2026 Cambridge IAS-Level Biology (9700) 模擬試題連答案詳解

Protons (\(\text{H}^+\)) are actively pumped out of the companion cell into the surrounding cell wall space using ATP. This generates a steep proton electrochemical gradient. Protons diffuse back into the companion cell through cotransporter carrier proteins, bringing sucrose molecules with them against the concentration gradient of sucrose. As sucrose accumulates and moves into the sieve tube element via plasmodesmata, the water potential decreases. Water enters the sieve tube element by osmosis from nearby xylem vessels, which increases the hydrostatic pressure at the source.

A is correct [1].
B is incorrect as sucrose movement into the companion cell requires a cotransport mechanism and sieve tube hydrostatic pressure increases.
C is incorrect as protons are pumped via active transport.
D is incorrect as protons move out by active transport and pressure increases.

A competitive inhibitor has a molecular structure similar to the substrate and competes with the substrate for the active site of the enzyme. At sufficiently high substrate concentrations, the substrate outcompetes the inhibitor, so the maximum rate of reaction (\(V_{\max}\)) remains unchanged. However, a higher substrate concentration is required to reach half the maximum rate (\(\frac{1}{2}V_{\max}\)), which means the value of \(K_m\) increases, indicating lower enzyme-substrate affinity.

A is correct: \(V_{\max}\) unchanged and \(K_m\) increased [1].
B is incorrect because \(K_m\) increases, not decreases.
C is incorrect because \(V_{\max}\) is unchanged and \(K_m\) increases (this row describes non-competitive inhibition).
D is incorrect because \(V_{\max}\) remains unchanged.

In the apoplast pathway, water moves through the interconnected network of cellulose cell walls and intercellular air spaces by mass flow and capillary action. When water reaches the endodermis, the impermeable suberin band of the Casparian strip blocks the apoplast pathway, forcing water to pass across the cell surface membrane and enter the symplast pathway.

B is correct [1].
A describes the symplast pathway.
C describes the vacuolar pathway.
D is incorrect as water is not actively pumped across suberin into xylem vessels.

In double-stranded DNA, complementary base pairing rules dictate that the percentage of cytosine (C) equals the percentage of guanine (G), and the percentage of adenine (A) equals the percentage of thymine (T).
Given \(\text{C} = 34\%\), then \(\text{G} = 34\%\).
Total \(\text{C} + \text{G} = 34\% + 34\% = 68\%\).
The remaining bases are adenine and thymine:
\(\text{A} + \text{T} = 100\% - 68\% = 32\%\).
Since \(\text{A} = \text{T}\), the percentage of adenine is \(\frac{32\%}{2} = 16\%\).

A is correct: 16% [1].
B is incorrect as 32% is the sum of adenine and thymine.
C is incorrect as 34% is the percentage of guanine.
D is incorrect as 68% is the sum of cytosine and guanine.

1. Find the distance represented by 2 stage micrometer divisions:
\(2 \times 0.01\text{ mm} = 0.02\text{ mm} = 20\ \mu\text{m}\).
2. Calculate the value of 1 eyepiece graticule unit:
\(1\text{ eyepiece unit} = \frac{20\ \mu\text{m}}{40} = 0.5\ \mu\text{m}\).
3. Calculate the actual diameter of the lymphocyte:
\(\text{Diameter} = 24 \times 0.5\ \mu\text{m} = 12.0\ \mu\text{m}\).

B is correct: \(12.0\ \mu\text{m}\) [1].
A is incorrect (corresponds to using 1 division instead of 2: \(24 \times 0.25 = 6.0\ \mu\text{m}\)).
C is incorrect (assuming 1 unit = \(1.0\ \mu\text{m}\)).
D is incorrect (multiplying by 2 instead of 0.5).

In actively respiring tissues, high carbon dioxide (\(\text{CO}_2\)) production leads to the diffusion of \(\text{CO}_2\) into red blood cells, where carbonic anhydrase catalyses the reaction \(\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3\). Carbonic acid dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). Protons (\(\text{H}^+\)) bind to oxyhaemoglobin to form haemoglobinic acid (\(\text{HHb}\)), which reduces the affinity of haemoglobin for oxygen and promotes the unloading of \(\text{O}_2\), shifting the oxygen dissociation curve to the right.

A is correct [1].
B is incorrect as respiring tissues have high \(\text{CO}_2\) and \(\text{O}_2\) affinity decreases.
C is incorrect because chloride ions enter the red blood cell (chloride shift), not leave it.
D is incorrect because \(\text{CO}_2\) production is increased and \(\text{O}_2\) affinity decreases.

Penicillin acts by inhibiting transpeptidase enzymes (glycoprotein peptidase) that cross-link peptidoglycan chains in bacterial cell walls during growth and division. This weakens the cell wall, causing lysis when water enters by osmosis. Viruses such as HIV do not possess a cell wall or peptidoglycan, and they lack metabolic enzymes targeted by antibiotics; hence penicillin has no effect on viruses.

A is correct [1].
B is incorrect because penicillin targets cell wall synthesis, not ribosomes (70S ribosome inhibition is characteristic of antibiotics like tetracycline/chloramphenicol).
C is incorrect as penicillin does not target DNA polymerase/replication directly.
D is incorrect as penicillin does not hydrolyse phospholipid bilayers.

During the mitotic cell cycle:
- DNA replicates semi-conservatively during the S (synthesis) phase of interphase.
- The nuclear envelope breaks down and disperses into vesicles during prophase.
- Centromeres divide and sister chromatids are pulled to opposite poles of the spindle during anaphase.

A is correct [1].
B is incorrect as DNA replication occurs in S phase, not \(\text{G}_1\), and nuclear envelope breakdown is in prophase.
C is incorrect as nuclear envelope breakdown occurs in prophase and chromatid separation occurs in anaphase.
D is incorrect as DNA replication is in S phase and sister chromatid separation occurs in anaphase.

1. Proton (\(\text{H}^+\)) pumps in the companion cell surface membrane use ATP to actively transport \(\text{H}^+\) ions out of the cell into the cell wall / apoplast.
2. Sucrose is loaded into the companion cell via sucrose-\( \text{H}^+ \) cotransporter proteins, moving \(\text{H}^+\) down its electrochemical gradient alongside sucrose.
3. The accumulation of sucrose lowers the water potential (\(\Psi\)) inside the sieve tube element, causing water to enter by osmosis down a water potential gradient from surrounding tissues.

A is correct [1].
B is incorrect because proton pumping requires active transport, and water movement is by osmosis.
C is incorrect because sucrose entry is by cotransport (secondary active transport), not simple facilitated diffusion.
D is incorrect because proton pumping is active and sucrose is symported with \(\text{H}^+\), not antiported with \(\text{Na}^+\).

Sunken stomata (1), epidermal hairs (2), and rolled leaves (3) all act by trapping a layer of moist, humid air next to the stomata, which increases the thickness of the boundary layer and therefore increases the diffusion path length for water vapour exiting into the outside air. In contrast, a thick waxy cuticle (4) provides an impermeable hydrophobic barrier that prevents water loss directly across epidermal cell walls (reducing cuticular transpiration), rather than increasing the diffusion distance for stomatal transpiration.

B is correct (features 1, 2 and 3 only) [1].
A, C and D are incorrect because feature 4 reduces permeability across the cuticle rather than increasing the diffusion distance of water vapour through the boundary layer.

