2025 CCEA AS-Level Biology 1010 Practice Paper with Answers

Glycogen is the main storage polysaccharide in animal cells, formed from many α-glucose molecules joined by glycosidic bonds; it is stored mainly in the liver and muscle cells.
Final answer: glycogen.

Correct answer — glycogen [1].

The substrate binds to a specific region of the enzyme called the active site, whose shape is complementary to the substrate.
Final answer: active site.

Correct answer — active site [1].

HIV is a retrovirus; it carries the enzyme reverse transcriptase, which uses the viral RNA as a template to synthesise a complementary strand of DNA inside the host cell.
Final answer: reverse transcriptase.

Correct answer — reverse transcriptase [1].

One example is the cellulose cell wall, which surrounds the cell surface membrane of a plant cell but is absent from an animal cell. Its function is to provide mechanical strength/support and to prevent the cell from bursting (lysing) when it takes up water and becomes turgid, by exerting an inward pressure that resists the outward pressure of the swollen cytoplasm.
Final answer: cellulose cell wall — provides structural support and prevents the cell bursting when turgid. (Other valid pairs, e.g. chloroplasts — site of photosynthesis; permanent vacuole — maintains turgor, are also acceptable.)

Valid structural feature named [1]; correct corresponding function [1].

A cell that is not actively dividing is in interphase, the phase between successive mitotic divisions in which the cell grows and its organelles replicate. DNA replication specifically occurs during the S phase (synthesis phase) of interphase, which lies between the G1 and G2 phases.
Final answer: interphase; S phase.

Correct answer — interphase [1]; correct answer — S phase (synthesis phase) [1].

Two molecules of α-glucose are joined together in a condensation reaction: a hydroxyl (–OH) group on carbon 1 of one glucose molecule reacts with a hydroxyl group on carbon 4 of the second glucose molecule. In this reaction a molecule of water is released, and a covalent glycosidic bond forms between the two glucose molecules, producing the disaccharide maltose.
Final answer: condensation reaction; a 1,4-glycosidic bond forms, with the release of a water molecule.

Correct identification of a condensation reaction [1]; correct reference to a hydroxyl group from each glucose molecule reacting, with release of water [1]; correct naming of the bond formed as a glycosidic bond [1].

The ion binds to a specific binding site on the carrier protein, on the side of the membrane where the ion is present. ATP is hydrolysed to ADP and inorganic phosphate, releasing energy; this energy causes the carrier protein to change shape (a conformational change). This change in shape moves the ion across the membrane, releasing it on the other side, even though this is against (up) its concentration gradient — hence the process requires metabolic energy, unlike facilitated diffusion.
Final answer: as stated above — ion binds to carrier protein; ATP hydrolysis provides energy for a shape change that moves the ion across the membrane against its concentration gradient.

Correct reference to the ion binding to a specific carrier protein [1]; correct reference to ATP being hydrolysed to provide energy [1]; correct explanation that this causes a change in shape of the carrier protein, moving the ion across against its concentration gradient [1].

When an enzyme catalyses a reaction, the substrate binds to the enzyme's active site to form an enzyme-substrate complex. This provides an alternative reaction pathway that has a lower activation energy than the uncatalysed reaction — the minimum energy required for the reaction to proceed is reduced, for example because binding to the active site puts strain on particular bonds within the substrate, making them easier to break. Because the activation energy is lower, a greater proportion of substrate (or reactant) molecules already possess enough kinetic energy to react at a given temperature, so more successful collisions occur per unit time, and the reaction proceeds at a faster rate than the uncatalysed reaction.
Final answer: as stated above — the enzyme lowers the activation energy required, so more molecules can react, increasing the rate.

Correct reference to formation of an enzyme-substrate complex providing an alternative pathway [1]; correct statement that this pathway has a lower activation energy [1]; correct link to a greater proportion of molecules having sufficient energy to react, increasing rate [1].

During prophase, the chromosomes condense (supercoil) and become visible as thread-like structures, each consisting of two identical sister chromatids joined at a centromere; the nuclear envelope breaks down and disappears; centrioles migrate to opposite poles of the cell, and spindle fibres begin to form between them. During metaphase, the chromosomes move to, and line up individually along, the equator of the cell (the metaphase plate), each attached to spindle fibres at its centromere, with one chromatid of each chromosome facing each pole.
Final answer: as stated above.

Prophase — correct description of chromosome condensation into sister chromatids [1]; correct reference to nuclear envelope breakdown and/or spindle formation [1]. Metaphase — correct description of chromosomes lining up at the equator/metaphase plate [1]; correct reference to attachment to spindle fibres at the centromere [1].

(a) A purple biuret result but a negative (blue, unchanged) Benedict's result identifies sample W as containing protein without reducing sugar.
(b) A purple biuret result together with a positive (brick-red) Benedict's result identifies sample Y as containing both protein and reducing sugar.
(c) A positive (purple/violet) biuret test indicates that the sample contains peptide bonds — specifically, at least two peptide bonds, as found in a protein or polypeptide (a single amino acid, with no peptide bond, does not give a full positive result). The Cu2+ ions in the reagent form a coloured complex with the peptide bonds under alkaline conditions.
Final answer: (a) W; (b) Y; (c) presence of (at least two) peptide bonds, i.e. protein/polypeptide, complexing with Cu2+ ions in alkaline conditions.

(a) Correct answer — W [1]. (b) Correct answer — Y [1]. (c) Correct reference to peptide bonds/protein present [1]; correct additional detail — at least two peptide bonds and/or Cu2+ complexing in alkaline conditions [1].

(a) Triglyceride P contains unsaturated fatty acids (fatty acid chains containing one or more C=C double bonds).
(b) Each C=C double bond in an unsaturated fatty acid chain introduces a 'kink' (bend) in the otherwise straight hydrocarbon tail. These kinks prevent the triglyceride molecules from packing closely together in a regular, ordered arrangement. Because the molecules cannot pack tightly, the (van der Waals) intermolecular forces between neighbouring molecules are weaker/fewer, so less energy is needed to overcome them, resulting in a lower melting point (liquid at room temperature, i.e. an oil).
Final answer: (a) unsaturated; (b) C=C double bonds cause kinks that prevent close packing, weakening intermolecular forces and lowering the melting point.

(a) Correct answer — unsaturated [1]. (b) Correct reference to C=C double bonds causing kinks/bends in the fatty acid chain [1]; correct explanation that this prevents close packing of molecules [1]; correct link to weaker intermolecular forces and lower melting point [1].

