MetadataPastPaper.sampleTitle

Part (a)
- Starch consists of \(\alpha\)-glucose monomers, whereas cellulose consists of \(\beta\)-glucose monomers.
- In \(\beta\)-glucose, the hydroxyl (-OH) group on carbon-1 is positioned above the ring, whereas in \(\alpha\)-glucose, it is below the ring.

Part (b)
- Heat the sample with Benedict's reagent first; if there is no colour change (remains blue), this suggests reducing sugars are absent or in low concentration.
- Take a fresh sample, add dilute hydrochloric acid (\(\text{HCl}\)) and heat in a water bath to hydrolyse any glycosidic bonds present.
- Neutralise the acid by adding sodium hydrogencarbonate (\(\text{NaHCO}_3\)) until it stops effervescing.
- Re-test by adding Benedict's reagent and heating in a water bath.
- A positive result is a colour change from blue to green, yellow, orange, or brick-red.

Part (c)
- Initially, maltose is a reducing sugar, so there are 150 reducing sugar molecules.
- Hydrolysis of one maltose molecule produces two monomers of \(\alpha\)-glucose, which are also reducing sugars.
- Total reducing molecules after complete hydrolysis = \(150 \times 2 = 300\) molecules of glucose.
- Percentage increase = \(\frac{300 - 150}{150} \times 100\% = 100\%\).

Part (a) [Max 2 marks]
- 1 mark: Starch is made of \(\alpha\)-glucose whereas cellulose is made of \(\beta\)-glucose.
- 1 mark: In \(\beta\)-glucose, the -OH group on C1 is above the plane/ring (or vice versa for \(\alpha\)-glucose).

Part (b) [Max 5 marks]
- 1 mark: Heat sample with Benedict's reagent first to confirm no/low reducing sugar.
- 1 mark: Heat a fresh sample with dilute hydrochloric acid / \(\text{HCl}\).
- 1 mark: Neutralise with sodium hydrogencarbonate / \(\text{NaHCO}_3\).
- 1 mark: Add Benedict's reagent and heat again.
- 1 mark: Red / orange / yellow / green precipitate/colour change indicates a positive result.

Part (c) [3.7 marks]
- 1 mark: Identifies that maltose is initially a reducing sugar (so 150 molecules at start).
- 1 mark: Identifies that complete hydrolysis produces 300 molecules of glucose (reducing sugar).
- 1.7 marks: Correct calculation showing a 100% increase (allow 1.7 marks for correct final answer with working; allow 1 mark for correct method of calculation but minor arithmetic slip).

Part (a)
- Competitive inhibitors bind to the active site of the enzyme, as they have a similar complementary shape to the substrate.
- Non-competitive inhibitors bind to a site other than the active site (the allosteric site).
- Competitive inhibitors do not affect the maximum rate of reaction (\(V_{\max}\)) because increasing substrate concentration can overcome the inhibition.
- Non-competitive inhibitors decrease the \(V_{\max}\) because they permanently alter the shape of the active site, making it non-functional regardless of substrate concentration.

Part (b)
- The active site has a specific tertiary structure determined by the unique primary sequence of amino acids (folding of polypeptide chains).
- This tertiary structure is highly specific, meaning only a substrate with a complementary shape can fit into the active site to form an enzyme-substrate complex.

Part (c)
- Rate decrease = 65%, which means the remaining rate is \(100\% - 65\% = 35\%\) of the original rate.
- New rate = \(1.2 \times 10^{-3} \times 0.35\)
- New rate = \(4.2 \times 10^{-4} \text{ mol dm}^{-3}\text{ s}^{-1}\) (or \(0.00042 \text{ mol dm}^{-3}\text{ s}^{-1}\)).

Part (a) [Max 4 marks]
- 1 mark: Competitive inhibitor binds to active site;
- 1 mark: Non-competitive inhibitor binds to allosteric site / site other than active site;
- 1 mark: Competitive inhibitor does not change \(V_{\max}\) (or can be overcome by high substrate concentration);
- 1 mark: Non-competitive inhibitor decreases/reduces \(V_{\max}\) (cannot be overcome).