A competitive inhibitor has a shape complementary to the active site and binds directly to it. Increasing the substrate concentration overcomes competitive inhibition, so \(V_{\max}\) remains the same, but a higher substrate concentration is required to reach \(\frac{1}{2}V_{\max}\), so \(K_{\text{m}}\) increases (Inhibitor \(\mathbf{X}\)).
A non-competitive inhibitor binds to an allosteric site (a site away from the active site), changing the tertiary structure of the enzyme and reducing turnover, thus lowering \(V_{\max}\). The affinity of unmodified enzyme active sites remains unchanged, so \(K_{\text{m}}\) is unchanged (Inhibitor \(\mathbf{Y}\)).

A is correct [1].
B, C, and D are incorrect based on the definition and kinetic effects of competitive versus non-competitive inhibition on \(V_{\max}\) and \(K_{\text{m}}\).

Immobilising enzymes inside an alginate matrix holds the protein molecules firmly in place, restricting excessive vibrational movement of the polypeptide chain at elevated temperatures. This stabilizes the tertiary structure and hydrogen/ionic bonds, making the active site more resistant to thermal denaturation.

C is correct [1].
A is incorrect because alginate is an inert support matrix, not an allosteric activator.
B is incorrect because alginate beads are porous and permit substrate entry at high temperatures.
D is incorrect because immobilisation involves non-covalent entrapment; it does not alter the primary structure or add disulfide bridges.

1. Double-stranded DNA of 1200 base pairs has \(1200 \times 2 = 2400\) total bases.
2. Adenine (A) = \(28\%\), so Thymine (T) = \(28\%\). Total A + T = \(56\%\).
3. Total G + C = \(100\% - 56\% = 44\%\). Cytosine (C) = \(22\%\).
4. Number of cytosine bases = \(0.22 \times 2400 = 528\).
5. Each linear single strand has 1200 nucleotides, so it has \(1200 - 1 = 1199\) phosphodiester bonds. For two strands: \(1199 \times 2 = 2398\) phosphodiester bonds.

A is correct [1].
B is incorrect because it forgets to subtract 1 bond per linear strand.
C is incorrect because it calculates cytosine per single strand rather than in the entire double-stranded molecule.
D is incorrect because 672 is the number of adenine bases, and 1199 is the number of phosphodiester bonds in only one strand.

1. Value of 16 stage micrometer divisions = \(16 \times 0.01\text{ mm} = 0.16\text{ mm} = 160\,\mu\text{m}\).
2. 40 eyepiece graticule units (epu) = \(160\,\mu\text{m}\).
3. 1 epu = \(\frac{160\,\mu\text{m}}{40} = 4.0\,\mu\text{m}\).
4. Actual length of cell = \(25\text{ epu} \times 4.0\,\mu\text{m} = 100\,\mu\text{m}\).

C is correct (
\(100\,\mu\text{m}\)) [1].
A is incorrect due to a power-of-ten calculation error.
B is incorrect from multiplying 16 by 4.
D is incorrect because \(160\,\mu\text{m}\) is the total length of 16 stage micrometer divisions.

At higher partial pressures of carbon dioxide (and lower pH caused by \(\text{H}^+\) formation from carbonic acid), haemoglobin binds \(\text{H}^+\) to form haemoglobinic acid, which causes an allosteric change lowering its affinity for oxygen. This causes haemoglobin to release (dissociate from) oxygen more readily at any given partial pressure of oxygen in active tissues.

A is correct [1].
B is incorrect because higher \(p\text{CO}_2\) decreases affinity for \(\text{O}_2\), not increases it.
C is incorrect because the Bohr shift occurs at higher \(p\text{CO}_2\), not lower \(p\text{CO}_2\).
D is incorrect because lower affinity promotes unloading of oxygen, not loading.

In the hybridoma technique, plasma cells (sensitised B-lymphocytes) from the spleen of an immunised animal provide the gene to produce a specific antibody, but they have a limited lifespan and cannot divide indefinitely in vitro. Myeloma cells (cancerous plasma cells) are fused with these B-lymphocytes to confer immortality and the ability to divide continuously in culture.

B is correct [1].
A is incorrect because myeloma cells do not produce the specific antigen-targeted antibody; the B-lymphocyte does.
C is incorrect because myeloma cells are fused to supply the capacity for indefinite cell division, not to stimulate T-lymphocytes.
D is incorrect because myeloma cells are used for indefinite division, whereas antibody specificity comes from the isolated B-lymphocyte.

Water and mineral ions moving along the apoplast pathway travel through the cell walls and intercellular spaces until they reach the Casparian strip of the endodermis. The Casparian strip contains waterproof suberin, which blocks the apoplast pathway, forcing all water and solutes to cross the selectively permeable cell surface membrane and enter the symplast pathway.

B is correct [1].
A is incorrect because the Casparian strip is located in the endodermis, not the pericycle, and blocks the apoplast pathway rather than the symplast pathway.
C is incorrect because the Casparian strip is made of suberin, not lignin, and forces movement into the symplast.
D is incorrect because plasmodesmata connect the cytoplasm of adjacent cells in the symplast pathway, not the apoplast pathway.

A competitive inhibitor binds reversibly to the active site of the enzyme, competing directly with the substrate. Because increasing the substrate concentration can outcompete the inhibitor, the maximum rate of reaction (\(V_{\text{max}}\)) remains the same. However, a higher substrate concentration is required to reach half \(V_{\text{max}}\), which corresponds to an increase in the Michaelis-Menten constant (\(K_{\text{m}}\)) and a lower enzyme affinity for the substrate.

C is correct [1].
A is incorrect because non-competitive inhibitors decrease \(V_{\text{max}}\).
B is incorrect because an increased \(K_{\text{m}}\) indicates a lower enzyme affinity for the substrate, not higher.
D is incorrect because the inhibitor binds reversibly to the active site, not an allosteric site.

Using the formula:
\[\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}}\]
Convert image size to micrometres:
\[36\text{ mm} = 36\,000\,\mu\text{m}\]
\[\text{Actual length} = \frac{36\,000\,\mu\text{m}}{24\,000} = 1.5\,\mu\text{m}\]

B is correct [1].
A is incorrect due to a power-of-ten conversion error (giving \(0.15\,\mu\text{m}\)).
C is incorrect due to an incorrect conversion factor (giving \(15.0\,\mu\text{m}\)).
D is incorrect because it is the inverse calculation.

An increase in carbon dioxide concentration leads to higher production of hydrogen ions (\(\text{H}^+\)) via the action of carbonic anhydrase and dissociation of carbonic acid. These hydrogen ions bind to oxyhaemoglobin to form haemoglobinic acid, displacing oxygen. This shifts the oxygen dissociation curve to the right (the Bohr shift), lowering haemoglobin's affinity for oxygen and promoting oxygen unloading in respiring tissues.

A is correct [1].
B is incorrect because the curve shifts to the right, not to the left.
C is incorrect because hydrogen ions bind to haemoglobin (forming haemoglobinic acid), not directly increasing oxygen binding.
D is incorrect because the affinity for oxygen decreases, which increases dissociation, not decreases it.

1. In a double-stranded linear DNA molecule of \(N\) nucleotides, each strand of \(N/2\) nucleotides has \((N/2 - 1)\) phosphodiester bonds. Total phosphodiester bonds = \(2 \times (1200 - 1) = 2398\).
2. According to Chargaff's rules, \(A = T = 360\).
Total \(A + T = 720\).
Total \(G + C = 2400 - 720 = 1680\).
Since \(G = C\), the number of cytosine bases = \(1680 / 2 = 840\).
Percentage of cytosine = \((840 / 2400) \times 100\% = 35\%\).