(a) \( \text{percentage change in mass} = \frac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100\% \)
\( = \frac{4.53 - 4.20}{4.20} \times 100\% = \frac{0.33}{4.20} \times 100\% = 7.857...\% \approx +7.9\% \)
Check by a second route: a 7.9% increase on 4.20 g is \( 4.20 \times 0.0786 = 0.330 \) g, and \( 4.20 + 0.330 = 4.53 \) g, which matches the given final mass.
(b) Since the potato cylinders gained mass, water must have moved into the cells by osmosis; water moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential, so the sucrose solution must have had a higher (less negative) water potential than the cells inside the potato tissue.
Final answer: (a) +7.9% (increase); (b) the solution had a higher (less negative) water potential than the potato cells.

(a) Correct equation for percentage change [1]; correct substitution [1]; correct answer +7.9% (increase clearly indicated) [1].
(b) Correct conclusion (solution has higher/less negative water potential than the cells) with valid reason (mass gained) [1].

(a) \( \psi = \psi_s + \psi_p = (-700) + (+200) = -500\ \text{kPa} \)
Check by a second route: rearranging, \( \psi_s = \psi - \psi_p = -500-200=-700 \) kPa, which matches the given solute potential.
(b) Pure water has a water potential of 0 kPa, which is higher than the cell's water potential (−500 kPa), so water will move into the cell, by osmosis, down the water potential gradient. As water enters, the cell contents push outward against the cell wall, increasing the pressure potential (ψp). This continues, and ψp continues to rise, until the cell's overall water potential rises to 0 kPa (equal to the surrounding pure water) and net water movement stops — at this point ψp will have risen to +700 kPa, exactly balancing ψs (−700 kPa).
Final answer: (a) −500 kPa; (b) pressure potential increases (to +700 kPa) as water enters by osmosis, until ψcell = 0.

(a) Correct equation ψ = ψs + ψp [1]; correct answer −500 kPa [1].
(b) Correct statement that ψp increases as water enters by osmosis [1]; correct explanation that this continues until ψcell = 0 (equal to pure water), i.e. ψp rises to +700 kPa [1].

\( \text{mitotic index} = \frac{\text{number of cells with visible chromosomes (in mitosis)}}{\text{total number of cells counted}} \times 100\% \)
\( = \frac{14}{280} \times 100\% = 5.0\% \)
Check by a second route: \( 5.0\% \text{ of } 280 = 0.05 \times 280 = 14 \), which matches the given number of cells in mitosis.
The mitotic index represents the proportion (percentage) of cells in a tissue sample that are undergoing mitosis (cell division) at the moment the sample was taken/fixed — it gives a measure of how actively the tissue is dividing/growing.
Final answer: mitotic index = 5.0%.

Correct equation for mitotic index [1]; correct substitution [1]; correct answer 5.0% [1]. Correct statement of what mitotic index represents [1].

Assuming the proportion of cells observed in a phase reflects the proportion of the total cell cycle time spent in that phase:
\( \text{proportion of cells in G1} = \frac{180}{400} = 0.45 \)
\( \text{time in G1} = 0.45 \times 20\ \text{hours} = 9.0\ \text{hours} \)
Check by a second route: \( \frac{9.0}{20} \times 100\% = 45\% \), and 45% of 400 cells = 180 cells, which matches the given data.
Assumption: that the sample of 400 cells was taken at random and is representative of the whole tissue, and that the proportion of cells in a given phase at any instant is proportional to the fraction of the cycle time spent in that phase.
Final answer: 9.0 hours.

Correct calculation of the proportion of cells in G1 (0.45 or 45%) [1]; correct method (proportion × total cycle time) [1]; correct answer 9.0 hours [1]. Valid assumption stated [1].

(a) Phloem sieve tube elements are living cells at maturity (unlike xylem vessels, which are dead).
(b) Xylem transports water and dissolved mineral ions in one direction only, upwards from the roots to the leaves and other parts of the plant.
(c) Each sieve tube element is closely linked to an adjacent companion cell by numerous plasmodesmata (cytoplasmic connections) passing through the cell walls between them. The companion cell retains its nucleus and organelles (e.g. many mitochondria) and carries out the metabolic activities — such as protein synthesis and the production of ATP needed for active loading of solutes — that keep the sieve tube element alive and functioning, effectively acting on its behalf.
Final answer: (a) Yes (living); (b) upwards (roots to rest of plant, one direction only); (c) companion cells, linked by plasmodesmata, carry out metabolism on behalf of the sieve tube elements.

(a) Correct answer — Yes/living [1]. (b) Correct answer — one direction, upwards (roots to shoot) [1]. (c) Correct reference to companion cells linked via plasmodesmata [1]; correct explanation that companion cells carry out metabolic functions on behalf of the sieve tube element [1].

(a) \( \text{actual surface area} = 100\ \text{cm}^2 \times 30 = 3000\ \text{cm}^2 \)
Check by a second route: \( 3000 / 30 = 100\ \text{cm}^2 \), which matches the flat-sheet area given.
(b) A large surface area increases the rate at which digested food molecules (such as amino acids and monosaccharides) can be absorbed by diffusion and active transport into the blood capillaries and lacteals within the villi, since the rate of diffusion is directly proportional to the surface area available. This allows the large quantity of nutrients released by digestion to be absorbed efficiently and quickly, within the limited length of the small intestine, meeting the body's metabolic demands.
Final answer: (a) 3000 cm²; (b) a greater surface area increases the rate of absorption of digested food by diffusion/active transport, as rate is proportional to surface area.

(a) Correct answer 3000 cm² [1].
(b) Correct reference to a larger surface area increasing the rate of absorption/diffusion [1]; correct reference to specific molecules absorbed and/or the route (blood capillaries, lacteals) [1]; correct link to meeting metabolic demand efficiently [1].

\( \text{mean rate} = \frac{\text{change in volume of O}_2}{\text{change in time}} = \frac{22.0 - 4.0}{80 - 20} = \frac{18.0}{60} = 0.30\ \text{cm}^3\,\text{s}^{-1} \)
Check by a second route: at a constant rate of 0.30 cm³ s⁻¹ over 60 s, the volume produced would be \( 0.30 \times 60 = 18.0\ \text{cm}^3 \), which matches the calculated change in volume (22.0 − 4.0 = 18.0 cm³).
Controlled variable: temperature must be kept constant (e.g. using a water bath), since temperature also affects enzyme activity and could confound the effect of enzyme concentration; pH and substrate concentration should also be standardised.
Final answer: mean rate = 0.30 cm³ s⁻¹; controlled variable — temperature (or pH, or substrate concentration).

Correct equation (change in volume / change in time) [1]; correct substitution [1]; correct answer 0.30 cm³ s⁻¹ [1]. Valid controlled variable named (e.g. temperature, pH, substrate concentration, volume of enzyme) [1].