Part (b) [Max 3 marks]
- 1 mark: Specific tertiary structure of the active site / protein;
- 1 mark: Determined by primary structure / amino acid sequence;
- 1 mark: Only substrate with a complementary shape can bind / form enzyme-substrate complexes.

Part (c) [3.7 marks]
- 1 mark: Showing method of finding remaining percentage (e.g., multiplying by 0.35 or calculating 65% as \(7.8 \times 10^{-4}\));
- 2.7 marks: Correct final answer of \(4.2 \times 10^{-4} \text{ mol dm}^{-3}\text{ s}^{-1}\) (accept 0.00042, ignore lack of units, but penalise incorrect units).

Part (a)
- Active transport is the movement of substances against their concentration gradient (from a region of lower concentration to a region of higher concentration).
- It requires energy in the form of ATP, which is hydrolysed to ADP and inorganic phosphate.
- Specific carrier proteins spanning the cell membrane bind to the solute / molecule.
- The binding and hydrolysis of ATP causes a conformational (shape) change in the carrier protein, which pumps the solute across the membrane.

Part (b)
- The phospholipid bilayer has a hydrophobic core composed of fatty acid tails.
- Polar molecules and ions are hydrophilic / water-soluble and cannot easily interact with or pass through this hydrophobic/non-polar region.
- They are repelled by the hydrophobic fatty acid tails, resulting in an extremely low rate of simple diffusion.

Part (c)
- Change in mass = \(4.14\text{ g} - 4.50\text{ g} = -0.36\text{ g}\).
- Percentage change = \(\frac{-0.36}{4.50} \times 100\% = -8.0\%\) (or a decrease of 8.0%).
- Since the potato tissue lost mass, net osmosis of water occurred out of the potato cells into the sucrose solution.
- Therefore, the water potential of the sucrose solution must be lower than that of the potato cells.

Part (a) [Max 4 marks]
- 1 mark: Movement against a concentration gradient;
- 1 mark: Requires ATP / energy from respiration;
- 1 mark: Involves specific carrier proteins;
- 1 mark: Protein changes shape to transport the molecule.

Part (b) [Max 3 marks]
- 1 mark: Bilayer has a hydrophobic / non-polar fatty acid tail core;
- 1 mark: Polar molecules/ions are hydrophilic / polar;
- 1 mark: Repelled by the hydrophobic core / cannot dissolve in the lipid tail region.

Part (c) [3.7 marks]
- 1 mark: For calculating change in mass (\(-0.36\text{ g}\));
- 1 mark: For calculating percentage change as \(-8.0\%\) (or 8.0% decrease);
- 1.7 marks: For stating that the water potential of the sucrose solution is lower (than the potato cells) because water moved out by osmosis.

Part (a)
- DNA helicase breaks the hydrogen bonds between the complementary base pairs of the two parental strands, unwinding the double helix and separating the strands.
- DNA polymerase aligns free activated DNA nucleotides against the exposed template strands through complementary base pairing.
- DNA polymerase catalyses the formation of phosphodiester bonds between adjacent nucleotides in a condensation reaction, forming the new sugar-phosphate backbone.

Part (b)
- DNA contains the pentose sugar deoxyribose, whereas RNA contains ribose.
- DNA contains the nitrogenous base thymine, whereas RNA contains uracil instead.
- DNA is typically double-stranded (double helix), whereas RNA is single-stranded.

Part (c)
- In double-stranded DNA, according to complementary base pairing rules, Adenine (A) = Thymine (T).
- Since \(A = 24\%\), then \(T = 24\%\).
- Total \(A + T = 48\%\).
- Therefore, the remaining bases, Guanine (G) and Cytosine (C), must make up \(100\% - 48\% = 52\%\).
- Since \(G = C\), the percentage of Guanine is \(\frac{52\%}{2} = 26\%\).

Part (a) [Max 4 marks]
- 1 mark: DNA helicase breaks hydrogen bonds between bases;
- 1 mark: DNA helicase separates the two strands / unwinds double helix;
- 1 mark: DNA polymerase joins nucleotides together / forms phosphodiester bonds (via condensation);
- 1 mark: DNA polymerase works in a 5' to 3' direction / complementary base pairing.

