water.
- More water moves out of the collecting duct filtrate into the hypertonic renal medulla by osmosis and is reabsorbed into blood, restoring normal water potential.

(a) (max 4 marks)
- high hydrostatic pressure in glomerulus / difference in lumen diameter between afferent arteriole (wider) and efferent arteriole (narrower); [1]
- fluid / small molecules forced out through pores / fenestrations in endothelial cells; [1]
- basement membrane acts as a selective filter / molecular sieve; [1]
- prevents passage of large proteins / blood cells / \(M_r > 68000\); [1]
- filtrate passes between filtration slits / foot processes / pedicels of podocytes; [1]

(b) (max 3 marks)
- \(\text{Na}^+\) ions actively pumped out of PCT cells into blood / tissue fluid (by \(\text{Na}^+/\text{K}^+\) pump); [1]
- creates a concentration / electrochemical gradient for \(\text{Na}^+\) between lumen and cytoplasm; [1]
- \(\text{Na}^+\) and glucose enter PCT cells together by co-transport / symport; [1]
- glucose moves out of PCT cell into blood / capillaries by facilitated diffusion; [1]

(c) (max 3 marks)
- osmoreceptors in hypothalamus detect fall in blood water potential; [1]
- posterior pituitary secretes ADH into blood; [1]
- ADH binds to receptors on cell surface membranes of collecting duct / DCT cells; [1]
- causes vesicles containing aquaporins to fuse with luminal membrane; [1]
- increases permeability of collecting duct to water so more water reabsorbed by osmosis (into medullary capillaries); [1]

(a)
1. An action potential arrives at the presynaptic motor neurone terminal, depolarising the presynaptic membrane.
2. Voltage-gated calcium ion channels open, causing \(\text{Ca}^{2+}\) ions to diffuse into the synaptic knob.
3. Influx of \(\text{Ca}^{2+}\) causes synaptic vesicles containing acetylcholine (ACh) to move towards and fuse with the presynaptic membrane, releasing ACh into the synaptic cleft by exocytosis.
4. ACh diffuses across the cleft and binds to specific **receptors (chemically/ligand-gated \(\text{Na}^+\) channels)** on the sarcolemma.
5. These channels open, allowing **\(\text{Na}^+\) ions to rush into the muscle fibre, causing depolarisation and generating an action potential along the sarcolemma.

(b)**
- **Role of \(\text{Ca}^{2+}\)**: Action potentials travel down T-tubules triggering the release of \(\text{Ca}^{2+}\) from the sarcoplasmic reticulum. \(\text{Ca}^{2+}\) binds to troponin, causing a conformational change that pulls tropomyosin away from the myosin-binding sites on the actin filament, exposing them.
- Role of ATP: Myosin heads bind to exposed actin sites to form cross-bridges; the power stroke occurs when ADP and Pi are released. ATP binds to the myosin head, breaking the cross-bridge (causing detachment from actin). Hydrolysis of ATP to ADP and Pi by myosin ATPase provides energy to reset/re-cock the myosin head to its high-energy conformation.

(c)
Differences:
1. Transmission pathway: Nervous signals travel as electrical impulses along neurones, whereas hormonal signals are chemical messengers transported in the bloodstream.
2. Speed and duration: Nervous responses are rapid and short-lived, whereas hormonal responses are typically slower and longer-lasting (or nervous is localised/targeted to specific effectors whereas hormonal is widespread).

(a) (max 4 marks)
- action potential arrives at presynaptic membrane / terminal; [1]
- voltage-gated \(\text{Ca}^{2+}\) channels open and \(\text{Ca}^{2+}\) enters synaptic knob; [1]
- synaptic vesicles containing acetylcholine (ACh) fuse with presynaptic membrane / ACh released by exocytosis; [1]
- ACh diffuses across synaptic cleft; [1]
- ACh binds to receptors / ligand-gated channels on sarcolemma; [1]
- \(\text{Na}^+\) channels open and \(\text{Na}^+\) enters muscle cell, depolarising sarcolemma; [1]