(a) The lytic cycle is a type of viral replication in which the virus takes over the host cell to make many new virus particles, ultimately destroying (lysing) the host cell in the process of releasing them.
(b) The bacteriophage first attaches to specific complementary receptor molecules on the surface of the bacterial cell wall. It then injects its nucleic acid (DNA) into the bacterial cell (the protein coat remains outside). Inside the cell, the phage DNA takes over the host's metabolic machinery (enzymes, ribosomes, nucleotides) to replicate multiple copies of the phage DNA and to synthesise phage proteins (e.g. coat proteins). These components then self-assemble into many new, complete phage particles. Finally, the bacterial cell lyses (its cell wall/membrane bursts open), releasing the new phage particles, which can then go on to infect further bacterial cells.
Final answer: as stated above.

(a) Correct definition — replication cycle in which the host cell is destroyed/lysed to release new virus particles [1].
(b) Correct reference to attachment to the host cell and injection of DNA [1]; correct reference to replication of phage DNA and synthesis of phage proteins using host machinery [1]; correct reference to assembly of new phages and lysis of the host cell to release them [1].

Indicative content:
• The cell surface membrane is composed mainly of a phospholipid bilayer: each phospholipid has a hydrophilic (polar) phosphate head and two hydrophobic (non-polar) fatty acid tails; in water, the phospholipids arrange themselves as a bilayer with the hydrophilic heads facing the aqueous cytoplasm and extracellular fluid on each side, and the hydrophobic tails facing inward, away from water.
• The 'fluid' aspect: the phospholipids (and proteins) are not fixed in position but can move sideways within their own layer, giving the membrane a flexible, fluid quality, allowing it to change shape, self-seal and fuse with other membranes (e.g. during endo-/exocytosis).
• The 'mosaic' aspect: a variety of protein molecules are embedded within, and scattered throughout, the phospholipid bilayer at different densities, giving a patchwork/mosaic appearance. These include intrinsic (integral) proteins spanning the whole bilayer, and extrinsic (peripheral) proteins attached to only one surface.
• Types of intrinsic protein include channel proteins (forming water-filled pores for facilitated diffusion of ions/polar molecules) and carrier proteins (which change shape to transport specific molecules by facilitated diffusion or active transport).
• Cholesterol molecules are also interspersed within the bilayer, regulating fluidity by preventing the membrane from becoming too fluid at high temperatures or too rigid at low temperatures.
• Glycoproteins and glycolipids (proteins/lipids with attached carbohydrate chains) project from the outer surface and are involved in cell signalling/recognition (e.g. as antigens or receptors).
• Relating structure to function: the hydrophobic core of the phospholipid bilayer prevents the free passage of water-soluble/polar substances and ions, while allowing small, non-polar/lipid-soluble molecules (e.g. O2, CO2) to diffuse directly through — this gives the membrane its selective permeability. Larger or polar/charged substances can only cross via specific channel or carrier proteins embedded in the membrane, which allow the cell precise control over exactly which substances (and how much) enter or leave.
Final answer: a well-organised account of the phospholipid bilayer, fluidity, the mosaic arrangement of intrinsic/extrinsic proteins (including channel and carrier proteins), cholesterol and glycoproteins, clearly linked to the concept of selective permeability.

Level 3 (7–9 marks): Detailed, accurate description of the phospholipid bilayer (hydrophilic heads/hydrophobic tails), the fluid and mosaic aspects (including intrinsic/extrinsic proteins, channel/carrier proteins, cholesterol, glycoproteins); clear, well-developed explanation linking this structure to selective permeability; answer well organised with fluent, accurate use of specialist terms.
Level 2 (4–6 marks): Reasonable description of the phospholipid bilayer and some correct reference to membrane proteins and/or cholesterol/glycoproteins, though possibly incomplete; some explanation of selective permeability given but not fully developed; mostly appropriate use of specialist terms.
Level 1 (1–3 marks): Basic, list-like points about membrane structure (e.g. phospholipids, proteins named) with little clear organisation or explanation of selective permeability; limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

Indicative content:
• Many substances — for example ions (e.g. Na+, Cl−) and polar/water-soluble molecules (e.g. glucose, amino acids) — cannot pass directly through the hydrophobic core of the phospholipid bilayer, so they require specific membrane proteins to cross.
• Channel proteins are intrinsic proteins that span the membrane and form water-filled pores/channels; the channel is often specific to a particular ion (e.g. a sodium channel), and its opening may be permanently open, gated by voltage, or gated by a ligand; ions move through the channel, by facilitated diffusion, down their concentration (electrochemical) gradient — this requires no metabolic energy (ATP).
• Carrier proteins are also intrinsic proteins that bind to a specific molecule/ion on one side of the membrane; binding causes the carrier protein to change shape, moving the bound substance across the membrane and releasing it on the other side.
• When carrier proteins move a substance down its concentration gradient, no energy is required — this is facilitated diffusion (used for larger/polar molecules such as glucose).
• When carrier proteins move a substance against its concentration gradient, energy from the hydrolysis of ATP is required to drive the change in shape of the carrier protein — this is active transport (e.g. the sodium-potassium pump).
• Because each channel and carrier protein is specific to a particular substance, the cell can precisely control which substances cross the membrane, and in which direction, independently of the phospholipid bilayer itself.
Final answer: channel proteins provide gated/selective pores for facilitated diffusion of ions down their gradient; carrier proteins bind a specific substance and change shape to move it across, either by facilitated diffusion (down gradient, no ATP) or active transport (against gradient, using ATP).

Level 3 (5–6 marks): Accurate explanation of both channel proteins (pores, ion-specific, facilitated diffusion) and carrier proteins (binding and shape change), with a clear, correct distinction between facilitated diffusion (down gradient, no ATP) and active transport (against gradient, requires ATP); fluent use of specialist terms.
Level 2 (3–4 marks): Reasonable explanation of channel and/or carrier proteins, with some correct reference to facilitated diffusion and/or active transport, though not fully complete or with minor inaccuracies; appropriate use of some specialist terms.
Level 1 (1–2 marks): Basic, limited points made (e.g. protein names only, or vague reference to 'proteins help move substances') without clear mechanism; limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

Species richness is simply the number of different species present within a habitat or community, without regard to how many individuals of each species are present. Species evenness is a measure of the relative abundance of each species — how equally the total number of individuals is distributed among the different species present (a habitat with similar numbers of each species has high evenness; one dominated by a few species has low evenness).
Final answer: as stated above.

Correct definition of species richness (number of different species) [1]; correct definition of species evenness (relative abundance/equal distribution of individuals among species) [1].

Physiological adaptation: e.g. the kidneys produce a small volume of highly concentrated urine (associated with an unusually long loop of Henle), minimising water loss in excretion.
Behavioural adaptation: e.g. being nocturnal, remaining inactive in a cool burrow during the heat of the day and only being active at night, reducing water loss through sweating/evaporation.
Final answer: as stated above.

