2023 CCEA AS-Level Biology 1010 模擬試題連答案詳解

(a) Magnesium is a component of chlorophyll.
(b) Iron is a component of haemoglobin (in the haem prosthetic group).
Final answer: (a) chlorophyll; (b) haemoglobin.

(a) Chlorophyll [1]. (b) Haemoglobin [1].

Primary structure is the specific sequence of amino acids in a polypeptide chain, linked together by peptide bonds. Secondary structure is the way the polypeptide chain folds locally, forming an α-helix or β-pleated sheet, held in shape by hydrogen bonds between different parts of the chain.
Final answer: as stated above.

Correct description of primary structure (amino acid sequence, peptide bonds) [1]; correct description of secondary structure (α-helix/β-pleated sheet, hydrogen bonds) [1].

(a) Active transport — movement against a concentration gradient, using a carrier protein and requiring energy from ATP.
(b) Simple diffusion — lipid-soluble molecules can pass directly through the phospholipid bilayer, down their concentration gradient, without a carrier protein.
Final answer: (a) active transport; (b) (simple) diffusion.

(a) Active transport [1]. (b) (Simple) diffusion [1].

Phagocytosis is the engulfing of solid particles or material (for example, bacteria) by the cell surface membrane, forming a phagocytic vesicle. Pinocytosis is the engulfing of liquid or dissolved substances by the cell surface membrane, forming small vesicles.
Final answer: phagocytosis — solid material; pinocytosis — liquid/dissolved material.

Correct description of phagocytosis (solid particles) [1]; correct description of pinocytosis (liquid/dissolved substances) [1].

Water potential (ψ) is a measure of the tendency of water molecules to move out of a solution (or cell) by osmosis, measured in units of pressure (e.g. kPa). Pure water at atmospheric pressure is defined as having a water potential of zero; water moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential.
Final answer: as stated above.

Correct core definition — tendency of water to move by osmosis [1]; correct reference to pure water = 0 and/or units of pressure [1].

Structural differences include: a bacteriophage contains DNA as its genetic material, bounded only by a protein coat, whereas HIV contains RNA, bounded by a protein coat and additionally surrounded by a lipid bilayer envelope (containing glycoprotein); HIV also contains the enzyme reverse transcriptase, which a phage does not.
Final answer: any two valid differences, e.g. nucleic acid type (RNA vs DNA), presence of a lipid envelope, presence of reverse transcriptase.

Any two valid structural differences (e.g. RNA vs DNA; lipid envelope present/absent; reverse transcriptase present/absent) [1 mark each, max 2].

HIV invades helper T-cells (a type of lymphocyte). This weakens the immune system, since these cells become progressively depleted/destroyed as the virus replicates within them.
Final answer: helper T-cells; weakens the immune system.

Correct host cell (helper T-cell/lymphocyte) [1]; correct effect (weakens the immune system) [1].

Phages invade bacterial cells and use the host cell's machinery to replicate, producing many new phage particles inside the cell. This buildup of new phages eventually causes the bacterial cell to burst open (lyse), destroying it and releasing the new phages to infect further bacterial cells.
Final answer: replication inside the host produces many new phages, which causes the bacterial cell to burst (lyse).

Correct description of replication inside the host cell producing many new phages [1]; correct description of the cell bursting/being destroyed (lysis) as a result [1].

Goblet cells secrete mucus, which lubricates and protects the epithelium as food passes over it. Microvilli are folds of the cell surface membrane of the epithelial cells that greatly increase the surface area available for absorption of digested food.
Final answer: goblet cells — secrete mucus; microvilli — increase surface area for absorption.

Correct role of goblet cells (secrete mucus) [1]; correct role of microvilli (increase surface area for absorption) [1].

The palisade mesophyll consists of tightly packed, column-shaped cells containing many chloroplasts, positioned near the upper surface of the leaf for maximum absorption of light for photosynthesis. The spongy mesophyll consists of loosely arranged, irregularly shaped cells with air spaces between them, which allow diffusion of gases (CO2 and O2) to and from the palisade cells and the stomata.
Final answer: as stated above.

Correct role of palisade mesophyll (light absorption for photosynthesis, many chloroplasts) [1]; correct role of spongy mesophyll (air spaces for gas diffusion) [1].

Prokaryotic cells possess naked, circular DNA that is not enclosed within a nuclear envelope (unlike the linear, histone-associated chromosomes enclosed in a nucleus in eukaryotic cells), and may also possess plasmids (small circular loops of additional DNA), which are not found in typical eukaryotic cells.
Final answer: naked circular DNA (no nuclear envelope); plasmids.

Any two valid features, e.g. naked/circular DNA not enclosed in a nuclear envelope; plasmids [1 mark each, max 2].

Magnification is the number of times larger an image is compared with the actual (real) size of the object. Resolution is the ability to distinguish between two points that are very close together as being two separate points, rather than one blurred point (i.e. the minimum distance apart that two points can be while still being seen as distinct).
Final answer: as stated above.

Correct definition of magnification (image size relative to actual size) [1]; correct definition of resolution (ability to distinguish two close points as separate) [1].

A cofactor is a non-protein component that some enzymes require, in addition to their protein structure, in order to function/catalyse their reaction (for example, by helping to bind the substrate or by taking part in the reaction). Example: a metal ion, such as Zn²⁺ or Cl⁻.
Final answer: cofactor — a non-protein component required for enzyme activity; example — a metal ion such as Zn²⁺.

Correct role of a cofactor (non-protein component required for the enzyme to function) [1]; valid example given [1].

DNA replication occurs during the S (synthesis) phase of interphase.
Final answer: S phase.