D is correct [1].
A is incorrect because \(2398\) bonds exist and cytosine is \(35\%\), not \(15\%\).
B is incorrect because there are \(2398\) phosphodiester bonds (two separate sugar-phosphate backbones), not \(2400\).
C is incorrect because the number of bonds is \(2398\) and the percentage of cytosine is \(35\%\).

Telomeres are non-coding repetitive DNA sequences located at the ends of eukaryotic chromosomes. They protect the ends of chromosomes from degradation and prevent the cell's DNA repair systems from recognising chromosome ends as double-strand breaks. They also ensure that essential genes near chromosome ends are not lost during successive rounds of DNA replication.

C is correct [1].
A is incorrect because attachment to spindle microtubules is the role of the centromere/kinetochore.
B is incorrect because telomeres do not hold sister chromatids together.
D is incorrect because telomeres do not act as origins of replication for lagging-strand synthesis.

Transfer of maternal antibodies across the placenta (IgG) and in colostrum/breast milk (IgA) is a natural process that does not involve the infant's own immune system producing the antibodies or generating memory cells. Therefore, this provides natural passive immunity.

B is correct [1].
A is incorrect because active immunity requires the baby's own lymphocytes to produce antibodies and memory cells.
C is incorrect because artificial passive immunity involves the injection of antibodies (e.g. antivenom).
D is incorrect because artificial active immunity involves vaccination with antigens.

Penicillin acts by inhibiting transpeptidase enzymes (glycopeptide transpeptidase), preventing the cross-linking of peptidoglycan chains during bacterial cell wall synthesis. Viruses are non-cellular structures consisting of genetic material enclosed within a protein coat (capsid) and lack cell walls and peptidoglycan. Thus, penicillin has no target in viruses.

C is correct [1].
A is incorrect because penicillin targets peptidoglycan cross-linking in cell wall synthesis, not 70S ribosomes.
B is incorrect because viruses do not possess peptidoglycan cell walls.
D is incorrect because penicillin inhibits peptidoglycan cross-linking, not viral reverse transcriptase or envelope formation.

Protons (\(H^+\)) are actively pumped out of companion cells into the cell wall by ATP-driven proton pumps, creating a steep proton gradient. Protons then diffuse back into the companion cell through cotransporter proteins, bringing sucrose molecules with them against the sucrose concentration gradient. Sucrose then diffuses down its concentration gradient from the companion cell into the sieve tube element via plasmodesmata.

A is correct [1]. B is incorrect because \(H^+\) is moved out via active transport, not facilitated diffusion. C is incorrect because sucrose is loaded via cotransport, not direct active transport into sieve tubes. D is incorrect because sucrose enters companion cells via secondary active cotransport with \(H^+\).

All three features are essential: (1) breakdown of end walls creates continuous open tubes reducing resistance to mass flow; (2) rigid lignification provides mechanical strength to prevent vessels collapsing under negative hydrostatic pressure (tension); (3) hydrophilic cellulose and lignin cell wall components permit adhesion with polar water molecules, preventing cavitation and maintaining a continuous water column.

A is correct [1] (statements 1, 2 and 3 are all correct functional adaptations). Options B, C and D are incorrect because they omit valid features.

A competitive inhibitor binds reversibly to the active site, competing with the substrate; this increases \(K_m\) (decreases affinity) but does not change \(V_{\max}\) because high substrate concentrations outcompete the inhibitor (Inhibitor X). A non-competitive inhibitor binds to an allosteric site, altering enzyme conformation; this reduces the number of functional active sites and lowers \(V_{\max}\), but leaves the affinity of remaining unaltered active sites unchanged, so \(K_m\) remains the same (Inhibitor Y).

A is correct [1]. B is incorrect as it reverses the types of inhibition. C is incorrect because increasing substrate concentration cannot overcome non-competitive inhibition (Inhibitor Y). D is incorrect because an increased \(K_m\) indicates lower affinity.

Statements 1, 2, and 4 are valid advantages of immobilising enzymes: the matrix traps the enzyme allowing simple separation and reuse, matrix trapping confers resistance to high temperature/pH changes, and downstream processing costs are reduced because product contamination is avoided. Statement 3 is incorrect because immobilisation does not deliberately alter the active site to increase affinity; in fact, diffusion limitations through alginate can slightly increase apparent \(K_m\).

A is correct [1] (statements 1, 2 and 4 only). B and C are incorrect as they include false statement 3. D is incorrect because it omits statement 2.

1. Number of adenine bases = \(0.28 \times 1200 = 336\).
2. By complementary base pairing, thymine \(\text{T} = 336\).
3. Total \(\text{A} + \text{T} = 336 + 336 = 672\).
4. Total \(\text{G} + \text{C} = 1200 - 672 = 528\).
5. Number of cytosine bases = \(528 / 2 = 264\).
6. Number of \(\text{A-T}\) base pairs = 336 (each has 2 hydrogen bonds) \(\rightarrow 336 \times 2 = 672\) hydrogen bonds.
7. Number of \(\text{G-C}\) base pairs = 264 (each has 3 hydrogen bonds) \(\rightarrow 264 \times 3 = 792\) hydrogen bonds.
8. Total hydrogen bonds = \(672 + 792 = 1464\).

A is correct [1] (264 cytosine bases and 1464 hydrogen bonds). Options B, C and D contain calculation errors in base pairing percentage or number of hydrogen bonds per pair.

1. 1 division on stage micrometer = \(0.01\text{ mm} = 10\ \mu\text{m}\).
2. 10 divisions on stage micrometer = \(10 \times 10\ \mu\text{m} = 100\ \mu\text{m}\).
3. 40 eyepiece graticule units (epu) = \(100\ \mu\text{m}\).
4. 1 epu = \(100\ \mu\text{m} / 40 = 2.5\ \mu\text{m}\).
5. Actual length of cell = \(18\text{ epu} \times 2.5\ \mu\text{m} = 45.0\ \mu\text{m}\).

B is correct [1] (\(45\ \mu\text{m}\)). A is incorrect by a factor of 10. C is incorrect from multiplying 18 by 4. D is incorrect from miscalculating the scale value.

In actively respiring tissues, high rates of aerobic respiration produce high \(p\text{CO}_2\). Carbon dioxide diffuses into erythrocytes where carbonic anhydrase catalyses its conversion to carbonic acid (\(\text{H}_2\text{CO}_3\)), which dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). \(\text{H}^+\) ions bind to oxyhaemoglobin to form haemoglobinic acid (\(\text{HHb}\)), buffering the cytoplasm and causing haemoglobin to release \(\text{O}_2\) (lower affinity). This is the Bohr effect, represented by a shift of the oxygen dissociation curve to the right.

A is correct [1]. B is incorrect because affinity for \(O_2\) decreases, not increases, and curve shifts to right. C is incorrect because \(p\text{CO}_2\) is high. D is incorrect because \(p\text{CO}_2\) is high and curve shifts to right.

During S phase of interphase, DNA replicates so the DNA mass doubles from \(4.0\text{ pg}\) to \(8.0\text{ pg}\). However, each chromosome now consists of two genetically identical sister chromatids joined at the centromere, so the chromosome number remains 16. At anaphase, sister chromatids separate into individual daughter chromosomes. After telophase and cytokinesis, each newly formed daughter cell receives 16 chromosomes and half of the duplicated DNA, returning the DNA mass to \(4.0\text{ pg}\).