Valid physiological adaptation [1]; valid behavioural adaptation [1].

In-situ conservation involves protecting a species (and its genetic diversity) within its natural habitat, for example by establishing a nature reserve or Site of Special Scientific Interest (SSSI). Ex-situ conservation involves conserving a species outside its natural habitat, for example in a zoo, botanic garden, seed bank, or captive breeding programme, often as a last resort or to support reintroduction.
Final answer: as stated above.

Correct definition of in-situ conservation with valid example [1]; correct definition of ex-situ conservation with valid example [1].

Surface area is proportional to (length)², whereas volume is proportional to (length)³. As an organism's linear dimensions increase, its volume therefore increases proportionally much faster than its surface area, so the ratio of surface area to volume (SA:V) becomes smaller as size increases.
Final answer: volume increases faster than surface area (as the cube rather than the square of linear size), so SA:V decreases with increasing size.

Correct reference to surface area increasing as the square of length and volume as the cube of length [1]; correct conclusion that volume increases faster, decreasing SA:V ratio [1].

The C-shaped rings of cartilage in the wall of the trachea provide support, holding the trachea permanently open so that it does not collapse when air pressure inside falls during inhalation; the incomplete (C-shaped, not full circle) rings also allow the adjacent oesophagus to expand when food is swallowed. Goblet cells, found in the epithelium lining the trachea, secrete mucus, which traps inhaled dust, debris and microorganisms before they can reach the lungs (this mucus is then moved upward by the beating of cilia on adjacent ciliated cells, towards the throat, to be swallowed or expelled).
Final answer: as stated above.

Correct role of cartilage rings (holding trachea open/preventing collapse) [1]; correct reference to the C-shape allowing oesophagus expansion [1]; correct role of goblet cells (secreting mucus to trap particles/pathogens) [1].

The left ventricle must pump blood at high pressure all the way around the body (the systemic circulation), which is a much longer circuit with higher overall resistance than the short pulmonary circulation from the right ventricle to the (nearby) lungs. To generate this higher pressure, the left ventricle needs to contract with much greater force, which requires a much thicker wall of cardiac muscle than the right ventricle (which only needs to generate enough pressure to push blood the short distance to the lungs, and low pressure is in fact needed there to avoid damaging the delicate capillaries of the alveoli).
Final answer: the left ventricle pumps blood at higher pressure over the much longer systemic circuit (whole body) compared with the short pulmonary circuit to the lungs, so it needs a thicker muscular wall to generate this greater force.

Correct reference to the left ventricle pumping blood around the whole body/systemic circulation (a longer/higher-resistance circuit) [1]; correct reference to the right ventricle only pumping to the (nearby) lungs [1]; correct link to the need for greater force/pressure requiring a thicker muscular wall in the left ventricle [1].

(a) As wind speed increases from 0 to 3 m s⁻¹, the rate of water uptake increases; the increase is approximately linear/proportional over this range (roughly a similar increase of about 9–10 mm³ min⁻¹ for each 1 m s⁻¹ increase in wind speed).
(b) Water vapour diffuses out of the stomata into the air immediately surrounding the leaf, building up a layer of relatively humid, still air close to the leaf surface. Without wind, this layer of humid air reduces the concentration gradient between the air spaces inside the leaf and the air just outside, slowing diffusion. Wind removes/disperses this layer of humid air as fast as it forms, maintaining a steep water vapour concentration gradient between the (saturated) air inside the leaf and the air outside, so that diffusion of water vapour out of the stomata — and therefore the rate of transpiration (and hence water uptake by the shoot, replacing the water lost) — continues at a faster rate.
Final answer: increased wind speed increases the water vapour concentration gradient at the leaf surface (by removing humid air), speeding diffusion and so transpiration rate.

(a) Correct description of a (roughly linear/proportional) positive relationship [1].
(b) Correct reference to wind removing/dispersing water vapour from around the stomata [1]; correct reference to this maintaining/steepening the diffusion (concentration) gradient [1]; correct link to a faster rate of diffusion/transpiration as a result [1].

At the source (e.g. a photosynthesising leaf), sucrose produced by photosynthesis is actively loaded into the sieve tube elements of the phloem (using companion cells and ATP). This lowers the water potential inside the sieve tube at the source, so water moves into the sieve tube from the adjacent xylem by osmosis, down the water potential gradient, raising the hydrostatic (turgor) pressure inside the sieve tube at the source end. At the sink (e.g. a growing root or storage organ), sucrose is actively unloaded/used up (e.g. in respiration or converted to starch for storage), which raises the water potential inside the sieve tube at the sink; water therefore leaves the sieve tube by osmosis at the sink, lowering the hydrostatic pressure there. This creates a hydrostatic pressure gradient, from high pressure at the source to low pressure at the sink, which drives a mass flow of phloem sap (water and dissolved sucrose together) through the sieve tubes from source to sink.
Final answer: as stated above.

Correct reference to active loading of sucrose at the source lowering water potential there [1]; correct reference to water entering by osmosis from the xylem, raising pressure at the source [1]; correct reference to sucrose being removed/used at the sink, raising water potential and causing water to leave [1]; correct conclusion that the resulting pressure gradient drives mass flow from source to sink [1].

(a) The maximum pressure reached during ventricular systole is called the systolic pressure.
(b) The aortic (semilunar) valve opens once the rising pressure in the left ventricle, generated by ventricular contraction (systole), exceeds the pressure in the aorta — this allows blood to be ejected from the ventricle into the aorta. As the ventricle then begins to relax (the start of diastole), its pressure falls rapidly; once ventricular pressure falls below the pressure remaining in the aorta, the aortic valve closes — this prevents blood from flowing backwards from the aorta into the (now lower-pressure) left ventricle.
Final answer: (a) systolic pressure; (b) the valve opens when ventricular pressure exceeds aortic pressure (during systole) and closes when ventricular pressure falls below aortic pressure (early diastole), preventing backflow.

(a) Correct answer — systolic pressure [1].
(b) Correct explanation of the valve opening when ventricular pressure exceeds aortic pressure [1]; correct explanation of the valve closing when ventricular pressure falls below aortic pressure [1]; correct reference to this preventing backflow of blood into the ventricle [1].

(a) The wide lumen of a vein offers relatively little resistance to the flow of blood through it. Since blood in the veins is at low, steady pressure (having lost most of its pressure passing through the arterioles and capillary networks), a wide lumen helps blood to still be returned to the heart at a reasonable flow rate despite this low pressure, aided also by the squeezing action of surrounding skeletal muscles and the presence of valves preventing backflow.
(b) A capillary wall consisting of a single layer of endothelial cells is extremely thin, giving a very short diffusion pathway/distance between the blood plasma and the surrounding tissue fluid/cells. This allows substances such as oxygen, glucose and carbon dioxide to diffuse rapidly across the capillary wall, which is essential given the limited time blood spends passing through each capillary.
Final answer: (a) a wide lumen reduces resistance to flow, compensating for low venous pressure; (b) a one-cell-thick wall gives a short diffusion distance, allowing rapid exchange of substances.