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

During peptide bond formation, the carboxyl (–COOH) group of one amino acid reacts with the amino (–NH2) group of a second amino acid; a molecule of water is released, and a covalent peptide bond forms between the two amino acids — this is a condensation reaction.
During digestion, this peptide bond is broken by hydrolysis: a molecule of water is added across the bond (catalysed by protease/peptidase enzymes), splitting the dipeptide back into its two separate amino acids.
Final answer: condensation (release of water, forms peptide bond) for synthesis; hydrolysis (addition of water, enzyme-catalysed) for breakdown.

Correct description of condensation reaction (reaction between –COOH and –NH2 groups) [1]; correct reference to release of water and formation of the peptide bond [1]; correct description of hydrolysis (addition of water) breaking the bond [1]; correct reference to enzyme catalysis (protease/peptidase) during digestion [1].

Because the two antiparallel strands of DNA have complementary base pairing (A pairs with T, C pairs with G), each strand can act as a template for the synthesis of a new, complementary strand. During replication, DNA helicase unwinds and separates the two strands of the double helix; DNA polymerase then synthesises a new complementary strand of nucleotides alongside each of the two original (template) strands, using the base-pairing rules to ensure the new strand is complementary. This produces two new DNA double helices, each containing one original (conserved) strand and one newly synthesised strand — this is why the process is described as semi-conservative.
Final answer: as stated above.

Correct explanation that complementary base pairing allows each strand to act as a template [1]; correct description of DNA helicase unwinding the strands and DNA polymerase synthesising new complementary strands [1]; correct explanation that each new molecule retains one original strand, making it semi-conservative [1].

Since the solution has a lower (more negative) water potential than the cell, water moves out of the cell, by osmosis, down the water potential gradient, through the partially permeable cell surface membrane. As water is lost, the cytoplasm and cell surface membrane shrink and pull away from the cellulose cell wall — the cell becomes flaccid and then plasmolysed (at incipient plasmolysis, the membrane has just started to pull away from the wall at the corners).
Final answer: water leaves the cell by osmosis down the water potential gradient, causing the cell to become plasmolysed (membrane pulling away from the cell wall).

Correct direction of water movement (out of the cell, by osmosis) [1]; correct reference to moving down the water potential gradient [1]; correct description of the resulting appearance (membrane pulling away from the wall / plasmolysis) [1].

Facilitated diffusion requires specific carrier or channel proteins in the cell membrane, whereas simple diffusion does not require any protein and occurs directly through the phospholipid bilayer. Facilitated diffusion is used to transport larger or water-soluble/polar substances (for example glucose or ions), which cannot easily pass through the hydrophobic phospholipid bilayer, whereas simple diffusion is used for small or lipid-soluble substances (for example O2 or CO2). Both processes, however, move substances down their concentration gradient and do not require metabolic energy from ATP.
Final answer: as stated above.

Correct distinction regarding protein involvement (facilitated diffusion uses carrier/channel proteins, simple diffusion does not) [1]; correct distinction regarding the type of substance transported [1]; correct point that both are passive processes not requiring ATP [1].

HIV is a retrovirus containing RNA as its genetic material. Once inside the host helper T-cell, the enzyme reverse transcriptase (carried by the virus) uses the viral RNA as a template to synthesise a complementary strand of viral DNA. This viral DNA is then used within the host cell to direct the production of new viral components (proteins and RNA), which are assembled into new HIV particles; these new viruses are eventually released, destroying the host helper T-cell in the process.
Final answer: as stated above.

Correct role of reverse transcriptase (RNA used as a template to synthesise viral DNA) [1]; correct link to the viral DNA directing production of new viral components [1]; correct outcome (new virus particles produced, host cell destroyed) [1].

Villi are finger-like projections of the mucosa that greatly increase the surface area of the ileum in contact with digested food, increasing the rate of absorption. Each villus contains a network of blood capillaries, which absorb amino acids and monosaccharides (and carry them away, maintaining a diffusion/concentration gradient), and a lacteal, which absorbs fats. The columnar epithelial cells lining the villi also have microvilli on their surface, further increasing the surface area for absorption, and contain numerous mitochondria, which provide ATP for the active transport of nutrients.
Final answer: as stated above.

Correct explanation that villi increase surface area for absorption [1]; correct description of blood capillaries and lacteals absorbing specific products of digestion [1]; correct reference to microvilli and/or mitochondria further aiding absorption [1].

According to the induced-fit hypothesis, the active site of the enzyme is not a rigid, fixed shape but is flexible. When the substrate binds to the active site, the active site changes shape slightly, moulding itself more closely around the substrate to form an enzyme-substrate complex. This improved fit puts strain on the bonds within the substrate, lowering the activation energy needed for the reaction to proceed, and so speeds up the rate of reaction compared with an uncatalysed reaction.
Final answer: as stated above.

Correct description of the active site changing shape to fit the substrate [1]; correct reference to the formation of an enzyme-substrate complex [1]; correct link to lowering the activation energy and increasing reaction rate [1].

(a) \( R_f = \frac{\text{distance travelled by spot}}{\text{distance travelled by solvent front}} = \frac{4.8}{6.0} = 0.80 \)
(b) The unknown amino acid's Rf value (0.80) matches the known Rf value for leucine (0.80) under the same conditions, so it is likely to be leucine.
Check by a second route: \( 0.80 \times 6.0 = 4.8 \) cm, which matches the given distance travelled by the spot.
Final answer: Rf = 0.80; likely to be leucine.

(a) Correct equation \( R_f = \text{distance moved by spot}/\text{distance moved by solvent} \) [1]; correct answer 0.80 (no unit) [1].
(b) Correct conclusion (likely leucine) with valid reason (matching Rf values) [1].

\( \text{percentage change in mass} = \frac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100\% \)
\( = \frac{4.68 - 5.20}{5.20} \times 100\% = \frac{-0.52}{5.20} \times 100\% = -10.0\% \)
Check by a second route: a 10.0% decrease from 5.20 g is \( 5.20 \times 0.10 = 0.52 \) g, and \( 5.20 - 0.52 = 4.68 \) g, which matches the given final mass exactly.
Final answer: −10.0% (a decrease), indicating the solution had a lower water potential than the potato cells, so water left the cells by osmosis.