A is correct [1] (Metaphase: 16 chromosomes, 8.0 pg; Daughter cell: 16 chromosomes, 4.0 pg). B is incorrect because chromatids attached at the centromere count as 1 chromosome. C is incorrect because DNA mass does double during S phase. D is incorrect because metaphase has 16 chromosomes with 8.0 pg DNA.

Sucrose loading involves active pumping of \(\text{H}^+\) ions (protons) out of companion cells into the surrounding cell wall space using ATP. This generates a steep proton concentration gradient. Protons then diffuse back into the companion cells through cotransporter (symporter) carrier proteins, bringing sucrose molecules with them against the sucrose concentration gradient. Sucrose then moves into the adjacent sieve tube elements via plasmodesmata down its concentration gradient.

A is correct [1]; B is incorrect as sucrose is not directly pumped by an ATP-driven pump without protons; C is incorrect as protons move down, not up, their gradient during cotransport; D is incorrect because loading into sieve tube elements occurs via companion cells and plasmodesmata.

A competitive inhibitor has a complementary shape to the active site and competes with the substrate for binding. At sufficiently high substrate concentrations, substrate outcompetes the inhibitor, so the maximum velocity \(V_{\text{max}}\) remains unchanged. However, a higher concentration of substrate is required to achieve half of \(V_{\text{max}}\), which means the affinity of the enzyme for its substrate appears reduced, resulting in an increased \(K_{\text{m}}\).

A is correct [1]; B is incorrect as \(V_{\text{max}}\) is unaffected and \(K_{\text{m}}\) increases; C is characteristic of non-competitive inhibition regarding \(V_{\text{max}}\); D is incorrect as \(K_{\text{m}}\) does not decrease.

1. At \(\times 100\) magnification:
- \(16\) divisions of stage micrometer \(= 16 \times 0.01\text{ mm} = 0.16\text{ mm} = 160\ \mu\text{m}\).
- \(40\) eyepiece graticule units (epu) \(= 160\ \mu\text{m}\).
- \(1\text{ epu} = \frac{160}{40} = 4.0\ \mu\text{m}\).

2. At \(\times 400\) magnification:
- Magnification is increased by a factor of \(\frac{400}{100} = 4\).
- Therefore, the actual distance represented by \(1\text{ epu}\) decreases by a factor of 4: \(1\text{ epu} = \frac{4.0\ \mu\text{m}}{4} = 1.0\ \mu\text{m}\).

3. Thickness of tunica media \(= 35\text{ epu} \times 1.0\ \mu\text{m/epu} = 35\ \mu\text{m}\).

A is correct (\(35\ \mu\text{m}\)) [1]; B is incorrect (fails to adjust for magnification change, giving \(35 \times 4.0 = 140\ \mu\text{m}\)); C is incorrect (multiplies by 4 instead of dividing); D is incorrect (divides by 4 twice).

The pressure in the left ventricle (\(14.2\text{ kPa}\)) is higher than in the left atrium (\(0.8\text{ kPa}\)), forcing the atrioventricular (bicuspid) valve closed to prevent backflow into the atrium. The pressure in the left ventricle (\(14.2\text{ kPa}\)) is also higher than the pressure in the aorta (\(10.5\text{ kPa}\)), which pushes the aortic semilunar valve open, allowing blood to flow from the left ventricle into the aorta.

A is correct [1]; B is incorrect as the bicuspid valve cannot be open when ventricular pressure exceeds atrial pressure; C is incorrect as ventricular pressure exceeds aortic pressure; D is incorrect because the bicuspid valve is closed.

1. Total nucleotides \(= 3000 \times 2 = 6000\).
2. Number of cytosine (C) bases \(= 0.22 \times 6000 = 1320\).
3. By complementary base pairing, guanine (G) \(= 1320\). Thus there are \(1320\) C–G base pairs.
4. Total C + G bases \(= 2640\). Remaining bases \(= 6000 - 2640 = 3360\).
5. Adenine (A) \(= \text{Thymine (T)} = \frac{3360}{2} = 1680\). Thus there are \(1680\) A–T base pairs.
6. Hydrogen bonds:
- From C–G pairs (3 H-bonds each): \(1320 \times 3 = 3960\)
- From A–T pairs (2 H-bonds each): \(1680 \times 2 = 3360\)
- Total hydrogen bonds \(= 3960 + 3360 = 7320\).
7. Purine bases (Adenine + Guanine) \(= 1680 + 1320 = 3000\) (or exactly \(50\%\) of the \(6000\) bases).

A is correct (\(7320\) hydrogen bonds, \(3000\) purines) [1]; B is incorrect (calculated purines as base pairs, 1500); C is incorrect (calculated H-bonds per single strand); D is incorrect (inverted base pairing rules).

The Casparian strip is a band of waterproof suberin deposited within the radial and transverse cell walls of the endodermal cells. This impermeable band blocks the apoplast pathway (movement through cell walls and intercellular spaces), compelling water and dissolved mineral ions to cross the selectively permeable cell surface membrane into the symplast pathway (cytoplasm).

A is correct [1]; B is incorrect because endodermal cells possess cell walls; C is incorrect because active transport of mineral ions lowers the water potential in the xylem, not increases it; D is incorrect because plasmodesmata mediate the symplast pathway, not the apoplast, and are not blocked by lignin.

Penicillin acts as an inhibitor of the bacterial enzyme transpeptidase (glycopeptide transpeptidase), which catalyses the cross-linking of peptidoglycan chains during cell wall synthesis in growing bacteria. The weakened cell wall cannot withstand internal turgor/osmotic pressure when water enters by osmosis, causing the bacterium to burst (lysis).

A is correct [1]; B is incorrect as penicillin does not target 70S ribosomes (that is the mechanism of other antibiotics like tetracycline/streptomycin); C is incorrect as it does not target phospholipids in the cell membrane; D is incorrect as it does not inhibit DNA polymerase.

Telomeres are repetitive non-coding DNA sequences at the ends of linear chromosomes. Their main function is to prevent the loss of vital coding genes from the ends of chromosomes during successive rounds of semi-conservative DNA replication, and to prevent chromosomes from fusing with each other. During anaphase of mitosis, centromeres split and sister chromatids (now individual chromosomes) are pulled by shortening spindle microtubules towards opposite poles.

A is correct [1]; B is incorrect as kinetochores/centromeres attach sister chromatids to spindle microtubules, not telomeres; C is incorrect because chromatid separation occurs in anaphase, not telophase; D is incorrect because histones are not synthesised by telomeres and chromatid separation does not occur in metaphase.

Water moves through the root cortex along two main pathways before reaching the vascular cylinder: the symplast pathway (via cytoplasm and plasmodesmata) and the apoplast pathway (through the porous cellulose cell wall matrix and intercellular spaces). When water reaches the endodermis, the presence of the Casparian strip, which is impregnated with the waterproof, hydrophobic substance suberin, blocks the apoplast pathway and forces water to cross the selectively permeable cell surface membrane into the symplast pathway.

1. apoplast / apoplastic (pathway); [1.0 mark]
2. suberin; [0.5 mark]

A apoplasm
R symplast / vacuolar
I lignin

The Michaelis-Menten constant (\(K_{\text{m}}\)) is defined as the substrate concentration at which an enzyme-catalysed reaction proceeds at half its maximum velocity (\(\frac{1}{2}V_{\text{max}}\)). It is an inverse measure of enzyme affinity for its substrate. A non-competitive inhibitor binds to an allosteric site rather than the active site, decreasing \(V_{\text{max}}\) without altering the affinity of the substrate for the active sites of unaffected enzyme molecules; therefore, \(K_{\text{m}}\) remains unchanged.