(a) Correct reference to a wide lumen reducing resistance to flow [1]; correct link to compensating for the low pressure in veins [1].
(b) Correct reference to a short diffusion distance/pathway [1]; correct link to allowing rapid diffusion/exchange of specific substances [1].

(a) \( \text{tidal volume} = \text{maximum volume} - \text{minimum volume} = 2.5 - 2.0 = 0.5\ \text{dm}^3 \)
(b) \( \text{pulmonary ventilation rate} = \text{tidal volume} \times \text{breathing rate} = 0.5 \times 15 = 7.5\ \text{dm}^3\,\text{min}^{-1} \)
Check by a second route: \( 7.5 / 15 = 0.5\ \text{dm}^3 \), which matches the tidal volume found in (a).
Final answer: (a) tidal volume = 0.5 dm³; (b) pulmonary ventilation rate = 7.5 dm³ min⁻¹.

(a) Correct answer 0.5 dm³ [1].
(b) Correct equation (tidal volume × breathing rate) [1]; correct substitution using own or given tidal volume [1]; correct answer 7.5 dm³ min⁻¹ [1].

(a) Both fields contain the same number of different species (6 each), so they have equal species richness — neither is higher.
(b) Field A has the higher species evenness. Species evenness measures how equally individuals are distributed among the species present in a community. In Field A, the individuals are described as being distributed roughly evenly among the 6 species, giving high evenness. In Field B, although 6 species are also present (equal richness), one species dominates numerically while the other five are only sparsely represented, meaning the individuals are very unequally distributed — giving low evenness, despite the equal species richness.
Final answer: (a) neither — equal richness; (b) Field A has higher evenness, as its individuals are distributed more equally among the species present.

(a) Correct answer — neither/equal [1].
(b) Correct answer — Field A [1]; correct explanation referring to more equal distribution of individuals among species in A [1]; correct contrast with Field B being dominated by one species [1].

The nitrate fertiliser entering the stream provides a large, additional source of a limiting mineral nutrient, causing rapid, excessive growth of algae and other plants near the point of entry (an algal bloom). This dense growth blocks light from reaching plants deeper in the water, causing these shaded plants to die. Saprobiotic (decomposer) microorganisms then break down this dead organic matter, and their aerobic respiration consumes large quantities of dissolved oxygen from the water — which explains why dissolved oxygen concentration falls sharply between 0 m and 100 m downstream. Further downstream (200–300 m), the nitrate has been diluted and much of the organic matter has already been decomposed, so the rate of oxygen-consuming decomposition decreases; oxygen is also continually replenished by diffusion from the air at the water surface and by photosynthesis of remaining plants/algae, so the dissolved oxygen concentration recovers/rises again further downstream.
Final answer: oxygen falls sharply near the point of entry due to increased aerobic decomposition of dead algae/plants (following an algal bloom triggered by the nitrate), then recovers further downstream as the nutrient is diluted and decomposition/oxygen demand decreases.

Correct reference to nitrate causing excessive algal/plant growth (algal bloom/eutrophication) [1]; correct reference to death of shaded plants and their subsequent decomposition [1]; correct reference to aerobic decomposers/microorganisms using up oxygen in respiration, explaining the fall in oxygen concentration [1]; correct explanation of the recovery of oxygen concentration further downstream (dilution/reduced decomposition, replenishment by diffusion/photosynthesis) [1].

(a) \( \text{percentage cleared} = \frac{\text{area cleared}}{\text{original area}} \times 100\% = \frac{900}{4500} \times 100\% = 20\% \)
Check by a second route: 20% of 4500 hectares = \( 0.20 \times 4500 = 900 \) hectares, which matches the area cleared given.
(b) Deforestation destroys the natural habitat of many forest-dwelling species, directly reducing the food, shelter and breeding sites available to them, which can reduce population sizes and, for specialist species unable to survive elsewhere, cause local extinction, reducing species diversity. It also fragments the remaining forest into smaller, more isolated patches; this reduces gene flow between separated populations (increasing inbreeding and reducing genetic diversity), and creates a larger proportion of 'edge habitat' exposed to different conditions (e.g. more light, wind, drier air), which many interior forest species cannot tolerate, further reducing biodiversity.
Final answer: (a) 20%; (b) direct habitat loss (reducing populations/causing local extinctions) and habitat fragmentation (reducing gene flow/genetic diversity and increasing edge effects).

(a) Correct equation [1]; correct answer 20% [1].
(b) One valid, well-explained mechanism (e.g. habitat loss reducing populations/causing extinction) [1–2 depending on development]; a second valid, well-explained mechanism (e.g. fragmentation reducing gene flow/genetic diversity) [1–2 depending on development]; max 3 marks.

(a) \( \text{percentage frequency} = \frac{\text{number of quadrats in which species is present}}{\text{total number of quadrats sampled}} \times 100\% = \frac{35}{50} \times 100\% = 70\% \)
Check by a second route: 70% of 50 = \( 0.70 \times 50 = 35 \), matching the given number of quadrats in which clover was present.
(b) Random sampling avoids the investigator (consciously or unconsciously) choosing sampling points that are unrepresentative (e.g. areas that look particularly clover-rich or convenient to reach), so the result obtained is more likely to give a reliable, unbiased estimate of the percentage frequency across the whole field.
(c) Clover leaves are held up on relatively long petioles (leaf stalks), allowing the leaves to be positioned above the immediate ground layer and closer to available light, helping the plant to compete with the shading effect of taller grass leaves.
Final answer: (a) 70%; (b) random sampling avoids bias, giving a more representative estimate; (c) e.g. long petioles raising leaves towards the light.

(a) Correct equation [1]; correct answer 70% [1].
(b) Valid reason referring to avoiding bias/unrepresentative sampling [1].
(c) Valid adaptation named [1]; correct explanation of how it aids competition for light [1].

\( N = 25+15+8+2 = 50 \)
\( \Sigma \left( \frac{n}{N} \right)^2 = \left(\frac{25}{50}\right)^2 + \left(\frac{15}{50}\right)^2 + \left(\frac{8}{50}\right)^2 + \left(\frac{2}{50}\right)^2 \)
\( = (0.50)^2 + (0.30)^2 + (0.16)^2 + (0.04)^2 = 0.2500 + 0.0900 + 0.0256 + 0.0016 = 0.3672 \)
\( D = 1 - 0.3672 = 0.6328 \approx 0.63 \)
Check by a second route: since D must lie between 0 (no diversity) and just under 1 (maximum diversity), and one species (limpet) makes up half of all individuals here, a moderate value of D such as 0.63 (rather than a value very close to 0 or to 1) is consistent with this degree of dominance.
Final answer: D ≈ 0.63.