Part (b) [Max 3 marks]
- 1 mark: Deoxyribose vs ribose;
- 1 mark: Thymine vs uracil;
- 1 mark: Double-stranded vs single-stranded.

Part (c) [3.7 marks]
- 1 mark: Deduces that \(T = 24\%\), so \(A + T = 48\%\);
- 1 mark: Deduces that \(G + C = 52\%\);
- 1.7 marks: For correct final answer of 26% Guanine (allow 1.7 marks for correct final answer with working; allow 1 mark for correct method but minor calculation error).

Part (a)
- Respiring tissues release carbon dioxide (\(\text{CO}_2\)), which dissolves in blood to form carbonic acid, lowering the pH.
- The lower pH alters the shape of hemoglobin slightly, reducing its affinity for oxygen.
- This shifts the oxygen dissociation curve to the right.
- Consequently, at any given partial pressure of oxygen (\(pO_2\)), hemoglobin is less saturated and releases (unloads) more oxygen to the respiring tissues.

Part (b)
- Tracheoles have thin walls (single cell layer) which provides a short diffusion pathway for gases.
- They are highly branched and extensively spread throughout tissues, providing a large surface area for gas exchange.
- The ends of tracheoles are filled with fluid; during intense activity, lactate accumulates in muscle cells, lowering water potential, causing fluid to move into muscle cells by osmosis. This exposes a larger surface area of the tracheole wall directly to air, increasing the rate of diffusion.

Part (c)
- At 96% saturation, the mean number of oxygen molecules bound is: \(4 \times 0.96 = 3.84\) molecules of \(O_2\).
- At 48% saturation, the mean number of oxygen molecules bound is: \(4 \times 0.48 = 1.92\) molecules of \(O_2\).
- Mean number of oxygen molecules released per hemoglobin molecule is: \(3.84 - 1.92 = 1.92\) molecules of \(O_2\).

Part (a) [Max 4 marks]
- 1 mark: Carbon dioxide produces carbonic acid / lowers pH;
- 1 mark: Lowers hemoglobin's affinity for oxygen;
- 1 mark: Shifts the oxygen dissociation curve to the right;
- 1 mark: More oxygen is unloaded/released at the same partial pressure of oxygen (\(pO_2\)).

Part (b) [Max 3 marks]
- 1 mark: Thin walls provide a short diffusion pathway;
- 1 mark: Highly branched network provides a large surface area;
- 1 mark: Fluid at the ends can be withdrawn during muscle activity, increasing gas diffusion rate.

Part (c) [3.7 marks]
- 1 mark: Calculating the change in saturation is 48% (or \(96\% - 48\% = 48\%\)) OR calculating the bound oxygen molecules at start (3.84) or end (1.92);
- 2.7 marks: Correct final answer of 1.92 molecules of \(O_2\) (allow 2.7 marks for correct final answer with working; allow 1 mark for correct method with a calculation error).

Part (a)
- It groups species based on evolutionary origins / evolutionary relationships / common ancestry.
- It uses a hierarchy of non-overlapping groups (taxa) where smaller groups are contained within larger groups.

Part (b)
- Serum / proteins (e.g., albumin) from species A are injected into a test animal (e.g., a rabbit).
- The rabbit produces antibodies specific to the antigens on species A's proteins.
- These antibodies are extracted from the rabbit's serum and mixed with the serum of other mammalian species.
- The antibodies bind to complementary antigens on the other species' proteins, forming a precipitate.
- The greater the amount of precipitate formed, the more similar the proteins (similar antigen shape), indicating a closer evolutionary relationship.

Part (c)
- Genus: *Canis*
- Species: *lupus*
- The binomial system is a universal naming system. It ensures that scientists worldwide use the same unique name for an organism, which avoids confusion caused by different common names in different languages or regions.

Part (a) [Max 3 marks]
- 1 mark: Based on evolutionary history / relationships / shared ancestors;
- 1 mark: Shows points of divergence / branching;
- 1 mark: Uses a hierarchy of non-overlapping groups (taxa).