Correct equation for percentage change [1]; correct substitution [1]; correct answer −10.0% (sign/decrease clearly indicated) [1].

(a) Convert to consistent units: 10 mm = 10 000 μm.
\( \text{magnification} = \frac{\text{scale bar length on image}}{\text{actual length represented}} = \frac{10\,000}{2} = 5000 \) (i.e. ×5000)
(b) \( \text{actual length} = \frac{\text{image length of mitochondrion}}{\text{magnification}} = \frac{15 \text{ mm}}{5000} = \frac{15\,000 \text{ μm}}{5000} = 3 \text{ μm} \)
Check by a second route: using the scale bar ratio directly, \( 15 \text{ mm} \times \frac{2\ \mu\text{m}}{10\ \text{mm}} = 15 \times 0.2 = 3\ \mu\text{m} \), which agrees; this is also a physically realistic size for a mitochondrion (typically 1–10 μm long).
Final answer: magnification = ×5000; actual length = 3 μm.

(a) Correct unit conversion and method [1]; correct answer ×5000 [1].
(b) Correct answer 3 μm with unit [1]. Accept ecf from (a).

This description — chromosomes condensing and becoming visible, nuclear envelope still intact, centrioles moving to opposite poles — corresponds to prophase, the first stage of mitosis.
Final answer: prophase.

Correct answer — prophase [1].

Mitosis produces two genetically identical daughter cells, each with the same chromosome number as the parent cell, so each daughter cell would contain 8 chromosomes.
Final answer: 8 chromosomes.

Correct answer — 8 [1].

Indicative content:
• Proteins are made of amino acid monomers, joined by peptide bonds formed in condensation reactions, to form polypeptide chains (primary structure).
• Secondary structure: hydrogen bonding between parts of the chain forms an α-helix or β-pleated sheet.
• Tertiary structure: further folding of the polypeptide, stabilised by hydrogen bonds, ionic bonds, disulfide bonds and hydrophobic interactions, giving the molecule its overall 3D shape.
• Quaternary structure: some proteins consist of more than one polypeptide chain associated together.
• Collagen (fibrous protein): several polypeptide chains wound together into a long, rope-like triple helix; this gives it high tensile strength, suited to its structural role (e.g. in tendons, skin, blood vessel walls) where strength rather than a specific binding site is required.
• An enzyme (globular protein): the polypeptide chain folds into a compact, roughly spherical (globular) shape, with the specific tertiary structure creating a precisely shaped active site; this specific 3D shape allows only a complementary substrate to bind, giving the enzyme its specificity for catalysing a particular reaction.
• Overall, the shape of a protein (determined by its sequence of amino acids and the resulting pattern of bonding) directly determines its function — a long, strong fibrous shape for structural roles, or a precisely folded globular shape with an active site for catalytic roles.
Final answer: a well-organised account covering the levels of protein structure and clearly relating collagen's fibrous shape to its structural function and an enzyme's globular shape (with active site) to its catalytic function.

Level 3 (6–7 marks): Clear, accurate account of primary, secondary, tertiary (and quaternary where relevant) protein structure; both collagen and an enzyme correctly linked to their structure (fibrous/triple helix → strength; globular/active site → specificity); answer well organised with fluent, accurate use of specialist terms.
Level 2 (3–5 marks): Reasonable account of protein structure with some detail missing; both examples mentioned but the structure-function link is only partially explained; mostly appropriate use of specialist terms.
Level 1 (1–2 marks): Basic, list-like points made about protein structure and/or the examples, with little clear structure-function linkage; limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

Indicative content:
• Mitosis follows the replication of DNA during the S phase of interphase, and is nuclear division producing two daughter nuclei that are genetically identical to the parent nucleus and to each other.
• Prophase: chromosomes condense and become visible as two identical sister chromatids joined at a centromere; the nuclear envelope begins to break down; centrioles move to opposite poles and spindle fibres begin to form.
• Metaphase: chromosomes line up individually along the equator (metaphase plate) of the spindle, attached to spindle fibres at their centromeres.
• Anaphase: the centromeres divide and the sister chromatids are pulled apart to opposite poles of the cell by the shortening of the spindle fibres.
• Telophase: chromatids (now individual chromosomes) arrive at the poles and uncoil; a nuclear envelope reforms around each set, forming two genetically identical nuclei.
• Cytokinesis then divides the cytoplasm, forming two separate daughter cells (differing slightly in mechanism between animal cells, where the cell membrane pinches inwards, and plant cells, where a new cell wall forms).
• Significance: mitosis maintains genetic constancy, producing daughter cells genetically identical to the parent cell, which is essential for the growth of a multicellular organism (increasing cell number) and for the repair and replacement of damaged or worn-out tissues.
• Loss of control over the cell cycle (e.g. checkpoints not functioning correctly) can lead to uncontrolled cell division, which is linked to the development of cancer.
Final answer: a well-organised, chronological account of prophase, metaphase, anaphase and telophase, with a clear explanation of the significance of mitosis for growth, tissue repair and genetic constancy, and its link to cancer when uncontrolled.

Level 3 (6–8 marks): Accurate, ordered description of all four stages of mitosis (prophase, metaphase, anaphase, telophase) including chromosome behaviour; clear explanation of significance (genetic constancy, growth, repair) with reference to the cell cycle/cancer link; answer well organised with fluent, accurate use of specialist terms.
Level 2 (3–5 marks): Most stages of mitosis described with some correct detail, though possibly incomplete or with minor inaccuracies; some explanation of significance given; reasonably organised with appropriate use of some specialist terms.
Level 1 (1–2 marks): Only 1–2 stages or basic facts about mitosis given; little or no explanation of significance; limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

Xylem vessels transport water and dissolved mineral ions from the roots to the rest of the plant (they also provide structural support). Phloem sieve-tubes translocate (transport) organic solutes, such as sucrose, around the plant, in a two-way flow between sources and sinks.
Final answer: xylem — water/ion transport; phloem — translocation of organic solutes.