Correct calculation of N = 50 [1]; correct calculation of Σ(n/N)² = 0.3672 (or equivalent working shown) [1]; correct final subtraction, D = 1 − 0.3672 [1]; correct final answer 0.63 (accept 0.6 to 0.633) [1].

\( \text{volume of water taken up} = \pi r^2 h = \pi \times (0.5)^2 \times 48 = \pi \times 0.25 \times 48 = \pi \times 12 = 37.70\ \text{mm}^3 \)
\( \text{rate} = \frac{\text{volume}}{\text{time}} = \frac{37.70}{2} = 18.85\ \text{mm}^3\,\text{min}^{-1} \)
Check by a second route: over 2 minutes at 18.85 mm³ min⁻¹, volume = \( 18.85 \times 2 = 37.70 \) mm³, and dividing by \( \pi r^2 = \pi \times 0.25 = 0.785 \) mm² gives a distance of \( 37.70/0.785 = 48.0 \) mm, matching the given bubble movement.
Final answer: rate of water uptake ≈ 18.85 mm³ per minute (this measures water uptake by the shoot, not transpiration directly, since some water is used within the plant and is not necessarily lost by transpiration).

Correct calculation of volume using \( \pi r^2 h \) [1]; correct substitution of values (r = 0.5 mm, h = 48 mm) [1]; correct volume 37.70 mm³ (or equivalent) [1]; correct final rate 18.85 mm³ min⁻¹ [1].

Indicative content:
• Transpiration is the loss of water vapour from the aerial parts of a plant (mainly through stomata in the leaves) by evaporation.
• Water evaporates from the moist cell walls of mesophyll cells into the intercellular air spaces within the leaf; water vapour then diffuses out through open stomata, down a concentration gradient, into the surrounding air, which is normally less humid than the air spaces inside the leaf.
• This creates the transpiration pull: as water is lost, water potential in the mesophyll cells falls, drawing water by osmosis from adjacent cells, and ultimately from the xylem in the leaf veins; because water molecules are cohesive (hydrogen bonding between water molecules) and adhesive (attraction to the walls of the xylem vessels), a continuous, unbroken column of water can be pulled up the xylem from the roots (the cohesion-tension theory).
• Light intensity: higher light intensity causes stomata to open wider (to allow CO2 in for photosynthesis), increasing the rate of water vapour loss.
• Temperature: higher temperature increases the kinetic energy of water molecules, increasing the rate of evaporation from cell surfaces, and increases the water vapour concentration gradient between the leaf's air spaces and the outside air, increasing transpiration rate.
• Humidity: lower relative humidity of the surrounding air produces a steeper water vapour concentration gradient between the leaf's air spaces and the outside air, increasing the rate of diffusion of water vapour out of the stomata; high humidity reduces the gradient and slows transpiration.
• Wind speed: greater air movement removes the layer of humid air that builds up around the leaf surface, maintaining a steep concentration gradient and increasing transpiration rate; still air allows humid air to accumulate, reducing the gradient and slowing transpiration.
Final answer: a comprehensive account of evaporation from mesophyll cells, diffusion of water vapour out via stomata, its link to the cohesion-tension mechanism pulling water up the xylem, and how light, temperature, humidity and wind speed each affect the rate by their effect on stomatal aperture and/or the water vapour concentration gradient.

Level 3 (7–9 marks): Accurate, detailed description of transpiration (evaporation from mesophyll cell walls, diffusion of water vapour through stomata) with correct reference to the transpiration pull/cohesion-tension mechanism; clear, well-developed explanation of at least three environmental factors (light, temperature, humidity, wind), correctly linked to stomatal aperture and/or the concentration gradient; fluent, accurate use of specialist terms.
Level 2 (4–6 marks): Reasonable description of transpiration with some correct detail; at least two environmental factors explained with some correct reasoning, though possibly incomplete; appropriate use of some specialist terms.
Level 1 (1–3 marks): Basic, list-like points about transpiration and/or factors affecting it, with little explanation of mechanism; limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

Indicative content:
• Habitat destruction: e.g. deforestation for agriculture or urban development directly destroys the habitat of many species, reducing available food/shelter/breeding sites and leading to population decline or local extinction.
• Habitat fragmentation: breaking large habitats into smaller, isolated patches (e.g. by roads or agricultural land) reduces gene flow between populations, increasing inbreeding and reducing genetic diversity, and increases vulnerability to edge effects.
• Pollution: e.g. agricultural nitrate/phosphate run-off causing eutrophication of waterways, leading to algal blooms, oxygen depletion and the death of aquatic organisms; or pesticide use directly killing non-target species (e.g. insecticides killing pollinators) or bioaccumulating through food chains.
• Overexploitation: e.g. overfishing reducing fish stocks faster than they can reproduce, or overhunting, directly reducing population sizes and potentially causing extinction.
• Introduction of non-native/invasive species: e.g. a non-native species introduced (deliberately or accidentally) may outcompete native species for resources, or prey on them, reducing native biodiversity.
• Climate change (from the burning of fossil fuels, increasing atmospheric CO2): changing temperature and rainfall patterns can alter/destroy habitats faster than some species can adapt or migrate, reducing biodiversity.
Final answer: named specific mechanisms (habitat destruction/fragmentation, pollution/eutrophication, overexploitation, invasive species, climate change) each linked with a plausible worked example and to a reduction in species number/genetic diversity.

Level 3 (5–6 marks): At least three distinct human activities/mechanisms explained accurately, each with a valid named example, clearly linked to a reduction in biodiversity; fluent use of specialist terms.
Level 2 (3–4 marks): At least two mechanisms explained with reasonable accuracy and example(s), though possibly with some lack of development; appropriate use of some specialist terms.
Level 1 (1–2 marks): Basic, undeveloped point(s) made (e.g. "pollution reduces biodiversity") without clear mechanism or example; limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

Add an equal volume of Benedict's reagent (blue) to the food sample (in solution) in a test tube, then heat the tube in a boiling water bath for about 5 minutes. A positive result for a reducing sugar is shown by a colour change from blue, through green, yellow and orange, to a brick-red precipitate, with the exact colour reached depending on the concentration of reducing sugar present.
Final answer: heat sample with Benedict's reagent in a boiling water bath; positive result = colour change from blue towards green/yellow/orange/brick-red precipitate.

Correct method — add Benedict's reagent and heat in a boiling water bath [1]; correct positive result — colour change from blue to green/yellow/orange/brick-red (precipitate) [1].