(b) (max 4 marks)
- \(\text{Ca}^{2+}\) released from sarcoplasmic reticulum; [1]
- \(\text{Ca}^{2+}\) binds to troponin; [1]
- causes tropomyosin to move / expose myosin-binding sites on actin; [1]
- ATP binds to myosin head, causing detachment from actin filament; [1]
- hydrolysis of ATP (by ATPase) cocks / resets myosin head / provides energy for power stroke; [1]

(c) (any 2 marks, comparative)
- nervous transmission is electrical impulses along neurones whereas hormonal is chemical transport in blood; [1]
- nervous response is fast / rapid whereas hormonal is slower; [1]
- nervous response is short-lived / temporary whereas hormonal is long-lasting; [1]
- nervous response is localised / specific to single effector whereas hormonal is widespread / affects multiple target organs; [1]

(a)
(i) Codominance is a condition in which both alleles in a heterozygous organism are fully and simultaneously expressed, contributing to the phenotype.
(ii) Recessive epistasis (or epistasis), because homozygous recessive alleles \(aa\) mask the phenotypic expression of the \(B/b\) gene.

(b)
(i)
- \(Aabb\): Has functional enzyme A to make pink pigment, but inactive enzyme B, so phenotype is pink.
- \(aaBb\): Lacks functional enzyme A so no pink precursor is made; phenotype is white.

(ii)
Cross: \(AaBb \times AaBb\)
Gametes for both: \(AB, Ab, aB, ab\)
Genotypic distribution out of 16 parts:
- \(9\) \(A\_B\_\) \(\rightarrow\) Purple (functional enzyme A and B)
- \(3\) \(A\_bb\) \(\rightarrow\) Pink (functional A, non-functional B)
- \(3\) \(aaB\_\) \(\rightarrow\) White (non-functional A)
- \(1\) \(aabb\) \(\rightarrow\) White (non-functional A)

Expected ratio = 9 purple : 3 pink : 4 white.

(c)
(i) Total offspring \(= 160\).
- Expected purple \(E = \frac{9}{16} \times 160 = 90\)
- Expected pink \(E = \frac{3}{16} \times 160 = 30\)
- Expected white \(E = \frac{4}{16} \times 160 = 40\)

$$\chi^2 = \frac{(92 - 90)^2}{90} + \frac{(28 - 30)^2}{30} + \frac{(40 - 40)^2}{40}$$
$$\chi^2 = \frac{4}{90} + \frac{4}{30} + 0 = 0.0444 + 0.1333 + 0 = 0.1778 \approx 0.18$$

(ii) Degrees of freedom \(= c - 1 = 3 - 1 = 2\).

(a)(i)
- both alleles are expressed in the phenotype of a heterozygote / both alleles contribute equally to the phenotype; [1]

(a)(ii)
- (recessive) epistasis / epistatic; [1]

(b)(i)
- \(Aabb\): pink; [1]
- \(aaBb\): white; [1]

(b)(ii)
- Punnett square / correct combination of gametes (\(AB, Ab, aB, ab\)); [1]
- identification of genotypes \(9\, A\_B\_\), \(3\, A\_bb\), \(3\, aaB\_\), \(1\, aabb\); [1]
- correct phenotypic ratio of 9 purple : 3 pink : 4 white; [1]

(c)(i)
- correct calculation of expected values: purple \(= 90\), pink \(= 30\), white \(= 40\); [1]
- correct \(\chi^2\) value \(= 0.18\) (accept \(0.177\) - \(0.18\)); [1]

(c)(ii)
- \(2\) (degrees of freedom \(= 3 - 1 = 2\)); [1]

(a)
- Directional selection is a mode of natural selection in which an extreme phenotype is favoured over other phenotypes, causing allele frequencies to shift continuously in one direction over generations.
- It occurs when there is a change in environmental conditions (e.g. climate cooling, introduction of an antibiotic/predator) or selection pressure.

(b)
(i)
- Frequency of homozygous recessive genotype \(q^2 = \frac{80}{500} = 0.16\).
- Frequency of recessive allele \(q = \sqrt{0.16} = 0.40\).
- Frequency of dominant allele \(p = 1 - q = 1 - 0.40 = 0.60\).

(ii)
- Frequency of heterozygotes \(2pq = 2 \times 0.60 \times 0.40 = 0.48\).
- Number of heterozygous snails \(= 0.48 \times 500 = 240\) snails.