Correct role of xylem (water/mineral ion transport) [1]; correct role of phloem (translocation of organic solutes) [1].

(a) The standard hierarchy is Kingdom, Phylum, Class, Order, Family, Genus, Species — so Order lies directly between Class and Family.
(b) This two-part naming system (genus + species) is called binomial nomenclature.
Final answer: (a) Order; (b) binomial nomenclature.

(a) Correct answer — Order [1]. (b) Correct answer — binomial nomenclature/system [1].

Promotes biodiversity: e.g. crop rotation, hedgerow conservation and maintenance, or polyculture (growing a variety of crops), all of which create a greater variety of habitats and food sources.
Reduces biodiversity: e.g. monoculture (growing a single crop species over a large area) or the use of pesticides, both of which reduce the variety of habitats and food sources available and can directly kill non-target species.
Final answer: as stated above.

Valid practice that promotes biodiversity [1]; valid practice that reduces biodiversity [1].

Behavioural adaptation: e.g. huddling together in groups to reduce heat loss, or migrating to a warmer area, or hibernating during the coldest months.
Morphological (structural) adaptation: e.g. a thick layer of fur or blubber for insulation, or a compact body shape with a small surface area to volume ratio (and small extremities), reducing heat loss.
Final answer: as stated above.

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

(a) \( \text{surface area} = 6 \times (\text{side length})^2 = 6 \times 20^2 = 6 \times 400 = 2400\ \mu\text{m}^2 \)
(b) \( \text{volume} = (\text{side length})^3 = 20^3 = 8000\ \mu\text{m}^3 \)
\( \text{SA:V ratio} = \frac{2400}{8000} = 0.3 \) (i.e. 0.3:1)
Check by a second route: for a cube, SA:V simplifies to \( 6/\text{side length} = 6/20 = 0.3 \), which agrees.
Final answer: surface area = 2400 μm²; volume = 8000 μm³; SA:V ratio = 0.3:1.

(a) Correct surface area 2400 μm² with unit [1].
(b) Correct volume 8000 μm³ and correct SA:V ratio 0.3:1 [1].

\( \text{surface area} = 6 \times 40^2 = 6 \times 1600 = 9600\ \mu\text{m}^2 \)
\( \text{volume} = 40^3 = 64\,000\ \mu\text{m}^3 \)
\( \text{SA:V ratio} = \frac{9600}{64\,000} = 0.15 \) (i.e. 0.15:1)
Comparison: this ratio (0.15:1) is exactly half the ratio found for the smaller cell (0.3:1), showing that doubling the length of a cube halves its surface area to volume ratio — confirming that as size increases, the SA:V ratio decreases.
Check by a second route: using the simplified rule \( SA:V = 6/\text{side length} = 6/40 = 0.15 \), which agrees.
Final answer: SA:V = 0.15:1, half the ratio of the smaller cell.

Correct SA:V ratio 0.15:1 with correct working shown [1]; correct comparison to the smaller cell's ratio (half/decreased as size increases) [1].

As an organism's size increases, its surface area increases less than its volume, so its surface area to volume (SA:V) ratio decreases. A large animal therefore has a relatively small surface area compared with the volume of metabolically active tissue it must supply; simple diffusion across its outer body surface would be far too slow (and the diffusion distances too great) to supply enough oxygen and nutrients to all its cells, or remove waste products, quickly enough. Large animals therefore need specialised exchange surfaces with a greatly increased surface area (for example, lungs containing millions of alveoli) to compensate, together with a mass transport (circulatory) system to carry substances rapidly over the large distances involved, maintaining diffusion gradients at the exchange surfaces. A small, single-celled organism has a large SA:V ratio, so its entire outer surface provides more than enough area for its (relatively low) metabolic demand, and diffusion distances within the cell are very short, so no specialised exchange surface or transport system is needed.
Final answer: as stated above.

Correct explanation that SA:V ratio decreases as size increases [1]; correct explanation of why large animals need specialised exchange surfaces and mass transport as a result [1]; correct contrast with a small organism (large SA:V, short diffusion distances, no need for such systems) [1].

Blood enters the right atrium from the vena cava; it passes through the tricuspid (right atrioventricular) valve into the right ventricle; the right ventricle pumps it through the pulmonary artery to the lungs, where gas exchange occurs; oxygenated blood returns via the pulmonary vein to the left atrium; it passes through the bicuspid (left atrioventricular/mitral) valve into the left ventricle; the left ventricle then pumps it out through the aorta to the rest of the body.
Final answer: as stated above.

Correct right-side pathway (vena cava → right atrium → right ventricle → pulmonary artery) [1]; correct passage through the lungs to the left atrium via the pulmonary vein [1]; correct left-side pathway (left atrium → left ventricle → aorta), referencing at least one valve [1].

At the start of ventricular systole, the left ventricle contracts, which increases the pressure of the blood inside it. Once the pressure in the left ventricle rises above the pressure in the left atrium, this pressure difference pushes the cusps of the bicuspid valve upwards and closes them, preventing blood from flowing backwards into the left atrium as the ventricle continues to contract.
Final answer: as stated above.

Correct reference to ventricular contraction increasing ventricular pressure [1]; correct statement that ventricular pressure exceeds atrial pressure [1]; correct link to this pressure difference closing the valve to prevent backflow [1].

The SA node, located in the wall of the right atrium, acts as the heart's natural pacemaker: it generates a wave of electrical excitation that spreads across both atria, causing them to contract together (atrial systole). This wave of excitation reaches the AV node, where it is delayed slightly, allowing the atria to finish contracting and emptying blood into the ventricles before the ventricles are stimulated. The excitation is then conducted rapidly down the bundle of His and along the Purkinje fibres, which spread through the septum and ventricle walls, causing the ventricles to contract (from the base of the heart upwards).
Final answer: as stated above.