Add an equal volume of sodium hydroxide solution to the food sample, followed by a few drops of dilute copper(II) sulfate solution, and mix gently (this test does not require heating). A positive result for protein is shown by a colour change from blue to purple/violet/lilac; if no protein is present, the mixture remains blue.
Final answer: add NaOH then dilute CuSO4 solution, mix (no heating); positive result = blue to purple/violet colour change.

Correct method — sodium hydroxide followed by dilute copper(II) sulfate, mixed without heating [1]; correct positive result — blue to purple/violet colour change [1].

Add a few drops of iodine solution (iodine dissolved in potassium iodide solution) directly to the food sample, at room temperature. A positive result for starch is shown by a colour change from the orange/yellow-brown colour of the iodine solution to a blue-black colour.
Final answer: add iodine (in potassium iodide) solution; positive result = orange-brown to blue-black colour change.

Correct method — add iodine (in potassium iodide) solution directly to the sample [1]; correct positive result — orange/brown to blue-black colour change [1].

Mix (shake) the food sample thoroughly with ethanol, allow any solid debris to settle, and then pour the ethanol layer into a test tube of water. A positive result for lipid is shown by the formation of a cloudy white emulsion in the water, because lipid dissolved in the ethanol comes out of solution as tiny droplets when added to water, scattering light.
Final answer: shake sample with ethanol, then add to water; positive result = cloudy white emulsion.

Correct method — mix/shake with ethanol, then add the ethanol layer to water [1]; correct positive result — cloudy white emulsion forms [1].

The description — an organelle bound by a double membrane, with the inner membrane folded into finger-like projections (cristae) extending into a granular interior (the matrix) — identifies this organelle as a mitochondrion. Its main function is to be the site of the later stages of aerobic respiration (the Krebs cycle and oxidative phosphorylation), producing most of the cell's ATP.
Final answer: mitochondrion; site of aerobic respiration/ATP production.

Correct identification — mitochondrion [1]; correct function — site of aerobic respiration/ATP production [1].

The description — a double-membrane-bound organelle containing stacks of flattened membranous discs (thylakoids, forming grana), embedded within the fluid stroma — identifies this organelle as a chloroplast. The dense round granules visible between the grana stacks are likely to be starch grains, representing temporary storage of the products of photosynthesis (glucose polymerised to starch) within the chloroplast.
Final answer: chloroplast; starch grains.

Correct identification — chloroplast [1]; correct identification of granules — starch grains [1].

\( \text{value of 1 eyepiece unit} = \frac{\text{number of stage micrometer divisions spanned} \times \text{value of 1 stage division}}{\text{number of eyepiece units spanned}} \)
\( = \frac{25 \times 10\ \mu\text{m}}{10} = \frac{250\ \mu\text{m}}{10} = 25\ \mu\text{m} \)
Check by a second route: 10 eyepiece units at 25 μm each = \( 10 \times 25 = 250\ \mu\text{m} \), and this should equal 25 stage divisions × 10 μm = 250 μm, which matches.
Final answer: 1 eyepiece unit = 25 μm (at ×40 objective magnification).

Correct method (total stage-micrometer distance spanned divided by number of eyepiece units) [1]; correct substitution [1]; correct answer 25 μm [1].

(a) \( \text{actual width} = \text{number of eyepiece units} \times \text{value of 1 eyepiece unit} = 6 \times 25\ \mu\text{m} = 150\ \mu\text{m} \)
Check by a second route: \( 150 / 25 = 6 \) eyepiece units, which matches the measurement given.
(b) Changing the objective lens changes the total magnification of the image, so the actual distance represented by each eyepiece graticule unit also changes (a higher-power objective magnifies the image more, so each eyepiece unit then represents a smaller actual distance). The calibration value obtained is therefore only valid for the specific objective lens used during calibration, and a fresh calibration is needed for any other objective lens.
Final answer: (a) 150 μm; (b) because the value of 1 eyepiece unit (in μm) depends on the magnification, which changes with the objective lens used.

(a) Correct method (eyepiece units × calibration value) [1]; correct answer 150 μm [1].
(b) Correct explanation that magnification (and therefore the actual distance per eyepiece unit) changes with the objective lens used [1].

Precautions include: (1) carefully aligning the zero line of the eyepiece graticule scale exactly with a line on the stage micrometer scale before starting to count divisions, to avoid a systematic offset error; and (2) taking the count of eyepiece units and stage micrometer divisions between two points where lines on both scales clearly coincide (rather than judging by eye partway along), and ideally repeating the calibration and calculating a mean value, to reduce random measurement error.
Final answer: any two valid precautions, e.g. careful zero-alignment of the two scales; taking readings between two points of exact coincidence; repeating and averaging.

Any two valid precautions relating to accurate calibration (e.g. careful alignment of zero lines, reading between two points of coincidence, repeating/averaging) [1 mark each, max 2].

\( \text{actual length} = 3.6 \times 25\ \mu\text{m} = 90\ \mu\text{m} \)
Check by a second route: \( 90 / 25 = 3.6 \) eyepiece units, matching the measurement given.
Final answer: 90 μm.

Correct method (eyepiece units × calibration value) [1]; correct answer 90 μm [1].

Two long tape measures are laid out at right angles along two adjacent sides of the field, to act as x and y coordinate axes. A random number generator (or random number table) is used to generate pairs of coordinates (an x value and a y value, each within the range of the tape measures). At each pair of coordinates, the quadrat is placed on the ground and the percentage cover of the target species within the quadrat is estimated (e.g. by eye against a standard percentage cover scale, or, more precisely, by counting how many of the smaller grid squares within a gridded quadrat contain the species and converting to a percentage). This process is repeated at a large number of randomly generated points (e.g. at least 20–30), and the percentage cover values obtained are averaged to give a reliable estimate of mean percentage cover for the field, while random sampling avoids bias in the choice of sampling positions.
Final answer: use random coordinates (from tape measures and random numbers) to position the quadrat at many random points, estimate percentage cover at each, and calculate the mean.

Correct method for generating random sampling points (e.g. tape measures as axes with random number coordinates) [1]; correct method for estimating percentage cover within a quadrat [1]; correct reference to using a sufficiently large sample of quadrats (repeats) [1]; correct final step — calculating a mean [1].

A measuring tape is laid out in a straight line, starting at the edge of the salt marsh, running perpendicular to the shoreline, so that it crosses the environmental gradient of interest (e.g. from more frequently tidally inundated to less frequently inundated ground). A quadrat is then placed at regular, fixed intervals along the tape (e.g. every 1–2 m, or a continuous belt using adjoining quadrats), and the percentage cover of the target species is recorded within each quadrat at each interval. To improve reliability, this procedure should be repeated using further transect lines at different, randomly chosen points along the length of the salt marsh, and the mean percentage cover at each distance calculated across the replicate transects. The resulting data (percentage cover plotted against distance from the marsh edge) shows how the abundance of the species changes along the environmental gradient.
Final answer: lay a transect line perpendicular to the gradient, sample percentage cover with a quadrat at regular intervals along it, and repeat with further transects (replicates) to obtain reliable mean values at each distance.