(iii)
Conditions (any two):
1. Large population size (no significant genetic drift).
2. Random mating.
3. No mutations occurring.
4. No immigration or emigration (isolated population / no gene flow).
5. No natural selection (all genotypes have equal reproductive fitness).

(c)
- A geographical barrier (e.g. river, mountain range) physically separates populations, preventing gene flow.
- The isolated populations experience different environmental conditions / selection pressures and accumulate different random mutations.
- Over time, allele frequencies change significantly until individuals from the two populations can no longer interbreed to produce fertile offspring (reproductive isolation).

(a)
- selection that favours one extreme phenotype / shifts mean phenotype towards an extreme; [1]
- caused by a change in environmental conditions / change in selection pressure; [1]

(b)(i)
- \(q^2 = 0.16\) so \(q = 0.40\) / \(0.4\); [1]
- \(p = 1 - 0.40 = 0.60\) / \(0.6\); [1]

(b)(ii)
- \(2pq = 2 \times 0.60 \times 0.40 = 0.48\); [1]
- number of snails \(= 0.48 \times 500 = 240\); [1]
(ecf allowed from (b)(i))

(b)(iii) (any 2 marks)
- large population / no genetic drift; [1]
- random mating; [1]
- no mutation; [1]
- no migration / gene flow / closed population; [1]
- no selection / all phenotypes equally fertile / viable; [1]

(c) (max 2 marks)
- geographic barrier prevents interbreeding / no gene flow between populations; [1]
- different selection pressures / different mutations in each population lead to divergence of gene pools / reproductive isolation; [1]

(a)
Differences between Archaea and Bacteria:
1. Membrane lipids: Archaea possess ether-linked branched hydrocarbon chains (phytanyl/isoprene units), whereas Bacteria possess ester-linked unbranched fatty acid chains.
2. Cell wall composition: Archaea cell walls lack peptidoglycan (often made of pseudopeptidoglycan or S-layer proteins), whereas Bacteria have cell walls containing peptidoglycan.
3. Ribosomal RNA / Transcription machinery: Archaea have RNA polymerase enzymes and transcription mechanisms more similar to eukaryotes (e.g. sensitive to different antibiotics, distinct rRNA sequences), whereas Bacteria have simpler, standard bacterial RNA polymerase.

(b)
(i)
- Sum of \(\left(\frac{n}{N}\right)^2 = 0.2025 + 0.0900 + 0.0225 + 0.0100 = 0.3250\).
- \(D = 1 - 0.3250 = 0.6750 \approx 0.68\).

(ii)
- Habitat X (\(D = 0.68\)) has a higher species diversity / higher species evenness than Habitat Y (\(D = 0.42\)).
- A higher index indicates that Habitat X is more ecologically stable, meaning it has more complex food webs and is more resilient to environmental perturbations or disease outbreaks compared to Habitat Y.

(c)
- IVF: Oocytes collected from endangered females are fertilised outside the body with cryopreserved sperm from genetically suitable males in a laboratory dish to increase genetic diversity.
- Embryo transfer: Successfully developing embryos are implanted into the uterus of a surrogate mother (often a common, non-endangered related species), allowing endangered females to produce more offspring without carrying full-term pregnancies.

(a) (any 3 marks, must be comparative)
- Archaea have ether links in membrane lipids whereas Bacteria have ester links; [1]
- Archaea have branched / isoprene hydrocarbon chains in lipids whereas Bacteria have unbranched fatty acids; [1]
- Archaea cell walls lack peptidoglycan whereas Bacteria cell walls contain peptidoglycan; [1]
- Archaea have RNA polymerase more complex / similar to eukaryotic RNA polymerase whereas Bacteria have simpler RNA polymerase; [1]
- Archaea have different rRNA sequences / sensitivity to diphtheria toxin / insensitive to antibiotics like penicillin; [1]

(b)(i)
- sum of \(\left(\frac{n}{N}\right)^2 = 0.325\); [1]
- \(D = 1 - 0.325 = 0.675\); [1]
- final answer rounded to \(0.68\); [1]

(b)(ii)
- Habitat X has higher diversity / greater evenness than Habitat Y; [1]
- Habitat X is more stable / more resilient to environmental change / less likely to be damaged if one species declines; [1]