Part (b) [Max 4 marks]
- 1 mark: Inject protein/albumin of species A into animal/rabbit to produce antibodies;
- 1 mark: Extract antibodies and mix with serum/proteins of other species;
- 1 mark: Precipitate forms when antibodies bind to complementary antigens;
- 1 mark: More precipitate = more closely related species (more similar antigen/protein structure).

Part (c) [3.7 marks]
- 1 mark: Genus is *Canis*;
- 1 mark: Species is *lupus*;
- 1.7 marks: The binomial system is universal / standardised worldwide, avoiding confusion caused by local common names / translation errors.

Part (a)
- Monocultures mean only one crop species is grown, which reduces the variety of plant species in an area.
- This reduces the variety of food sources and niches/habitats available for other organisms (such as insects or birds).
- Clearing hedgerows directly destroys habitats, food sources, and nesting sites.
- The use of chemical pesticides/herbicides directly kills pest species and weeds, further reducing biodiversity and disrupting food chains.

Part (b)
- A high value of \(d\) indicates a high species diversity and a more stable ecosystem / community. In a stable community, abiotic changes are less likely to lead to extinction of species because there are many alternative food sources.
- It is more useful than species richness because species richness only measures the number of different species, whereas the index of diversity takes into account the abundance / number of individuals of each species (preventing a community dominated by one species from appearing highly diverse).

Part (c)
- Total number of individuals, \(N = 15 + 25 + 10 = 50\).
- \(N(N-1) = 50 \times 49 = 2450\).
- For each species, calculate \(n(n-1)\):
- Species A: \(15 \times 14 = 210\)
- Species B: \(25 \times 24 = 600\)
- Species C: \(10 \times 9 = 90\)
- Sum of \(n(n-1) = 210 + 600 + 90 = 900\).
- Index of Diversity, \(d = \frac{2450}{900} \approx 2.72\) (or 2.7).

Part (a) [Max 4 marks]
- 1 mark: Monoculture reduces variety of plants;
- 1 mark: Hedgerow clearance removes habitats / nesting sites / shelters;
- 1 mark: Fewer food sources / niches/habitats for consumers/insects;
- 1 mark: Pesticides/herbicides kill non-target organisms / remove weeds, disrupting food webs.

Part (b) [Max 3 marks]
- 1 mark: High \(d\) means high diversity and high stability / less hostile environment;
- 1 mark: Species richness only counts the number of species (does not consider population sizes / evenness);
- 1 mark: Index of Diversity accounts for abundance/evenness of each species.

Part (c) [3.7 marks]
- 1 mark: Correct calculation of \(N = 50\) and \(N(N-1) = 2450\);
- 1 mark: Correct calculation of \(\sum n(n-1) = 900\) (showing working for individual species);
- 1.7 marks: Correct final answer of 2.72 (accept 2.7).

At 0 nM: \(\text{MI} = \frac{48}{800} \times 100 = 6\%\)
At 10 nM: \(\text{MI} = \frac{192}{800} \times 100 = 24\%\)
Percentage change: \(\frac{24 - 6}{6} \times 100 = \frac{18}{6} \times 100 = +300\%\) (or a 300% increase).

(b) Spindle fibres attach to the centromeres of chromosomes and are responsible for pulling sister chromatids to opposite poles of the cell during anaphase. This movement requires the shortening and disassembly of the spindle microtubules. If disassembly is prevented, chromatids cannot separate to opposite poles, preventing the completion of mitosis and causing the cell to undergo apoptosis (programmed cell death).

(c) Cancer cells divide rapidly and uncontrollably, meaning they undergo mitosis much more frequently than normal body cells. Consequently, they are much more likely to be in a phase of the cell cycle (mitosis) that is targeted by the drug. Normal body cells divide much less frequently, so they are less affected.

(d) In binary fission: 1. No spindle fibres are formed, whereas spindle fibres are essential in mitosis. 2. Circular DNA is replicated and attaches to the cell membrane, whereas in mitosis linear chromosomes condense and are separated after the nuclear envelope breaks down. 3. There is no packaging of DNA into histones.