1. Michaelis(-Menten) constant / \(K_{\text{m}}\); [1.0 mark]
2. unchanged / no effect / stays the same / remains constant; [0.5 mark]

A \(K_m\)
R increases / decreases

Centrioles are cylindrical structures present in animal cells (and some lower plants/fungi), consisting of a ring of nine groups of three fused microtubules (microtubule triplets). During interphase of the cell cycle, the centrioles replicate, and during prophase they move to opposite poles of the cell, where they act as microtubule organising centres (MTOCs) that help organise and polymerise the tubulin subunits into the spindle apparatus responsible for moving chromosomes.

1. centriole(s) / centrosome; [1.0 mark]
2. organisation / polymerisation / formation of spindle fibres / spindle microtubules / attachment to kinetochores / separation of chromatids; [0.5 mark]

R centromere
I cell division unqualified

Inside erythrocytes, carbon dioxide reacts rapidly with water to form carbonic acid (\(\text{H}_2\text{CO}_3\)) in a reaction catalysed by the enzyme carbonic anhydrase. The carbonic acid subsequently dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\). As hydrogencarbonate ions diffuse down their concentration gradient out of the erythrocyte into the plasma, chloride ions (\(\text{Cl}^-\)) move in from the plasma to maintain electrical neutrality, a phenomenon called the chloride shift.

1. carbonic anhydrase; [1.0 mark]
2. chloride (ion) / \(\text{Cl}^-\); [0.5 mark]

A correct spelling only for carbonic anhydrase
R chlorine (atom/molecule) / \(\text{Cl}_2\)

In a polynucleotide chain such as DNA, adjacent nucleotides are linked via condensation reactions creating strong covalent phosphodiester bonds between the 3'-OH of the pentose sugar and the 5'-phosphate group of the neighbouring nucleotide. During semi-conservative DNA replication, synthesis on the lagging strand is discontinuous, forming short Okazaki fragments. The enzyme DNA ligase catalyses the formation of phosphodiester bonds to join these discontinuous fragments together into a continuous strand.

1. phosphodiester (bond / linkage); [1.0 mark]
2. DNA ligase; [0.5 mark]

A covalent ester bond
R peptide / glycosidic / hydrogen
R DNA polymerase

Cholera is a water-borne diarrhoeal disease caused by the comma-shaped, Gram-negative bacterium Vibrio cholerae. Penicillin acts by inhibiting transpeptidase enzymes (also known as penicillin-binding proteins), preventing the cross-linking of glycan chains within the bacterial peptidoglycan (or murein) cell wall during synthesis, leaving the wall structurally weak and prone to osmotic lysis.

1. Vibrio cholerae (genus capitalized, species lowercase); [1.0 mark]
2. (peptidoglycan / murein) cell wall; [0.5 mark]

A V. cholerae if correctly formatted
R cell surface membrane / plasma membrane

1. Identify the values from the data table:
- Distance moved at $0.5\text{ m}\,\text{s}^{-1} = 16\text{ mm}$
- Distance moved at $4.0\text{ m}\,\text{s}^{-1} = 76\text{ mm}$

2. Calculate the difference in distance moved:
$$\text{Change} = 76 - 16 = 60\text{ mm}$$

3. Calculate the percentage increase relative to the baseline value at $0.5\text{ m}\,\text{s}^{-1}$:
$$\text{Percentage increase} = \left(\frac{76 - 16}{16}\right) \times 100 = \left(\frac{60}{16}\right) \times 100 = 375\%$$

- Correct calculation of change in distance or rate divided by the original value at $0.5\text{ m}\,\text{s}^{-1}$, e.g. $\frac{76 - 16}{16} \times 100$ OR $\frac{5.07 - 1.07}{1.07} \times 100$ ; [1.0]
- Correct final answer: $375$ (%) ; [0.5]

1. Identify the volume of oxygen gas produced in the initial 15-second interval:
- Total volume at $t = 15\text{ s}$ is $13.5\text{ cm}^3$
- Total volume at $t = 0\text{ s}$ is $0.0\text{ cm}^3$
$$\Delta V = 13.5 - 0.0 = 13.5\text{ cm}^3$$

2. Calculate the initial rate of reaction:
$$\text{Initial rate} = \frac{\Delta V}{\Delta t} = \frac{13.5\text{ cm}^3}{15\text{ s}} = 0.90\text{ cm}^3\,\text{s}^{-1}$$

- Correct method showing change in volume divided by time, e.g. $\frac{13.5}{15}$ ; [1.0]
- Correct calculated answer: $0.90$ / $0.9$ ($\text{cm}^3\,\text{s}^{-1}$) ; [0.5]

1. Proton pumps in the companion cell membrane actively pump \(\text{H}^+\) ions (protons) out of the companion cell into the surrounding cell wall spaces using ATP.
2. A steep proton gradient is established. Protons diffuse back into the companion cell down their concentration/electrochemical gradient through cotransporter (symport) proteins, transporting sucrose against its concentration gradient simultaneously.
3. Sucrose accumulates in high concentration within the companion cell and moves into the adjacent sieve tube element by diffusion through plasmodesmata.

1. \(\text{H}^+\) / protons actively pumped / transported out of companion cells (into cell wall) using ATP / energy ;
2. \(\text{H}^+\) ions move / diffuse down their concentration / electrochemical gradient via a cotransporter (protein), carrying sucrose with them against its concentration gradient ;
3. sucrose moves from companion cell into sieve tube element via plasmodesmata (by diffusion) ;

1. \(\text{CO}_2\) diffuses into the erythrocyte and combines with \(\text{H}_2\text{O}\) in a reaction catalysed by the enzyme carbonic anhydrase to form carbonic acid (\(\text{H}_2\text{CO}_3\)).
2. Carbonic acid dissociates spontaneously into hydrogen ions (\(\text{H}^+\)) and hydrogencarbonate ions (\(\text{HCO}_3^-\)).
3. \(\text{HCO}_3^-\)\ ions diffuse down a concentration gradient out of the erythrocyte into the blood plasma. Chloride ions (\(\text{Cl}^-\)) move into the erythrocyte to maintain electrochemical neutrality (chloride shift), while \(\text{H}^+\) ions are buffered by haemoglobin forming haemoglobinic acid (HHb).

1. \(\text{CO}_2\) reacts with \(\text{H}_2\text{O}\) to form \(\text{H}_2\text{CO}_3\) / carbonic acid, catalysed by carbonic anhydrase ;
2. \(\text{H}_2\text{CO}_3\) dissociates into \(\text{H}^+\) and \(\text{HCO}_3^-\) / hydrogencarbonate ions ;
3. \(\text{HCO}_3^-\) diffuses out of the erythrocyte / red blood cell into plasma, and chloride ions (\(\text{Cl}^-\)) diffuse in to maintain electrical neutrality / chloride shift [OR \(\text{H}^+\) binds to haemoglobin to form haemoglobinic acid / HHb] ;

1. DNA polymerase adds complementary DNA nucleotides to the exposed template strand in a 5' to 3' direction, forming discontinuous segments called Okazaki fragments.
2. DNA polymerase removes RNA primers and replaces them with corresponding DNA nucleotides.
3. DNA ligase forms phosphodiester bonds between the sugar-phosphate backbones of adjacent Okazaki fragments, creating a continuous lagging strand.