(c) (max 2 marks)
- IVF: fertilisation of egg and sperm occurs in vitro / outside body / in lab to produce embryos; [1]
- embryo transfer: developing embryo implanted into uterus of surrogate mother (related non-endangered species); [1]
- allows rapid increase in offspring / avoids risks of pregnancy in rare females / enables use of cryopreserved gametes; [1]

(a)
(i) Reverse transcriptase: Synthesises a single strand of complementary DNA (cDNA) using mature human mRNA as a template (ensuring no introns are present).
(ii) Restriction endonuclease: Recognises and cuts DNA at a specific palindromic base sequence, producing fragments with compatible sticky ends (or blunt ends) on both the plasmid vector and donor DNA.
(iii) DNA ligase: Catalyses the formation of covalent phosphodiester bonds between the sugar-phosphate backbones of the cut plasmid and the inserted human gene.

(b)
- Prokaryotic RNA polymerase requires a specific bacterial promoter region to recognise, bind, and initiate transcription.
- Without a compatible promoter, the bacterial transcription machinery will not transcribe the structural gene into mRNA, so no recombinant protein will be synthesised.

(c)
(i)
1. **Denaturation (\(94 - 95^{\circ}\text{C}\))**: High temperature breaks hydrogen bonds between base pairs, separating double-stranded DNA into single strands.
2. **Annealing (\(50 - 65^{\circ}\text{C}\))**: Lower temperature allows short single-stranded DNA primers to bind (anneal) via complementary base pairing to the \(3'\) ends of target DNA strands.
3. **Extension/Elongation (\(72^{\circ}\text{C}\))**: Optimum temperature for Taq polymerase to synthesise new complementary DNA strands by adding free deoxynucleotides (dNTPs).
(ii) Taq polymerase is isolated from thermophilic bacteria (Thermus aquaticus) and is thermostable; it does not denature at the high temperatures (\(95^{\circ}\text{C}\)) required for strand separation, avoiding the need to add fresh enzyme after each cycle.

(a)(i)
- synthesises (complementary / cDNA) DNA from (mature) mRNA template; [1]

(a)(ii)
- cuts DNA / plasmid / gene at a specific palindromic recognition site / sequence; [1]
- produces sticky ends / staggered cuts with complementary overhanging bases; [1]

(a)(iii)
- joins sugar-phosphate backbone / forms phosphodiester bonds between plasmid and gene; [1]

(b) (max 2 marks)
- promoter provides binding site for (bacterial) RNA polymerase / transcription factors; [1]
- ensures initiation of transcription / synthesis of mRNA (for the structural gene); [1]
- eukaryotic promoters are not recognised by bacterial RNA polymerase; [1]

(c)(i)
- \(94 - 95^{\circ}\text{C}\) / denaturation: breaks hydrogen bonds to separate double-stranded DNA into single strands; [1]
- \(50 - 65^{\circ}\text{C}\) / annealing: primers attach / anneal to complementary sequences on single strands; [1]
- \(72^{\circ}\text{C}\) / extension / elongation: Taq polymerase synthesises complementary strands / adds nucleotides; [1]

(c)(ii)
- Taq polymerase is thermostable / heat-resistant / does not denature at \(95^{\circ}\text{C}\); [1]

(a)(i)
Primers are short single-stranded sequences of DNA (typically 15–25 nucleotides) that bind to complementary base sequences at the $3'$ ends of the target DNA strands. They provide a free $3'\text{-OH}$ group required for DNA polymerase to initiate nucleotide addition.

(a)(ii)
Taq polymerase is thermostable (heat-resistant), meaning it does not denature at the high temperatures (approx. $95\,^{\circ}\text{C}$) used in the denaturation step to separate double-stranded DNA into single strands. This allows the PCR process to run continuously for multiple cycles without the need to replenish enzyme after every heating step.

(b)
- DNA fragments possess a net negative charge due to the presence of phosphate groups in their sugar-phosphate backbone.
- When an electrical potential difference (electric current) is applied across an agarose gel submerged in a buffer solution, DNA fragments migrate towards the positive electrode (anode).
- The agarose gel acts as a molecular sieve; shorter/smaller DNA fragments encounter less resistance and migrate faster / travel further through the pores of the gel than larger/longer fragments over a set time.