(a) [3 marks]
- Mitotic index at 0 nM is 6% AND at 10 nM is 24% (1 mark)
- Correct formula for percentage change: \(\frac{24-6}{6} \times 100\) (1 mark)
- Correct answer: +300% or 300% increase (1 mark). Accept '300%' without plus sign. Reject -300% or 75% increase.

(b) [3 marks]
- Spindle fibres align chromosomes/chromatids and/or pull them to opposite poles of the cell (1 mark)
- Disassembly/shortening of spindle fibres is required for sister chromatids to separate during anaphase (1 mark)
- Preventing disassembly prevents separation of chromosomes, stopping mitosis / inducing apoptosis (1 mark)

(c) [2.7 marks]
- Cancer cells undergo rapid/uncontrolled cell division / have a shorter cell cycle (1 mark)
- Therefore, a much higher proportion of cancer cells will be undergoing mitosis at any given time (1 mark)
- Normal cells divide slowly/infrequently / are mostly in interphase/G0, so are less affected (0.7 marks)

(d) [2 marks]
- No spindle fibres form in binary fission (unlike mitosis) (1 mark)
- Binary fission involves replication of a single circular DNA molecule, while mitosis involves replication and separation of multiple linear chromosomes (1 mark)

(a) Immediately after infection, the patient's immune system has not had enough time to undergo clonal selection, expansion, and differentiation of B cells into plasma cells to produce a high concentration of HIV-specific antibodies. The concentration of antibodies is below the limit of detection of the ELISA test. This is known as the 'window period'.

(b) Reverse transcriptase: Uses the viral single-stranded RNA as a template to synthesize a complementary DNA (cDNA) strand, converting the viral genetic material into double-stranded DNA. Integrase: Catalyzes the insertion (integration) of this viral double-stranded DNA into the host cell's nuclear DNA (genome), forming a provirus that can be transcribed by host enzymes.

(c) NNRTIs bind to an allosteric site (a site other than the active site) on the reverse transcriptase enzyme. This binding alters the tertiary structure of the enzyme, changing the shape of the active site. Consequently, the active site is no longer complementary to the viral RNA template or nucleotides, preventing the formation of enzyme-substrate complexes and halting cDNA synthesis.

(a) [3 marks]
- The body requires time to produce antibodies against HIV / undergo humoral response (1 mark)
- In the first two weeks, antibody levels are too low / below detection limit (1 mark)
- This is known as the window period / seroconversion has not yet occurred (1 mark)

(b) [4 marks]
- Reverse transcriptase uses viral RNA as a template (1 mark)
- To synthesize double-stranded viral DNA / cDNA (1 mark)
- Integrase inserts this viral DNA (1 mark)
- Into the host cell's DNA/genome (1 mark)

(c) [3.7 marks]
- NNRTIs bind to an allosteric site / site other than the active site (1 mark)
- This alters the tertiary structure of the enzyme (1 mark)
- This changes the shape of the active site (1 mark)
- Substrate can no longer bind / no enzyme-substrate complexes can form (0.7 marks)

(a) First, calculate the heart rate (HR) in beats per minute:
\(\text{HR} = \frac{60 \text{ seconds}}{0.75 \text{ seconds/beat}} = 80 \text{ beats per minute}\).
Next, calculate Cardiac Output (CO):
\(\text{CO} = \text{Stroke Volume} \times \text{Heart Rate}\)
\(\text{Stroke Volume} = 72 \text{ cm}^3 = 0.072 \text{ dm}^3\)
\(\text{CO} = 0.072 \text{ dm}^3 \times 80 \text{ min}^{-1} = 5.76 \text{ dm}^3\text{ min}^{-1}\).

(b) During ventricular systole, the ventricles contract, raising the intraventricular pressure. This pressure rapidly exceeds the pressure inside the atria, pushing the blood backwards and forcing the atrioventricular (AV) valves to close (preventing backflow). As ventricular contraction continues, the pressure inside the ventricle rises until it exceeds the pressure in the aorta/pulmonary artery. This pressure gradient forces the semi-lunar (SL) valves to open, allowing blood to be ejected into the arteries.