1. DNA polymerase adds (free / activated) DNA nucleotides (to 3' end / in 5' to 3' direction) to synthesise Okazaki fragments / short segments ;
2. DNA polymerase catalyses phosphodiester bond formation between adjacent nucleotides / replaces RNA primers with DNA nucleotides ;
3. DNA ligase joins / seals Okazaki fragments together by catalysing the formation of phosphodiester bonds (between adjacent fragments) ;

1. Substrate binding causes the active site of the enzyme to undergo a conformational (shape) change to fit the substrate more closely/tightly.
2. This tight fit places mechanical strain or distortion on particular chemical bonds in the substrate molecule, weakening them.
3. By destabilising bonds and orienting functional groups favorably, less energy is needed to reach the transition state, thus lowering the activation energy.

1. substrate binding induces a conformational / shape change in the active site / enzyme (to become fully complementary) ;
2. places strain / tension on specific bonds within the substrate / aligns catalytic residues / reactive groups favorably ;
3. weakens / destabilises bonds, reducing the energy barrier / activation energy needed to reach the transition state / form products ;

1. Neutrophil cell surface membrane receptors bind to bacterial antigens, and the membrane invaginates/extends pseudopodia to engulf the bacterium into an endocytic vesicle known as a phagosome.
2. Lysosomes inside the neutrophil move towards and fuse with the phagosome membrane, forming a phagolysosome.
3. Hydrolytic enzymes (e.g., lysozyme and proteases) from the lysosomes are discharged onto the bacterium, hydrolysing its cell wall and internal macromolecules.

1. cell surface membrane invaginates / extends pseudopodia to engulf the bacterium into a phagocytic vacuole / phagosome ;
2. lysosome(s) fuse with the phagosome to form a phagolysosome ;
3. hydrolytic / digestive enzymes / lysozymes hydrolyse / digest / break down the bacterial cell wall / macromolecules ;

1. Polypeptides synthesised on ribosomes enter the lumen of the rough endoplasmic reticulum (rER), where they fold into tertiary structure and are packaged into transport vesicles that bud off the rER.
2. Transport vesicles travel to and fuse with the cis face of the Golgi body; inside the Golgi cisternae, the protein is chemically modified (e.g. addition of carbohydrate chains) and sorted.
3. Secretory vesicles containing the mature protein bud from the trans face of the Golgi body, move along the cytoskeleton to the cell surface membrane, and fuse with it to release the protein outside the cell via exocytosis.

1. protein is packaged into (transport) vesicles which bud off from the rough endoplasmic reticulum / rER ;
2. transport vesicles fuse with the Golgi apparatus / body, where the protein is modified / processed / sorted ;
3. secretory vesicles bud from the Golgi apparatus, move to and fuse with the cell surface membrane to release protein by exocytosis ;

1. The centromere of each metaphase chromosome splits/divides, uncoupling the two sister chromatids.
2. The kinetochore spindle fibres/microtubules shorten by depolymerisation at the poles.
3. The separated chromatids (now individual daughter chromosomes) are pulled by their centromeres toward opposite poles/centrosomes of the cell, often assuming a characteristic V-shape.

1. centromeres divide / split, separating sister chromatids (into daughter chromosomes) ;
2. spindle fibres / microtubules shorten / depolymerise ;
3. (daughter) chromosomes are pulled / moved towards opposite poles of the cell (centromeres leading / V-shape) ;

1. Goblet cells (and submucosal mucous glands) synthesise and secrete sticky mucus onto the luminal surface of the airways to trap inhaled pathogens and dust particles.
2. Cilia on the apical membrane of ciliated epithelial cells beat rhythmically in a coordinated wave.
3. This mucociliary escalator carries the layer of mucus containing trapped pathogens upward towards the throat/pharynx, where it is swallowed and sterilised by hydrochloric acid in the stomach or coughed out.

1. goblet cells (and mucous glands) produce / secrete mucus which traps inhaled dust / pathogens / bacteria ;
2. cilia beat / waft in a coordinated / rhythmic wave ;
3. mucus (with trapped pathogens) is moved / swept upwards towards the pharynx / throat / larynx to be swallowed (into stomach acid) or expectorated ;

To gain full credit:
1. Draw a clear outline of a single cell using clean, continuous lines without any shading or sketchy lines. The nucleus must be drawn with a characteristic lobed shape (typically 3 to 5 distinct lobes joined by narrow strands or bridges of chromatin).
2. Use a straight, ruled label line touching the nuclear region to clearly identify the lobed nucleus.

1. Drawing quality & nucleus structure:
- sharp, continuous outline drawn without gaps, sketching or shading;
- nucleus clearly drawn as multi-lobed / segmented (with 2 to 5 interconnected lobes);
[1]

2. Label:
- straight / ruled label line clearly touching the nucleus and labelled 'nucleus' / 'lobed nucleus' / 'multilobed nucleus';
[1]

Additional guidance:
- Do not credit if more than one distinct separate cell is drawn.
- Reject label line that has an arrow head, crosses another line, or does not clearly touch the target structure.

1. Protons (H+ ions) are actively pumped out of companion cells into the apoplast / cell wall using ATP from respiration. 2. This creates a steep proton concentration / electrochemical gradient across the companion cell surface membrane. 3. Protons move back down their concentration gradient into the companion cell by facilitated diffusion via cotransporter / symporter proteins, co-transporting sucrose molecules against their concentration gradient. 4. Sucrose accumulates at high concentration inside companion cells and moves into sieve tube elements by simple diffusion through plasmodesmata.

1. Proton pumps actively transport / pump hydrogen ions / H+ / protons out of the companion cell (into the cell wall / apoplast) using ATP ; 2. Generates / establishes a proton / H+ concentration / electrochemical gradient (higher outside companion cell) ; 3. Protons / H+ move back / return into companion cell by facilitated diffusion through a co-transporter / symporter protein (bringing sucrose with them / against its concentration gradient) ; 4. Sucrose moves from companion cell into sieve tube element via / through plasmodesmata (by simple diffusion / down concentration gradient) ; [max 4]

To carry out a serial dilution that halves the concentration at each successive step (a two-fold dilution):

1. Halving the concentration:
- $1.0\% \div 2 = 0.50\%$
- $0.50\% \div 2 = 0.25\%$
- $0.25\% \div 2 = 0.125\%$
- $0.125\% \div 2 = 0.0625\%$

2. Calculating volumes for equal halving:
To have at least $10.0\text{ cm}^3$ of each solution remaining, mix $10.0\text{ cm}^3$ of the previous concentration with $10.0\text{ cm}^3$ of distilled water to give a total volume of $20.0\text{ cm}^3$. Then transfer $10.0\text{ cm}^3$ into the next tube, leaving $10.0\text{ cm}^3$ of that concentration.