(c)
1. Extraction: mRNA is extracted from both cancerous and non-cancerous cells.
2. cDNA synthesis & labelling: Reverse transcriptase is used to synthesize complementary DNA (cDNA) from the mRNA. The cDNA from cancerous cells and non-cancerous cells are labelled with distinct fluorescent tags (e.g., green for normal cells, red for cancer cells).
3. Hybridisation: The labelled cDNA samples are mixed and incubated with the microarray chip, which contains known single-stranded DNA probes in distinct wells/spots. Complementary cDNA sequences hybridise/bind to matching probes.
4. Washing & detection: Unbound cDNA is washed off. The microarray is scanned with a laser to detect the fluorescence emitted at each spot; the colour and intensity of fluorescence indicate relative levels of gene expression.

(a)(i)
- provides a starting point for DNA polymerase / attaches / binds / anneals to complementary base sequence to allow DNA elongation / replication to begin [1]

(a)(ii)
- Taq polymerase is thermostable / heat-tolerant / heat-resistant / does not denature at high temperatures / at $94\text{--}98\,^{\circ}\text{C}$; [1]
- so does not need to be replaced after every cycle / allows process to be automated / repeated for multiple cycles; [1]

(b)
- DNA (molecules / fragments) have a net negative charge (due to phosphate groups); [1]
- DNA fragments migrate / move towards the anode / positive electrode when an electric current is applied; [1]
- gel acts as a sieve / mesh / smaller or shorter fragments move faster / travel further (through gel) than larger or longer fragments; [1]

(c)
Any 4 from:
1. mRNA extracted / isolated from cancerous and non-cancerous cells; [1]
2. reverse transcriptase used to convert mRNA into (single-stranded) cDNA; [1]
3. cDNA from each cell type labelled with a different fluorescent dye / tag (e.g. red and green); [1]
4. cDNA samples mixed and allowed to hybridise / bind to (complementary single-stranded) probes on the microarray / chip; [1]
5. excess / unbound cDNA is washed away; [1]
6. microarray is scanned / illuminated with a laser (and colour / intensity of fluorescence recorded to measure relative expression); [1]

(a)(i)
The stroma (of the chloroplast).

(a)(ii)
Rubisco catalyses the fixation of carbon dioxide by reacting carbon dioxide with ribulose bisphosphate (RuBP, a 5-carbon sugar) to form an unstable 6-carbon intermediate, which immediately splits into two molecules of glycerate 3-phosphate (GP, a 3-carbon compound).

(b)
- Water is split into protons ($\text{H}^{+}$), electrons ($\text{e}^{-}$), and oxygen ($\text{O}_{2}$) using light energy associated with the oxygen-evolving complex at PSII.
- The electrons replace those lost from the reaction centre chlorophyll ($P_{680}$) of PSII when electrons are excited and passed along the electron transport chain.
- Protons accumulate in the thylakoid lumen, contributing to the electrochemical proton gradient used by ATP synthase to produce ATP, and are also used to reduce $\text{NADP}^{+}$ to reduced NADP (NADPH).
- Oxygen is released as a by-product.

(c)(i)
- As light intensity increases from $0$ to $80\,\text{a.u.}$, the rate of photosynthesis increases rapidly / steeply from $0.0$ to $18.4\,\text{cm}^{3}\,\text{h}^{-1}$.
- Above $80\,\text{a.u.}$ / between $80$ and $120\,\text{a.u.}$, the rate of increase levels off / plateaus, reaching a maximum rate of $20.0\,\text{cm}^{3}\,\text{h}^{-1}$ at $120\,\text{a.u.}$

(c)(ii)
At $100\,\text{a.u.}$, light intensity is no longer the limiting factor. The Calvin cycle / light-independent reaction involves enzyme-catalysed reactions (such as rubisco). At $25\,^{\circ}\text{C}$, enzymes and substrates possess greater kinetic energy, leading to more frequent successful collisions / more enzyme-substrate complexes formed per unit time, resulting in a higher rate of photosynthesis than at $15\,^{\circ}\text{C}$ where temperature is the limiting factor.