Correct role of the SA node as pacemaker, initiating excitation across the atria [1]; correct role of the AV node (delaying the wave) [1]; correct pathway (bundle of His, Purkinje fibres) conducting excitation to the ventricles [1].

Arteries carry blood away from the heart at high, fluctuating pressure, due to the pumping action of the heart. The thick layer of elastic fibres in their walls allows the artery to stretch and then elastically recoil as blood surges through with each heartbeat, helping to smooth out the pulse and maintain pressure between heartbeats; the smooth muscle allows vasoconstriction and vasodilation, controlling blood flow/supply to different organs. Veins carry blood back to the heart at much lower, steadier pressure, so they do not need to withstand or accommodate such high, fluctuating pressure; instead, they rely on valves (to prevent backflow) and their large lumen, aided by the squeezing action of surrounding skeletal muscles, to assist the low-pressure return of blood to the heart.
Final answer: as stated above.

Correct reasoning that arteries carry high-pressure blood needing elastic recoil to accommodate this [1]; correct role of smooth muscle in controlling blood flow (vasoconstriction/dilation) [1]; correct contrast with veins (low pressure, rely on valves/large lumen instead) [1].

Red blood cells transport oxygen, bound to haemoglobin, around the body. Polymorphs are a type of phagocytic white blood cell that engulf and destroy pathogens by phagocytosis, as part of the immune response. Platelets are cell fragments involved in blood clotting at a wound site.
Final answer: as stated above.

Correct role of red blood cells (oxygen transport) [1]; correct role of polymorphs (phagocytosis of pathogens) [1]; correct role of platelets (blood clotting) [1].

Features include: (1) a large surface area, since there are millions of spherical alveoli in the lungs, providing a large total area for gas exchange; (2) a very thin wall (a single layer of squamous epithelium, together with a thin capillary wall), giving a short diffusion path/distance for gases; (3) a moist inner surface, which allows gases to dissolve before diffusing across; and (4) a rich network of blood capillaries providing a good blood supply, which maintains steep diffusion (concentration) gradients for O2 and CO2 by constantly removing oxygen and delivering carbon dioxide.
Final answer: any three of the above (large surface area; thin wall/short diffusion path; moist surface; rich blood supply maintaining diffusion gradients).

Any three valid features, each with an appropriate reason [1 mark each, max 3].

The external intercostal muscles contract, pulling the ribcage upwards and outwards. At the same time, the diaphragm (a dome-shaped sheet of muscle) contracts and flattens, moving downwards. Together, these actions increase the volume of the thoracic cavity; this decreases the air pressure inside the lungs relative to the pressure of the atmosphere outside, so air flows into the lungs down this pressure gradient (from the higher pressure outside to the lower pressure inside).
Final answer: as stated above.

Correct role of the intercostal muscles (contract, raise ribcage) [1]; correct role of the diaphragm (contracts and flattens) [1]; correct explanation of the resulting volume increase/pressure decrease causing air to flow in [1].

Chemicals in cigarette smoke damage the lining of the alveoli and trigger an inflammatory response; this causes the release of enzymes (from white blood cells attracted to the site) that break down the elastin and connective tissue in the alveoli walls. As a result, the walls between adjacent alveoli break down, merging many small alveoli into fewer, much larger air spaces. This greatly reduces the total surface area available for gas exchange and reduces the elastic recoil of the lungs, making gas exchange much less efficient and causing breathlessness (this condition is emphysema).
Final answer: as stated above.

Correct mechanism (chemicals in smoke damage/break down the alveoli walls, e.g. via released enzymes) [1]; correct outcome (alveoli merge into fewer, larger air spaces) [1]; correct explanation of the effect on gas exchange (greatly reduced surface area, less efficient exchange) [1].

Transpiration — the evaporation of water from mesophyll cells and its diffusion out through open stomata — creates a negative pressure (tension) at the top of the xylem in the leaf. Because water molecules are cohesive (they form hydrogen bonds with one another), this tension is transmitted down the continuous, unbroken column of water in the xylem, pulling more water up from the roots to replace the water lost. Water molecules also adhere to the walls of the xylem vessels (adhesion), which helps to support the water column and resist it breaking under tension.
Final answer: as stated above.

Correct explanation that transpiration creates tension/negative pressure at the top of the xylem [1]; correct reference to cohesion between water molecules transmitting the pull down the column [1]; correct reference to adhesion to the xylem walls [1].

The cells of the endodermis have a band of waterproof, waxy material — the Casparian strip — running around their radial and transverse cell walls. This blocks the apoplast pathway (which runs through the cell walls) at the endodermis, so water and dissolved ions travelling via the apoplast pathway are forced to cross into the cytoplasm of an endodermal cell (through its cell surface membrane), joining the symplast pathway, in order to continue into the stele and reach the xylem. This allows the plant to exert control over which mineral ions are allowed to enter the xylem, since they must cross a selectively permeable cell membrane (which can use active transport) rather than simply diffusing through cell walls.
Final answer: as stated above.

Correct identification of the Casparian strip blocking the apoplast pathway [1]; correct explanation that water/ions must cross into the symplast pathway via the cell membrane to continue [1]; correct link to the plant being able to control/select ion uptake into the xylem [1].

Abiotic factors that could be measured include, for example: soil pH, light intensity, soil moisture content, or temperature.
Final answer: any two valid abiotic factors, e.g. soil pH and light intensity.

Any two valid abiotic factors [1 mark each, max 2].

Consequences include: (1) an algal bloom forms at the surface, blocking light from reaching plants growing below the surface, which then die from lack of light for photosynthesis; and (2) the increased decomposition of dead organic matter (algae and plants) by microorganisms uses up dissolved oxygen, increasing the biological oxygen demand (BOD) and leading to the death of oxygen-dependent organisms such as fish and invertebrates.
Final answer: as stated above.