Completed Table 1.1:

| Concentration of amylase / % | Volume of amylase solution transferred / $\text{cm}^3$ | Source of amylase solution | Volume of distilled water / $\text{cm}^3$ |
| :---: | :---: | :---: | :---: |
| $1.0$ | $20.0$ | stock A | $0.0$ |
| $0.50$ | $10.0$ | $1.0\%$ / stock A | $10.0$ |
| $0.25$ | $10.0$ | $0.50\%$ | $10.0$ |
| $0.125$ | $10.0$ | $0.25\%$ | $10.0$ |
| $0.0625$ | $10.0$ | $0.125\%$ | $10.0$ |

1. Correct concentrations: $0.5(0)\%$, $0.25\%$, $0.125\%$, and $0.0625\%$ (or $0.063\%$) ;
2. Correct dilution scheme: shows transfer of equal volume of amylase and distilled water (e.g. $10.0\text{ cm}^3$ amylase and $10.0\text{ cm}^3$ water) at each step AND correct source of solution stated for every row ;

To construct an acceptable Cambridge AS Level results table: 1. Draw ruled enclosing grid lines for all rows, columns, and headers. 2. Provide comprehensive column headings with units separated by a solidus (/): 'Percentage concentration of hydrogen peroxide / %' (or 'Concentration of substrate / %'), 'Time taken for disc to rise / s' for Trial 1 and Trial 2 (or under a shared header 'Time taken / s' with subheadings 'Trial 1' and 'Trial 2'), 'Mean time / s', and 'Rate of reaction / s^-1' (or '1000 / mean time / s^-1' or 'arbitrary units'). 3. Arrange the independent variable logically in ascending order (0.2, 0.4, 0.6, 0.8, 1.0) and record raw times as integers. 4. Calculate mean times correctly: 0.2% = 60 s; 0.4% = 35 s; 0.6% = 23 s; 0.8% = 18 s; 1.0% = 14 s. 5. Calculate rate values (1000 / mean time) to a consistent degree of precision (3 significant figures): 0.2% = 16.7; 0.4% = 28.6; 0.6% = 43.5; 0.8% = 55.6; 1.0% = 71.4.

1 mark: Table drawn with all borders and grid lines fully ruled, enclosing all headings and data cells. ; 1 mark: Fully descriptive column headings for independent variable AND dependent variable AND derived values, with units given after a solidus (e.g. 'Concentration of hydrogen peroxide / %' or 'Concentration / %' AND 'Time taken / s' AND 'Mean time / s' AND 'Rate / s^-1' or 'Rate / a.u.'); R units written in body of table. ; 1 mark: Independent variable values in ascending (0.2 to 1.0) or descending order AND all raw data recorded correctly as whole numbers. ; 1 mark: Mean time calculated correctly for all 5 concentrations (60, 35, 23, 18, 14). ; 1 mark: Rate calculated correctly for all 5 concentrations (16.7, 28.6, 43.5, 55.6, 71.4 or rounded consistently to 2 s.f.: 17, 29, 44, 56, 71) recorded to a consistent number of significant figures (2 or 3 s.f.).

The recorded time for unknown U is 98 s. Looking at the standard series:
- 120 s corresponds to \(0.4\,\text{g}\,\text{dm}^{-3}\)
- 82 s corresponds to \(0.6\,\text{g}\,\text{dm}^{-3}\)

Since 98 s lies between 82 s and 120 s, the estimated glucose concentration must lie between \(0.4\,\text{g}\,\text{dm}^{-3}\) and \(0.6\,\text{g}\,\text{dm}^{-3}\) (or approximately \(0.5\,\text{g}\,\text{dm}^{-3}\)).

1 mark: (concentration of U is) between 0.4 (g dm⁻³) and 0.6 (g dm⁻³) / any single value > 0.4 and < 0.6 (g dm⁻³) [with or without units] ;
[A] 0.5 (g dm⁻³)
[R] values ≤ 0.4 or ≥ 0.6

A negative control is used to confirm that the reaction is driven by biological enzyme activity rather than non-enzymatic spontaneous decomposition of hydrogen peroxide. Boiling the plant tissue denatures the catalase enzyme, rendering it inactive, while keeping all other physical and chemical factors constant.

1 mark: use boiled and cooled plant tissue (discs) / replace plant tissue (discs) with (an equal volume of) distilled water / buffer (while keeping all other conditions identical) ;
[A] autoclaved plant tissue / enzyme
[R] omitting hydrogen peroxide / replacing hydrogen peroxide with water

In practical investigations assessing reaction rates, experimental limitations and errors reduce the accuracy and reliability of measurements.

1. Temperature control:
- Error: The reaction was carried out on an open bench where ambient temperature could fluctuate between runs, altering kinetic energy and collision frequency.
- Improvement: Incubate test-tubes in a thermostatically controlled water bath at a constant temperature (e.g. \(35^\circ\text{C}\)) and pre-equilibrate solutions before mixing.

2. Sampling time resolution:
- Error: A 30-second interval is relatively wide, meaning the true end-point could occur anywhere between measurements (e.g. at 35 s or 55 s), leading to an overestimate or inaccuracy in time.
- Improvement: Reduce sampling intervals to every 10 or 15 seconds to narrow the window of uncertainty.

3. Subjective determination of colour end-point:
- Error: Judging when the blue-black colour has fully cleared by eye is subjective and prone to human error.
- Improvement: Compare against a standard colour reference tile or measure absorbance/transmission using a colorimeter at timed intervals.

4. Lack of replication:
- Error: Each concentration was tested only once, so anomalous results cannot be identified and mean values cannot be calculated.
- Improvement: Carry out at least three repeats for each starch concentration and calculate the mean time.

Award [1] mark for each correctly identified source of error and [1] mark for each corresponding workable improvement, up to a maximum of [4] marks.

[Max 2 marks for errors; Max 2 marks for improvements]:

1. Error: Temperature not controlled / room temperature fluctuates / tubes left on bench [1]
Improvement: Use a thermostatically controlled water bath / pre-incubate solutions to set temperature [1]

2. Error: Time interval (30 s) too long / end-point occurs between intervals / difficult to obtain exact time [1]
Improvement: Sample at shorter / more frequent intervals (e.g. every 10 s / 15 s) [1]

3. Error: Judging end-point / colour change is subjective / difficult to see when blue-black completely disappears [1]
Improvement: Use a colorimeter / compare against a set colour standard / reference tube [1]

4. Error: Only one trial per concentration / no repeats (cannot identify anomalies) [1]
Improvement: Repeat at least three times at each concentration and calculate a mean [1]

5. Error: Contamination between tests from using the same dropping pipette [1]
Improvement: Use a clean / separate pipette for each concentration or wash pipette between uses [1]

Note: Improvement MUST correspond directly to the stated error to gain credit.

To gain full marks on a Paper 3 bar chart question, the following criteria must be met:

1. Axes and Headings (1 mark):
- Horizontal axis (x-axis) is labelled with the independent variable category: 'Plant species' (or labels A, B, C, D, E clearly positioned centrally beneath each bar).
- Vertical axis (y-axis) is labelled with the full heading and units from the table: 'mean rate of water loss / \(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\)' (or '\(\text{mg}/\text{cm}^{2}/\text{h}\)').

2. Scale (1 mark):
- The y-axis scale must be linear, starting at 0, with easy-to-read scale markings (e.g., 0, 5, 10, 15, 20, 25, 30, where 20 mm / 2 cm represents 5 \(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\)).
- Plotted data must occupy more than 50% of the available grid vertically and horizontally.
- Each bar must have an identical width, and bars should be separated by equal spaces (or all touch with equal widths).

3. Plotting (1 mark):
- Heights of all 5 bars plotted precisely to within \(\pm 0.5\) of a small grid square:
- A: 14.5 \(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\)
- B: 28.0 \(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\)
- C: 8.5 \(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\)
- D: 22.0 \(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\)
- E: 17.5 \(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\)

4. Line Quality and Presentation (1 mark):
- Drawn with a sharp pencil and ruler.
- Continuous, thin lines with complete horizontal tops; no multiple/feathered lines and no erratic shading.