(c) The left ventricle must pump blood through the aorta to the systemic circulation (the entire body), which is a high-resistance, high-pressure system. A thicker muscular wall allows for a stronger contraction, generating the high hydrostatic pressure needed to overcome this resistance and transport blood over long distances. The right ventricle only pumps blood to the lungs (pulmonary circulation), which is close to the heart, low-resistance, and requires lower pressure to avoid damaging the delicate capillaries in the alveoli.

(a) [3 marks]
- Correct calculation of heart rate: \(60 / 0.75 = 80 \text{ bpm}\) (1 mark)
- Conversion of stroke volume to dm\(^3\): \(72 \text{ cm}^3 = 0.072 \text{ dm}^3\) (1 mark)
- Correct final answer with units: \(5.76 \text{ dm}^3\text{ min}^{-1}\) (1 mark). Reject \(5760\).

(b) [4.7 marks]
- During ventricular systole, ventricles contract, increasing ventricular pressure (1 mark)
- Pressure in ventricle becomes higher than in atrium, forcing AV valves to close (1 mark)
- This closure prevents backflow of blood into the atrium (1 mark)
- Ventricular pressure becomes higher than pressure in the aorta/arteries (1 mark)
- This forces the semi-lunar (SL) valves to open to allow ejection of blood (0.7 marks)

(c) [3 marks]
- Left ventricle pumps blood to the whole body (systemic circulation), whereas right ventricle only pumps to lungs (pulmonary circulation) (1 mark)
- Systemic circulation is longer / has higher resistance, requiring higher pressure (1 mark)
- Thicker muscle wall allows stronger contraction / generates higher pressure (1 mark)

(a) Hydrogen ions (\(\text{H}^+\)) are actively transported (pumped) out of the companion cells into the cell wall space using ATP. This establishes a proton concentration gradient. \(\text{H}^+\) ions then diffuse back down their electrochemical gradient into the companion cell through co-transporter proteins. As they do, they co-transport sucrose molecules against their concentration gradient into the companion cell. The high concentration of sucrose allows it to diffuse into the sieve tube element via plasmodesmata.

(b) The loading of sucrose into the sieve tube element lowers its water potential. Water moves from the xylem (and surrounding cells) into the sieve tube element by osmosis down a water potential gradient. This entry of water increases the hydrostatic pressure inside the sieve tube element at the source. At the sink, sucrose is unloaded, which increases the water potential in the sieve tube, causing water to leave by osmosis and lowering the hydrostatic pressure. This creates a hydrostatic pressure gradient from source to sink, driving the mass flow of phloem sap.

(c) Active transport requires ATP. If a metabolic inhibitor like dinitrophenol is applied, ATP synthesis is stopped. If translocation slows down or stops completely when the inhibitor is applied, it proves that the process requires energy/ATP (active transport of \(\text{H}^+\) ions) and is not purely passive diffusion or physical flow.

(a) [4.7 marks]
- Active transport of \(\text{H}^+\) (protons) out of companion cells into cell wall (1 mark)
- Requires energy from ATP hydrolysis (1 mark)
- Creates a proton gradient / higher \(\text{H}^+\) concentration in cell wall (1 mark)
- \(\text{H}^+\) diffuses back through co-transporter protein, bringing sucrose with it (1 mark)
- Sucrose diffuses into sieve tube elements via plasmodesmata (0.7 marks)

(b) [4 marks]
- Loading of sucrose into sieve tube lowers its water potential (1 mark)
- Water enters sieve tube from xylem by osmosis (1 mark)
- This generates high hydrostatic pressure at the source (1 mark)
- Unloading of sucrose at sink lowers hydrostatic pressure, creating a pressure gradient that drives mass flow (1 mark)

(c) [2 marks]
- Inhibitors stop ATP production / aerobic respiration (1 mark)
- This stops active transport of \(\text{H}^+\) / halts translocation, demonstrating that translocation requires metabolic energy (1 mark)

(a) 1. Chemical products of the pathogen attract the macrophage (chemotaxis). 2. Receptors on the cell surface membrane of the macrophage bind to foreign antigens on the pathogen. 3. The macrophage engulfs the pathogen via endocytosis, invaginating its cell membrane to form a vesicle called a phagosome. 4. Lysosomes within the macrophage fuse with the phagosome to form a phagolysosome. 5. Lysozymes and hydrolytic enzymes inside the lysosomes digest and destroy the pathogen. 6. The macrophage processes the pathogen's antigens and displays them on its cell surface membrane (becoming an antigen-presenting cell).