(a)(i)
- stroma (of chloroplast); [1]

(a)(ii)
- catalyses the fixation of carbon dioxide / reaction between $\text{CO}_{2}$ and ribulose bisphosphate / RuBP; [1]
- to produce (an unstable 6C intermediate that splits into two molecules of) glycerate 3-phosphate / GP; [1]

(b)
Any 3 from:
- splits water into electrons, protons / $\text{H}^{+}$ and oxygen; [1]
- (electrons) replace electrons lost by (reaction centre chlorophyll / $P_{680}$ in) photosystem II / PSII; [1]
- (protons) contribute to proton gradient across thylakoid membrane / used to synthesise ATP; [1]
- (protons) combine with $\text{NADP}^{+}$ / reduce NADP to form reduced NADP; [1]

(c)(i)
- rate of photosynthesis increases with increasing light intensity (from $0$ to $80\,\text{a.u.}$ / $100\,\text{a.u.}$); [1]
- rate of increase slows down / levels off / plateaus above $80\,\text{a.u.}$ / reaches maximum of $20.0\,\text{cm}^{3}\,\text{h}^{-1}$ at $120\,\text{a.u.}$; [1]
*(Note: Accept manipulated data quote comparing two intensities at $25\,^{\circ}\text{C}$ with units)*

(c)(ii)
- at $100\,\text{a.u.}$, light intensity is not the limiting factor / temperature is the limiting factor (at $15\,^{\circ}\text{C}$); [1]
- (higher temperature provides) greater kinetic energy to enzymes / substrates / rubisco leading to more frequent collisions / more enzyme-substrate complexes formed (per unit time); [1]

(a)(i)
Independent variable: Concentration of ethanol (\(\text{mol}\,\text{dm}^{-3}\)).

(a)(ii)
Dependent variable: Absorbance / percentage transmission of light through the solution measured using a colorimeter (which indicates the quantity of betalain pigment leaked).

(b)
Using the dilution formula \(C_1 V_1 = C_2 V_2\):
\[(10.0\,\text{mol}\,\text{dm}^{-3}) \times V_1 = (4.0\,\text{mol}\,\text{dm}^{-3}) \times (50.0\,\text{cm}^3)\]
\[V_1 = \frac{4.0 \times 50.0}{10.0} = 20.0\,\text{cm}^3\]
Measure \(20.0\,\text{cm}^3\) of the \(10.0\,\text{mol}\,\text{dm}^{-3}\) stock ethanol solution using a graduated pipette / burette, transfer to a container, and add \(30.0\,\text{cm}^3\) (\(50.0 - 20.0\)) of distilled water, mixing thoroughly.

(c)
1. Preparation of tissue: Use a cork borer to cut cylinders from the same beetroot and a ruler and scalpel to trim them to identical lengths (e.g., \(20\,\text{mm}\)) to ensure equal surface area and volume.
2. Rinsing: Wash the cylinders thoroughly in distilled water until the washings are completely clear and gently blot with a paper towel to remove any pigment released by damaged surface cells during cutting.
3. Standardising volumes and conditions: Place \(10.0\,\text{cm}^3\) of each prepared ethanol concentration (\(2.0, 4.0, 6.0, 8.0, 10.0\,\text{mol}\,\text{dm}^{-3}\)) into labelled test tubes. Place all tubes in a thermostatted water bath at a constant temperature (e.g., \(25\,^{\circ}\text{C}\)) for \(5\) minutes to equilibrate.
4. Incubation: Add one beetroot cylinder to each tube, insert a stopper to prevent ethanol evaporation, and incubate for a fixed period (e.g., \(30\,\text{minutes}\)).
5. Colorimeter measurement: Remove the beetroot cylinders, shake the tube to distribute pigment evenly, transfer a sample of each solution into a cuvette, and measure the absorbance (or percentage transmission) using a colorimeter with a blue/green filter (approx. \(520\text{--}540\,\text{nm}\)), zeroing/calibrating the colorimeter first with distilled water.
6. Reliability & Controls: Repeat the entire procedure at least \(3\) times (or set up \(3\) replicates per concentration) to calculate a mean and identify anomalies. Set up a negative control by incubating beetroot cylinders in pure distilled water instead of ethanol.
7. Risk Assessment: Ethanol is flammable, so keep it away from open naked flames / use a thermostatted electric water bath. Scalpels and cork borers are sharp, so always cut on a cutting tile away from hands.