Any two valid consequences, e.g. algal bloom blocking light; increased BOD depleting oxygen and killing aquatic organisms [1 mark each, max 2].

(a) The graph would be an S-shaped (sigmoid) curve.
(b) Trend 1: as pO2 increases, the percentage saturation of haemoglobin increases, e.g. from 5% at 1 kPa to 97% at 12 kPa. Trend 2: the rate of increase is greatest over the middle range of pO2 (for example, saturation rises steeply from 15% to 88% between 2 kPa and 8 kPa), but the curve levels off (plateaus) at higher pO2 values (rising only from 95% to 97% between 10 kPa and 12 kPa).
Final answer: sigmoid curve; saturation increases with pO2, steepest in the mid-range and levelling off at high pO2.

(a) Correct answer — S-shaped/sigmoid curve [1].
(b) Two valid trends identified, each referencing specific data values [1 mark each, max 2].

(a) Resting: \( 14 \times 0.5 = 7.0 \text{ dm}^3\text{/min} \)
Exercise: \( 32 \times 2.2 = 70.4 \text{ dm}^3\text{/min} \)
(b) Minute ventilation increases substantially (roughly 10-fold, from 7.0 to 70.4 dm³/min) during exercise, due to increases in both breathing rate and tidal volume, meeting the increased demand for oxygen (and increased removal of CO2) by the actively respiring muscles.
Check by a second route: \( 70.4/7.0 = 10.06 \), confirming approximately a 10-fold increase.
Final answer: resting = 7.0 dm³/min; exercise = 70.4 dm³/min; a large (approx. 10-fold) increase during exercise.

(a) Correct resting value 7.0 dm³/min [1]; correct exercise value 70.4 dm³/min [1].
(b) Valid description of the trend (substantial/large increase during exercise) [1].

(a) As wind speed increases, the rate of water uptake (transpiration) increases, but the rate of increase becomes smaller (the curve levels off) at higher wind speeds.
(b) Increased wind speed blows away the layer of humid air that would otherwise build up around the stomata and leaf surface. This maintains a steeper water vapour concentration (diffusion) gradient between the air spaces inside the leaf and the air immediately outside the stomata, so the rate of diffusion of water vapour out of the stomata (transpiration) increases. The effect levels off at higher wind speeds because the humid air layer has already been largely removed at lower wind speeds, so further increases in wind speed have progressively less additional effect on the diffusion gradient.
Final answer: as stated above.

(a) Correct trend described (increases, levelling off at higher wind speeds) [1].
(b) Correct explanation referring to wind removing the humid air layer [1]; correct link to a steeper diffusion gradient increasing the rate of water vapour loss [1].

Trend: as annual rainfall increases, the mean leaf surface area of the plant species also increases.
Reason: in drier habitats (lower rainfall), water conservation is more important, so plants have smaller leaves (a xerophytic adaptation) to reduce the surface area available for water loss by transpiration. Where rainfall is higher and water is less limiting, this pressure to conserve water is reduced, so larger leaves (which are more advantageous for photosynthesis, having a greater area to absorb light) can be supported.
Final answer: leaf surface area increases with rainfall; smaller leaves conserve water in drier conditions.

Correct trend described (leaf surface area increases with rainfall) [1]; valid reason linking smaller leaf area to reduced water loss in drier conditions [1].

Indicative content:
• Water uptake: water enters root hair cells from the soil by osmosis, down a water potential gradient (root hair cells have a lower/more negative water potential than the soil water); mineral ions are taken up separately by active transport.
• Movement across the root: water crosses the root cortex to the xylem via two pathways — the apoplast pathway (through the cellulose cell walls and intercellular spaces, without crossing any membranes) and the symplast pathway (through the cytoplasm of cells, connected by plasmodesmata, crossing cell membranes by osmosis).
• The endodermis contains a waterproof Casparian strip that blocks the apoplast pathway, forcing water into the symplast pathway before it can enter the stele and xylem, allowing the plant to control which ions enter the xylem.
• Movement up the xylem: explained by the cohesion-tension theory — transpiration from the leaf creates tension (negative pressure) that is transmitted down the continuous water column in the xylem due to cohesion between water molecules (hydrogen bonding), pulling more water up from the roots; adhesion of water to the xylem walls also helps support the column. (Root pressure may also make a minor contribution.)
• Transpiration: water evaporates from the moist cell walls of mesophyll cells into the air spaces of the leaf, then diffuses out through open stomata (the main route), down a water vapour concentration gradient, into the surrounding air (the cuticle is a minor alternative route).
• Factors affecting the rate of transpiration: internal factors include leaf surface area, stomatal density and cuticle thickness; external factors include light intensity (affecting stomatal aperture), temperature, humidity, air movement (wind) and soil water availability.
• Xerophytic adaptations to reduce water loss: e.g. a thickened, waxy cuticle (reduces evaporation through the epidermis); sunken stomata, often in pits or grooves (traps a layer of humid air, reducing the diffusion gradient for water vapour loss); rolled/curled leaves (traps humid air near the stomata); hairs on the leaf surface (trap a boundary layer of still, humid air); reduced leaf surface area, e.g. needle-like leaves or spines (reduces the area for water loss, while spines may also deter herbivores); succulent tissue (stores water for use during dry periods); and deep or extensive root systems (to access water from a larger volume of soil or deeper water tables).
Final answer: a well-organised, logically sequenced account covering uptake by root hairs, apoplast/symplast movement and the role of the endodermis, cohesion-tension movement up the xylem, transpiration through stomata and the factors affecting its rate, and a range of correctly explained xerophytic adaptations.