Correct description of laying the transect line perpendicular to the environmental gradient [1]; correct method of sampling at regular intervals along the transect using a quadrat [1]; correct reference to recording percentage cover at each interval [1]; correct reference to repeating with further transects/replicates for reliability [1].

(a) \( \text{population estimate} = \frac{n_1 \times n_2}{m_2} \), where \( n_1 \) = number marked and released initially, \( n_2 \) = total number caught in the second sample, \( m_2 \) = number of marked individuals recaptured in the second sample.
\( = \frac{40 \times 50}{8} = \frac{2000}{8} = 250 \)
Check by a second route: if the true population were 250, and 40 (a fraction 40/250 = 0.16) were marked, then in a second sample of 50 the expected number marked would be \( 0.16 \times 50 = 8 \), which matches the number recaptured.
(b) Any one valid assumption, e.g.: the population is closed (no significant immigration, emigration, births or deaths between sampling occasions); marked individuals redistribute themselves randomly throughout the population before the second sample is taken; marking does not affect an individual's survival, behaviour or chance of being recaptured; marks are not lost between sampling occasions.
Final answer: (a) 250; (b) any one valid assumption as above.

(a) Correct equation (Lincoln index) [1]; correct substitution [1]; correct answer 250 [1].
(b) Valid assumption stated [1].

(a) Potassium hydroxide solution absorbs the carbon dioxide gas produced by the respiring seeds as it is released. This means that any decrease in the volume (or pressure) of gas within the respirometer chamber that is measured can be attributed solely to the oxygen being consumed by the seeds during respiration, rather than being partly masked by carbon dioxide being released at the same time.
(b) The control tube, set up identically but containing inert glass beads (of similar size/thermal mass to the seeds) instead of respiring seeds, allows the student to detect and correct for any change in gas volume caused by changes in external conditions (e.g. atmospheric temperature or pressure) during the experiment, rather than by respiration itself. Any movement of the manometer fluid in the control tube is due to these external factors and can be subtracted from the reading in the experimental tube, isolating the change due to respiration alone.
Final answer: (a) absorbs CO2 released by respiration, so that gas volume change reflects O2 uptake only; (b) corrects for volume changes due to temperature/pressure changes rather than respiration.

(a) Correct reference to absorbing carbon dioxide [1]; correct explanation that this isolates the effect of oxygen consumption on the volume/pressure change [1].
(b) Correct reference to controlling for changes in temperature/pressure [1]; correct explanation that this allows correction of the experimental reading [1].

\( \text{volume of O}_2 \text{ consumed} = \text{cross-sectional area} \times \text{distance moved} = 0.05 \times 1.5 = 0.075\ \text{cm}^3 \)
\( \text{rate} = \frac{\text{volume}}{\text{time}} = \frac{0.075}{3} = 0.025\ \text{cm}^3\,\text{min}^{-1} \)
Check by a second route: over 3 minutes at 0.025 cm³ min⁻¹, volume = \( 0.025 \times 3 = 0.075 \) cm³, and dividing by the cross-sectional area (0.05 cm²) gives a distance of \( 0.075/0.05 = 1.5 \) cm, matching the given fluid movement.
Final answer: rate of oxygen consumption = 0.025 cm³ min⁻¹ (25 mm³ min⁻¹).

Correct equation (volume = cross-sectional area × distance moved) [1]; correct volume 0.075 cm³ [1]; correct method (volume ÷ time) [1]; correct final answer 0.025 cm³ min⁻¹ (or equivalent, e.g. 25 mm³ min⁻¹) [1].

(a) The result at 30°C (14 units) is anomalous, as it is lower than both neighbouring readings (17 units at 20°C and 35 units at 40°C) and breaks the otherwise smooth, continuously rising trend expected up to the optimum. Excluding this point, the trend would show a smooth, continuous increase in rate from 10°C to 40°C (rather than the dip seen at 30°C), followed by a sharp decrease from 40°C to 60°C.
(b) The highest recorded rate (35 units) occurs at 40°C, and the rate falls sharply above this temperature, so the optimum temperature is approximately 40°C. Above the optimum, rising temperature causes increasing vibration of the atoms within the enzyme molecule; this breaks the hydrogen bonds (and other bonds, e.g. ionic and hydrophobic interactions) that hold the enzyme's tertiary structure — including the precise shape of the active site — in place. This changes (denatures) the shape of the active site so that it is no longer complementary to the substrate and can no longer bind (or effectively catalyse) the reaction, so the rate falls sharply as temperature rises further above the optimum.
Final answer: (a) 30°C is anomalous; excluding it gives a smooth rising trend to the peak; (b) optimum ≈ 40°C; above this, denaturation of the enzyme's tertiary structure (including the active site) causes the sharp fall in rate.

(a) Correct identification of the anomalous result — 30°C (14 units) [1]; correct description of the expected corrected trend (smooth continuous rise to the peak) [1].
(b) Correct optimum temperature identified (≈40°C, from the highest rate recorded) [1]; correct explanation of denaturation — increased vibration breaking bonds maintaining tertiary/active site structure, so substrate can no longer bind [1].

(a) A correlation between two variables shows only that they vary together (are associated); it does not, on its own, demonstrate that a change in one variable (light intensity) directly causes the change in the other (moss cover). It remains possible that a third, confounding variable, which happens to also vary alongside light intensity across the sampling points, is actually responsible for the change in moss cover, or that both variables are affected by some other underlying factor.
(b) A plausible confounding factor is substrate/soil moisture (or relative humidity). In a woodland, areas with high light intensity are often gaps in the canopy, which also tend to be drier (more exposed to direct sun and wind, less sheltered) and less humid than shaded areas beneath a dense canopy. Since mosses lack a waterproof cuticle and true roots, and rely on a consistently moist surface for water uptake and for reproduction, it may be this lower moisture/humidity associated with higher light intensity, rather than the light intensity itself, that is the true cause of reduced moss cover.
Final answer: (a) correlation shows association, not causation, as a confounding variable could be responsible; (b) e.g. substrate moisture/humidity, which tends to be lower in high-light gaps and to which moss growth is very sensitive.

(a) Correct statement that correlation shows association, not causation [1]; correct reference to a possible confounding variable/other explanation [1].
(b) Valid confounding factor named (e.g. moisture/humidity) [1]; correct explanation of how it could plausibly vary with light intensity and affect moss cover [1].