(b) A specific helper T cell with a complementary receptor binds to the presented antigen on the antigen-presenting macrophage. This binding, alongside chemical signals (cytokines) released by the macrophage, activates the helper T cell. Once activated, the helper T cell divides rapidly by mitosis (clonal expansion). These activated helper T cells release cytokines that stimulate specific cytotoxic T cells (which destroy infected host cells by releasing perforin) and B cells.

(c) Active immunity is acquired by exposure to an antigen (either via infection or vaccination), leading to the production of antibodies and memory cells by the individual's own immune system. This provides long-term protection. Passive immunity is acquired by receiving antibodies produced by another organism (e.g., across the placenta or via injection). It does not produce memory cells and only provides short-term protection as the antibodies are broken down.

(a) [5 marks]
- Chemotaxis / macrophage attracted to pathogen by chemicals (1 mark)
- Receptor binding / macrophage binds to foreign antigens on pathogen (1 mark)
- Engulfment / endocytosis to form a phagosome (1 mark)
- Fusion of lysosomes with phagosome / release of lysozymes/hydrolytic enzymes (1 mark)
- Digestion of pathogen AND presentation of antigens on the cell surface membrane (1 mark)

(b) [3.7 marks]
- Specific helper T cell with complementary receptor binds to presented antigen (1 mark)
- Helper T cell is activated and divides by mitosis / undergoes clonal expansion (1 mark)
- Helper T cells release cytokines (1 mark)
- Cytokines activate/stimulate cytotoxic T cells to destroy infected cells (0.7 marks)

(c) [2 marks]
- Active: individual produces own antibodies/memory cells AND is long-term (1 mark)
- Passive: antibodies introduced from outside / no memory cells produced AND is short-term (1 mark)

(a) Increased methylation occurs at the promoter region of the tumor suppressor gene. This prevents transcription factors from binding to the promoter and causes chromatin to become more tightly packed (heterochromatin). As a result, transcription of the gene is silenced, and the specific tumor suppressor protein is not produced. Without this protein (which normally repairs DNA, regulates the cell cycle, or induces apoptosis), cell division becomes rapid and uncontrolled, leading to a tumor.

(b) Proto-oncogenes code for proteins that stimulate cell division. A mutation converts a proto-oncogene into an oncogene, which is a 'gain-of-function' mutation. This leads to an overactive protein or overproduction of protein that constantly stimulates cell division, even if only one allele is mutated (dominant). In contrast, tumor suppressor genes code for proteins that inhibit cell division. A mutation in a tumor suppressor gene is a 'loss-of-function' mutation. If only one allele is mutated, the remaining normal allele can still produce enough functional protein to control cell division. Therefore, both alleles must be mutated (recessive) to completely lose control of cell division.

(c) 1. Malignant tumors grow rapidly, whereas benign tumors grow slowly. 2. Malignant tumors are invasive and can infiltrate surrounding tissues, whereas benign tumors are surrounded by a fibrous capsule and remain localized. 3. Malignant tumor cells can undergo metastasis (spread via blood/lymph to form secondary tumors elsewhere), whereas benign tumors do not metastasize.

(a) [4 marks]
- Methyl groups added to promoter region of tumor suppressor gene (1 mark)
- Prevents binding of transcription factors / RNA polymerase OR causes DNA to wind tightly/form heterochromatin (1 mark)
- Gene is silenced / not transcribed/translated (1 mark)
- Absence of tumor suppressor protein leads to uncontrolled cell division / lack of cell cycle control / no apoptosis (1 mark)

(b) [3.7 marks]
- Oncogene mutation is a 'gain-of-function' / active stimulation of cell division (1 mark)
- Only one mutated allele is needed to continuously stimulate division (dominant) (1 mark)
- Tumor suppressor mutation is a 'loss-of-function' / failure to inhibit cell division (1 mark)
- If one normal allele remains, it still produces functional inhibitor protein (recessive) (0.7 marks)