Level 4 (12–15 marks): Comprehensive, accurate and logically sequenced account covering water uptake by root hairs (osmosis), apoplast and symplast pathways with the role of the endodermis/Casparian strip, the cohesion-tension theory for movement up the xylem, transpiration through stomata with at least two correctly explained influencing factors, AND at least three correctly explained xerophytic adaptations; fluent, accurate use of specialist terms throughout.
Level 3 (8–11 marks): Good coverage of most stages (uptake, movement through the plant, transpiration) with reasonable accuracy, and at least two xerophytic adaptations described, though one area may be under-developed or contain minor inaccuracies; mostly appropriate use of specialist terms.
Level 2 (4–7 marks): Some correct points made about water movement and/or xerophytic adaptations, but the answer is incomplete, only partially accurate, or lacks clear logical sequencing; some appropriate use of specialist terms.
Level 1 (1–3 marks): Only basic, fragmentary, list-like points made; little coherent explanation; limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

Add a few drops of sodium hydroxide solution to the food sample, followed by a few drops of dilute copper sulfate solution (or add Biuret reagent directly), and mix. If protein is present, the solution changes colour from blue to purple/lilac (mauve). If the solution remains blue, the test is negative (no protein present).
Final answer: as stated above.

Correct reagents/method described [1]; correct positive result (blue to purple/lilac) [1]; correct negative result comparison (remains blue) [1].

(a) Iodine test: add iodine solution (iodine in potassium iodide solution) to the sample. If starch is present, the colour changes from orange-brown to blue-black.
(b) Benedict's test: add Benedict's reagent to the sample and heat in a water bath (e.g. at around 80–100°C for a few minutes). If a reducing sugar is present, the solution changes colour from blue to green, then yellow, then to a brick-red/orange precipitate.
Final answer: as stated above.

(a) Correct method and positive result (blue-black) [1].
(b) Correct method including heating [1]; correct positive result (brick-red precipitate) [1].

Cut off a few millimetres from the tip of a rapidly growing root (e.g. onion or garlic). Place the tip in warm dilute hydrochloric acid for a few minutes to macerate/soften the tissue (hydrolysing the middle lamella so the cells separate more easily), then rinse. Place the root tip on a microscope slide and add a stain, such as acetic orcein or toluidine blue, to stain the chromosomes. Cover with a coverslip and gently squash the tissue (e.g. tapping with a mounted needle, or pressing with a thumb, protected by blotting paper) to spread the cells into a single layer, before viewing under the light microscope.
Final answer: as stated above.

Correct root tip maceration step (warm dilute HCl) [1]; correct staining step (e.g. acetic orcein) [1]; correct squashing technique to spread cells into a single layer [1].

Mix the amylase enzyme with starch solution and start a stopwatch. At regular time intervals, remove a small sample of the reaction mixture and add it to a well/tube containing iodine solution, then measure the colour intensity using a colorimeter fitted with an appropriate filter (e.g. a red filter, since the starch-iodine complex is blue-black) to record the absorbance (or % transmission). As the reaction proceeds, starch is progressively hydrolysed by amylase into shorter sugars, so the intensity of the blue-black colour formed with iodine decreases, and the absorbance recorded by the colorimeter falls over time.
Final answer: as stated above.

Correct sampling method using iodine at regular time intervals [1]; correct use of the colorimeter (appropriate filter, measuring absorbance/transmission) [1]; correct interpretation (absorbance decreases as starch is broken down) [1].

First, identify the major blood vessels on the outside of the heart — the thick-walled aorta, the pulmonary artery (more anterior, thinner-walled than the aorta), and the vena cavae entering the right atrium. Make an incision down the front of the heart, cutting through the wall of the right atrium and continuing into the right ventricle (following the direction of blood flow), to expose the tricuspid valve, its chordae tendinae and papillary muscles. Make a similar incision on the left side of the heart to expose the left atrium, left ventricle, and the bicuspid (mitral) valve. Compare the thickness of the muscular walls of the two ventricles — the wall of the left ventricle is much thicker than that of the right, since it must pump blood at a higher pressure around the whole body (rather than just to the lungs) — and identify the interventricular septum separating the two ventricles.
Final answer: as stated above.

Correct identification of major vessels externally [1]; correct dissection method exposing the right side/tricuspid valve [1]; correct dissection method exposing the left side/bicuspid valve [1]; correct comparison of ventricle wall thickness and/or identification of the septum [1].

Peel a thin layer of epidermis from the inner (concave) surface of an onion bulb scale using forceps. Place it flat on a microscope slide, add a drop of water (or the test solution), and lower a coverslip carefully at an angle using a mounted needle, to avoid trapping air bubbles, before viewing under the light microscope.
To investigate plasmolysis: prepare a range of sucrose solutions of different, known concentrations (e.g. from 0.0 to 1.0 mol/dm³, in equal increments) using a serial dilution. Place separate pieces of onion epidermis into each solution for a fixed period of time (e.g. 10 minutes), to allow water potential equilibrium to be approached. Mount each piece on a slide, as described above, and observe under the microscope; count the number of cells (out of a set sample size, e.g. 50 cells) showing visible plasmolysis in each solution, and calculate the percentage of plasmolysed cells for each concentration.
Final answer: as stated above.

Correct slide preparation technique (thin epidermis peel, avoiding air bubbles) [1]; correct use of a range of known sucrose concentrations (serial dilution) [1]; correct use of a fixed immersion time [1]; correct method of assessing results (counting % of cells plasmolysed from a set sample) [1].

(a) The organelle is a mitochondrion.
(b) The cristae (the folded inner membrane) — the site of the electron transport chain and ATP synthase (oxidative phosphorylation); and the matrix (the granular fluid interior) — the site of the Krebs cycle enzymes (and mitochondrial DNA/ribosomes).
(c) The cristae greatly increase the surface area of the inner membrane available for the attachment of electron carriers and ATP synthase enzymes, increasing the rate of ATP production during aerobic respiration.
(d) Conventions include: using clear, continuous (unbroken) lines with no shading, colouring or sketchy/feathery lines; drawing structures proportionally accurate to their actual relative sizes, with a scale or magnification stated; using thin, ruled label lines that do not cross one another and that touch (but do not enter) the structure being labelled; and including an appropriate title.
Final answer: as stated above.