(a)(i)
- concentration of ethanol ; [1]

(a)(ii)
- absorbance / transmission / optical density (of the surrounding solution) / rate of pigment leakage ; [1]
Reject: 'amount of color' without reference to colorimeter / absorbance.

(b)
- \(20.0\,\text{cm}^3\) of \(10.0\,\text{mol}\,\text{dm}^{-3}\) ethanol ; [1]
- \(30.0\,\text{cm}^3\) of distilled water (and mix) ; [1]

(c) Any six from:
1. standardised beetroot size / surface area: use cork borer and ruler/scalpel to produce cylinders of identical dimensions / mass ; [1]
2. washing beetroot pieces thoroughly in distilled water until rinse water is clear AND blotting dry before use ; [1]
3. temperature control: keep test tubes in a thermostatted water bath (state a specific value, e.g. \(20\text{--}25\,^{\circ}\text{C}\)) ; [1]
4. constant volume of solution: state a specific volume of ethanol solution placed into each tube (e.g. \(10\,\text{cm}^3\)) ; [1]
5. constant time: incubate for a specified, timed period (e.g. \(20\text{--}30\,\text{minutes}\)) ; [1]
6. colorimeter use: use of colorimeter with blue/green filter (\(520\text{--}540\,\text{nm}\)) / zeroing with water/blank ; [1]
7. replicates: repeat at least \(3\) times at each concentration to calculate mean (and identify anomalies) ; [1]
8. control: set up a tube containing distilled water instead of ethanol under identical conditions ; [1]
9. safety: ethanol is flammable (use electric water bath / avoid naked flames) OR scalpel/cork borer can cut skin (cut onto white tile / cut away from fingers) ; [1]
(Maximum 6 marks for part c)

(a)
DCPIP acts as an artificial electron acceptor (accepting electrons released by the photolysis of water / electron transport chain at photosystem I/II).
It replaces NADP (or \(\text{NADP}^+\)).

(b)(i)
Keeping the buffer ice-cold slows down the activity of hydrolytic / proteolytic enzymes released during cell disruption that would otherwise degrade chloroplast membrane proteins and thylakoid complexes.

(b)(ii)
An isotonic buffer ensures there is no net movement of water into or out of the chloroplasts by osmosis, preventing chloroplasts from swelling and bursting (lysis) or shrinking (crenation).

(c)(i)
Rate at \(10\,\text{cm} = 0.084\,\text{a.u.}\min^{-1}\)
Rate at \(40\,\text{cm} = 0.021\,\text{a.u.}\min^{-1}\)
\[\text{Decrease} = 0.084 - 0.021 = 0.063\]
\[\text{Percentage decrease} = \left(\frac{0.063}{0.084}\right) \times 100\% = 75\%\]

(c)(ii)
As light intensity increases (or as distance decreases), the rate of DCPIP reduction increases (they are directly related / positively correlated).

(c)(iii)
The standard deviations are very small across all distances (e.g., \(0.002\text{--}0.004\)), indicating that the replicate data points are close to the mean and have high precision / repeatability.

(c)(iv)
1. Dark control: A tube containing isolated chloroplasts, buffer, and DCPIP wrapped in aluminium foil / kept in complete darkness (proves light is required for the reaction).
2. Inactivated / No chloroplast control: A tube containing boiled (denatured) chloroplasts + DCPIP in the light OR buffer + DCPIP without chloroplasts in the light (proves intact, active enzymes/chloroplasts are required).