1. x-axis labelled 'plant species' / 'A, B, C, D, E' AND y-axis labelled 'mean rate of water loss / \(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\)' ; [1]
[A: 'mean rate of water loss (\(\text{mg}\,\text{cm}^{-2}\,\text{h}^{-1}\))' or '\(\text{mg}/\text{cm}^2/\text{h}\)']
[R: missing units or unit without slash/bracket]

2. Scale on y-axis is linear, with values at least every 20 mm, uses more than half the grid along both axes AND bars are of equal width with equal spacing between each bar (or touching bars of equal width) ; [1]
[R: non-linear scale / awkward scale subdivisions like 3 or 7]

3. Plotting: all five bars plotted accurately to within \(\pm 0.5\) small square of the true values (14.5, 28.0, 8.5, 22.0, 17.5) ; [1]

4. Quality of drawing: all vertical and horizontal bar lines ruled, sharp, thin, and continuous with no overlapping lines, breaks, or messy shading ; [1]

To produce an accurate low-power plan diagram of a dicotyledonous root section:
1. Quality of drawing: Use a sharp pencil to draw single, clear, unbroken lines without any sketching, gaps, or overlapping ends. Do not include any shading or draw individual cells.
2. Size: The diagram must occupy at least half the available space provided.
3. Tissue layers & shape: Draw a defined sector (one-quarter of the circular cross-section) showing the outer layer (epidermis/exodermis), the broad region of parenchyma (cortex), the circular inner boundary (endodermis/pericycle enclosing the stele), and the vascular core.
4. Proportions: The cortex should be considerably wider than the central vascular cylinder (stele radius occupies less than half the total radius of the sector).
5. Vascular tissue arrangement: The xylem forms a distinct central, radiating/star-shaped lobed core with discrete phloem areas situated between the xylem arms.
6. Labelling: Use a ruler to draw a single, straight label line that directly touches the xylem region, labelled neatly as 'xylem'.

1. Clear, sharp, continuous lines drawn with a sharp pencil AND no shading AND diagram occupies at least half the space provided ;
2. Sector / one quarter of the root section drawn showing outer layer (epidermis) and central vascular cylinder (stele) enclosed by endodermis ;
3. Correct proportion: cortex drawn significantly wider than the central stele (stele radius is less than half the total root radius) ;
4. Correct tissue distribution within the stele showing central radiating / star-shaped xylem with distinct phloem areas alternating between the arms of xylem ;
5. One label line ruled with a straight edge directly touching the xylem tissue AND labelled 'xylem' ;

Guidance notes:
- `R` if any individual cells are drawn.
- `R` feathery, overlapping, or broken pencil lines.
- `R` label line with arrowheads or lines that do not touch the target tissue.

To gain maximum marks for this high-power cell drawing:
1. Quality of lines: Use a sharp HB pencil to draw clear, smooth, unbroken lines without shading, stippling, feathering, or gaps.
2. Size: The drawing must occupy at least 50% of the available space on the page.
3. Cell arrangement & number: Draw exactly four complete, touching cortex cells showing realistic polyhedral/rounded shapes and proportional sizes.
4. Cell wall detail: Cell walls between adjacent cells must be drawn as double lines to represent wall thickness, showing junctions between neighbouring cells correctly.
5. Label: Draw a single straight line using a ruler, ensuring the line touches the cell wall directly without crossing it or ending in an arrowhead, labelled clearly as 'cell wall'.

1. Clarity of lines & no shading: Sharp, continuous, single lines used throughout AND no shading or sketchy/feathered lines anywhere in the drawing ; [1]
2. Size: Drawing occupies greater than 50% of the available space provided ; [1]
3. Cell representation & grouping: Exactly four adjacent, touching cortex cells drawn AND showing correct relative shapes/proportions ; [1]
4. Cell wall structure: Cell walls represented by two distinct continuous lines (double line) to show thickness, including appropriate three-cell or four-cell junctions ; [1]
5. Label & label line: One ruled label line with a ruler touching the cell wall AND correctly labelled cell wall ; [1]

Marking notes:
- `R` if any cell is incomplete or open-ended.
- `R` if label line has an arrowhead or does not touch the target structure.
- `I` internal organelles/inclusions if correctly positioned without shading.

A suitable completed table of observable differences:

$$\begin{array}{|l|l|l|}
\hline
\textbf{Feature} & \textbf{Stem in Fig. 1.1 (terrestrial plant)} & \textbf{Stem on slide M1 (aquatic plant)} \\
\hline
\text{Air spaces in cortex / ground tissue} & \text{Small / few intercellular air spaces} & \text{Large / extensive aerenchyma / air spaces present} \\
\hline
\text{Distribution / arrangement of vascular tissue} & \text{Vascular bundles arranged in a distinct ring} & \text{Vascular tissue clustered in a central core / stele} \\
\hline
\text{Xylem vessels} & \text{Many / large, thick-walled vessels} & \text{Few / small / reduced xylem vessels} \\
\hline
\text{Pith / central region} & \text{Large central parenchymatous pith present} & \text{No distinct central pith / centre occupied by vascular tissue} \\
\hline
\end{array}$$

(Any three valid comparative structural differences)

1. Table drawn with complete ruled lines AND fully labelled column/row headings including 'feature / structural characteristic', 'stem in Fig. 1.1' (or 'terrestrial stem'), and 'stem on slide M1' (or 'aquatic stem') ; [1]
2. Feature column / row provided AND only observable structural differences recorded (R functional interpretations / non-observable statements, e.g. 'carries more water', 'adapted for buoyancy') ; [1]
3. Any two correct paired comparative observations directly comparing the same feature for Fig. 1.1 and slide M1 ; [1]
4. Third correct paired comparative observation ; [1]

Acceptable comparative points include:
- Air spaces / aerenchyma: Fig. 1.1 has small / few / normal intercellular spaces AND M1 has large / wide / extensive air chambers / aerenchyma ;
- Vascular bundle arrangement: Fig. 1.1 has vascular bundles arranged in a ring / circle AND M1 has vascular tissue in a single central group / core / no ring ;
- Xylem development / vessel size: Fig. 1.1 has numerous / large / prominent / thick-walled xylem vessels AND M1 has few / small / reduced / thin-walled xylem vessels ;
- Central region / pith: Fig. 1.1 has a large central pith / parenchyma core AND M1 has central pith absent / reduced / replaced by central vascular cylinder ;
- Mechanical tissue / sclerenchyma (caps): Fig. 1.1 has bundle caps / sclerenchyma / collenchyma present AND M1 has no sclerenchyma / mechanical tissue absent ;

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

Step 2: Determine the total distance covered by 18 divisions:
\[18 \times 10\ \mu\text{m} = 180\ \mu\text{m}\]

Step 3: Calculate the value of 1 eyepiece graticule unit (epgu):
\[1\text{ epgu} = \frac{180\ \mu\text{m}}{40} = 4.5\ \mu\text{m}\]

(b)
Step 1: Multiply the number of eyepiece graticule units by the calibrated value of one unit:
\[\text{Actual diameter} = 15 \times 4.5\ \mu\text{m} = 67.5\ \mu\text{m}\]
(Accept rounded to 68 \(\mu\text{m}\))

(a)
- MP1: Calculation showing conversion to \(\mu\text{m}\) (e.g. \(18 \times 10 = 180\ \mu\text{m}\) or \(0.18\text{ mm}\)) AND division by \(40\) [1]
- MP2: Correct answer of \(4.5\) (\(\mu\text{m}\)) [1]

(b)
- MP3: Multiplies \(15\) by candidate's calculated value from (a) (allow ecf) [1]
- MP4: Correct answer of \(67.5\) or \(68\) (\(\mu\text{m}\)) [1]