(a) Correct answer — mitochondrion [1].
(b) Correct naming and function of cristae [1]; correct naming and function of matrix [1].
(c) Correct function — increases surface area for ATP synthase/electron transport chain, increasing ATP production [1].
(d) Any two valid drawing conventions (e.g. clear unbroken lines, no shading; proportionally accurate with scale stated; ruled non-crossing label lines; title given) [1 mark each, max 2].

(a) \( \text{actual size} = \frac{\text{size on drawing}}{\text{magnification}} = \frac{45\text{ mm}}{1500} = 0.03\text{ mm} = 30\ \mu\text{m} \)
(b) \( \text{size on drawing} = \text{actual size} \times \text{magnification} = 30\ \mu\text{m} \times 750 = 22\,500\ \mu\text{m} = 22.5\text{ mm} \)
Check by a second route: since the magnification has halved (1500 → 750), the drawn size for the same actual width should also halve: \( 45\text{ mm} / 2 = 22.5\text{ mm} \), which agrees.
Final answer: actual width = 30 μm; width on the second drawing = 22.5 mm.

(a) Correct equation with substitution [1]; correct answer 30 μm with unit [1].
(b) Correct equation with substitution [1]; correct answer 22.5 mm with unit [1]. Accept ecf from (a).

Lay out two long measuring tapes at right angles along two edges of each area to create a grid of coordinates. Use a random number generator (or random number table) to generate pairs of coordinates for each sampling point, and place the quadrat at each randomly generated position. Record the percentage cover of daisies within the quadrat at each position. Repeat this for a large number of quadrats (e.g. at least 20) within each area, to obtain a representative sample, then calculate the mean percentage cover of daisies for each area, allowing a fair comparison between the two.
Final answer: as stated above.

Correct method of generating random coordinates to avoid bias [1]; correct procedure of placing the quadrat and recording % cover at each point [1]; correct point about repeating with a sufficient number of quadrats and calculating a mean [1].

A suitable table would have the following columns: 'Quadrat number' (rows numbered 1 to 10, one row per quadrat position), 'Percentage cover of daisies in mown area (%)', and 'Percentage cover of daisies in unmown area (%)'. A final row could be added labelled 'Mean', to record the mean percentage cover calculated for each area once all 10 readings have been taken.
Final answer: a table with quadrat number (1–10) as rows, and columns for % cover of daisies in each area, with clear units and headings, plus a mean row.

Correct column headings for quadrat number and % cover in each area, with units [1]; correct number of rows for repeats (10 per area) [1]; clear, logical table layout (e.g. including a mean row) [1].

If the student deliberately chose quadrat positions with the most daisies, this would introduce bias into the results — the sample would not represent the true, average distribution of daisies across the whole area, and could exaggerate any real difference (or create an apparent difference) between the two areas, leading to misleading or invalid conclusions. Random sampling ensures that every position within the area has an equal chance of being selected, giving a representative, unbiased estimate of the true percentage cover in each area.
Final answer: as stated above.

Correctly identifies the issue of bias with non-random selection [1]; correct explanation of how this gives an unrepresentative/misleading result [1]; correct statement that random sampling avoids this by giving every position an equal chance of selection [1].

Lay a measuring tape in a straight line starting in the mown area, running across the boundary, and continuing into the unmown area. Starting at one end (0 m), place a quadrat alongside the tape at regular, fixed intervals (for example, every 2 m) along its entire length. At each position, record the percentage cover of daisies within the quadrat. Continue along the whole length of the transect, recording results at every interval. This method shows how the percentage cover of daisies changes with distance/position along the transect, allowing any pattern in cover across the boundary between the two areas to be identified.
Final answer: as stated above.

Correct set-up (measuring tape laid across the boundary/area of interest) [1]; correct method (quadrats placed at regular, fixed intervals along the tape) [1]; correct recording (% cover recorded at each position) [1]; correct explanation of what this reveals (change in cover with distance/across the boundary) [1].

The iodine test result (no colour change, remains orange-brown) is negative, so starch is absent. The Benedict's test result (brick-red precipitate on heating) is positive, so a reducing sugar is present. The Biuret test result (remains blue) is negative, so protein is absent. The ethanol emulsion test result (no white emulsion) is negative, so lipid is absent.
Final answer: reducing sugar is present; starch, protein and lipid are all absent.

Correct identification that a reducing sugar is present, with reference to the Benedict's test result [1]; correct identification that starch is absent, with reference to the iodine test result [1]; correct identification that both protein and lipid are absent, with reference to the Biuret and ethanol emulsion test results [1].

Add ethanol to the food sample and shake well to dissolve any lipid present, then pour the resulting mixture into a test tube of water. If lipid is present, a cloudy white emulsion forms (as the dissolved lipid comes out of solution and disperses as tiny droplets in the water).
Final answer: as stated above.

Correct method (ethanol added and shaken, then poured into water) [1]; correct positive result (cloudy white emulsion forms) [1].

The initial negative result shows that no reducing sugar was present in the sample in its original form. However, boiling the sample with dilute hydrochloric acid hydrolyses the glycosidic bond in any non-reducing sugar present (such as sucrose), breaking it down into its constituent monosaccharides, which are reducing sugars. Once the acid is neutralised, these reducing sugars give a positive result with Benedict's reagent (brick-red precipitate), showing that the original sample contained a non-reducing sugar.
Final answer: the sample contains a non-reducing sugar, which is hydrolysed into reducing sugars by boiling with acid, giving the later positive result.

Correct explanation of the initial negative result (no reducing sugar present) [1]; correct explanation of acid hydrolysis of a non-reducing sugar producing reducing sugars, accounting for the later positive result [1].