(a)
- terminal / final electron acceptor (accepts \(\text{H}^+\) / electrons from photolysis/ETC) ; [1]
- NADP / \(\text{NADP}^+\) ; [1]

(b)(i)
- reduces / slows enzyme activity / prevents autolysis / prevents degradation of thylakoid proteins / membrane ; [1]

(b)(ii)
- equal water potential / prevents net osmosis so chloroplasts do not burst / lyse / crenate ; [1]

(c)(i)
- \(\frac{0.084 - 0.021}{0.084} \times 100\) (or \(0.063 / 0.084\)) ; [1]
- \(75\,\%\) ; [1]
(Note: 2 significant figures required)

(c)(ii)
- as light intensity increases, the rate of DCPIP reduction increases / positive correlation ; [1]

(c)(iii)
- small SDs indicate data points are close to the mean / low variation between repeats / high repeatability / precision ; [1]

(c)(iv) Any two from:
- chloroplasts + DCPIP kept in the dark / wrapped in foil ; [1]
- boiled / heat-treated / denatured chloroplasts + DCPIP in the light ; [1]
- DCPIP + buffer with no chloroplasts in the light ; [1]
(Maximum 2 marks for part iv)

(a)(i)
1. Place the stage micrometer onto the microscope stage and focus on the micrometer scale using the chosen objective lens.
2. Align the zero line of the eyepiece graticule with a known line on the stage micrometer scale.
3. Count the number of eyepiece graticule units that correspond to a known distance (e.g., \(100\,\mu\text{m}\) or number of stage micrometer subdivisions).
4. Calculate the calibration factor: \(\text{value of } 1 \text{ graticule unit} = \frac{\text{distance on stage micrometer}}{\text{number of graticule units}}\).

(a)(ii)
- Align the eyepiece graticule perpendicular to the long axis of the stoma and measure the pore opening at its widest point.
- Select stomata randomly from multiple different fields of view (e.g., \(20\text{--}30\) stomata per peel) and calculate a mean value to avoid bias.

(b)(i)
\[\text{Lower limit} = 3.8 - (2 \times 0.20) = 3.8 - 0.40 = 3.4\,\mu\text{m}\]
\[\text{Upper limit} = 3.8 + (2 \times 0.20) = 3.8 + 0.40 = 4.2\,\mu\text{m}\]
Range: \(3.4\text{--}4.2\,\mu\text{m}\).

(b)(ii)
- The difference is not statistically significant.
- Reasoning: The \(95\%\) confidence intervals for \(20\,\mu\text{mol}\,\text{dm}^{-3}\) (\(3.6\text{--}4.6\,\mu\text{m}\)) and \(40\,\mu\text{mol}\,\text{dm}^{-3}\) (\(3.4\text{--}4.2\,\mu\text{m}\)) overlap (in the range \(3.6\text{--}4.2\,\mu\text{m}\)), indicating there is more than a \(5\%\) probability (\(p > 0.05\)) that the observed difference is due to chance alone.

(c)
1. Epidermal peeling disrupts the natural leaf anatomy and may mechanically damage subsidiary/epidermal cells, altering turgor pressure and cell-to-cell signalling.
2. In intact plants, ABA is delivered via the xylem and responds to internal signals from mesophyll cells and roots, whereas bathing peels exposes all guard cells directly and uniformly from the external surface.

(a)(i)
- focus stage micrometer and align / line up graticule scale with stage micrometer scale ; [1]
- count number of graticule units corresponding to a known distance / number of micrometer divisions ; [1]
- divide known stage distance by number of graticule units to calculate value of one graticule division ; [1]

(a)(ii) Any two from:
- measure pore width at widest point ; [1]
- random selection of stomata / measure across multiple fields of view ; [1]
- measure a large sample (minimum \(\ge 15\) stomata) to calculate mean ; [1]
(Maximum 2 marks for part ii)

(b)(i)
- \(3.4\text{--}4.2\) (\(\mu\text{m}\)) ; [1]

(b)(ii)
- not (statistically) significant ; [1]
- confidence intervals overlap (between \(3.6\) and \(4.2\,\mu\text{m}\)) / there is a greater than \(5\%\) probability that the difference is due to chance (\(p > 0.05\)) ; [1]

(c) Any two from:
- peeling damages epidermal / guard / subsidiary cells (changing mechanical support / turgor) ; [1]
- lack of interaction / signalling from other leaf tissues (e.g. mesophyll / vascular bundle) ; [1]
- bathing in liquid solution is unnatural compared to aerial leaf / gas exchange / transpiration stream in xylem ; [1]
- exogenous ABA concentration in solution may not reflect physiological concentration inside guard cell symplast ; [1]
(Maximum 2 marks for part c)