(c) [3 marks]
- Rate of growth: Malignant is rapid, benign is slow (1 mark)
- Invasion/Capsulation: Malignant is unencapsulated/invasive, benign is encapsulated/localized (1 mark)
- Metastasis: Malignant can spread to form secondary tumors, benign does not (1 mark)

(a) 1. The cholera bacterium releases a protein toxin (cholera toxin) in the small intestine. 2. The toxin binds to receptors on the epithelial cells of the intestinal wall. 3. This binding activates an enzyme (adenylate cyclase), which increases the concentration of cyclic AMP (cAMP) inside the cells. 4. High cAMP levels cause protein channels (CFTR) in the cell membrane to open. 5. Chloride ions (\(\text{Cl}^-\)) are actively transported / secreted out of the epithelial cells into the lumen of the intestine. 6. The accumulation of chloride ions in the lumen lowers its water potential. 7. Water moves out of the epithelial cells and blood into the lumen by osmosis down a water potential gradient, causing severe watery diarrhoea.

(b) Drinking pure water is ineffective because the water potential of the intestinal lumen is extremely low due to the high concentration of ions. Pure water is not absorbed and may worsen diarrhoea. ORS contains a balanced mixture of glucose and sodium ions (\(\text{Na}^+\)). The presence of both allows them to be transported together into the epithelial cells via sodium-glucose co-transporter proteins. This increases the solute concentration inside the epithelial cells, lowering their water potential. As a result, water is drawn out of the lumen and into the cells/blood by osmosis, successfully rehydrating the patient.

(c) 1. Providing clean, chlorinated drinking water to prevent ingestion of the bacteria. 2. Safe and proper disposal of sewage/human waste to prevent contamination of water sources. 3. Good personal hygiene practices, such as washing hands with clean water and soap before preparing food.

(a) [5 marks]
- Toxin binds to receptors on epithelial cells and increases cAMP levels (1 mark)
- This causes chloride ion channels to open (1 mark)
- Chloride ions (\(\text{Cl}^-\)) move/are secreted into the lumen of the intestine (1 mark)
- This lowers the water potential of the lumen (1 mark)
- Water moves out of epithelial cells/blood into the lumen by osmosis down a water potential gradient (1 mark)

(b) [3.7 marks]
- Drinking pure water does not work because water cannot be absorbed against the osmotic gradient / may worsen diarrhoea (1 mark)
- ORS contains both sodium ions and glucose (1 mark)
- Sodium and glucose are co-transported into the epithelial cells (1 mark)
- This lowers the water potential inside the cells, drawing water back into the body by osmosis (0.7 marks)

(c) [2 marks]
- Treatment of water supply / chlorination (1 mark)
- Proper sewage treatment / separation of drinking water from waste (1 mark)
- Good personal hygiene/handwashing (1 mark) (accept any two)

(a) CA1P binds competitively to the active site of RuBisCO, preventing carbon dioxide from combining with RuBP. Because carbon dioxide fixation is blocked, the rate of GP production falls, leading to a decrease in its concentration as existing GP is still reduced to TP. Conversely, RuBP is still being regenerated from the existing pool of TP, but since it cannot be used to fix CO2, it accumulates, leading to an increased concentration.

(b) The change in GP concentration is \(420 - 85 = 335\) a.u.
Percentage decrease = \(\frac{335}{420} \times 100 = 79.7619\%\).
Rounded to three significant figures, this is \(79.8\%\).

(c) A decrease in GP concentration means there is less substrate available for reduction. Consequently, the rate of triose phosphate (TP) production will fall, even if ATP and reduced NADP from the light-dependent reactions are abundant. Since TP is required to regenerate RuBP (with 5 out of every 6 TP molecules recycled to keep the Calvin cycle running), a shortage of TP will subsequently limit and reduce the rate of RuBP regeneration, eventually halting the cycle.

Part (a) [4 marks]:
- 1 mark: CA1P inhibits RuBisCO so carbon dioxide is not fixed/cannot combine with RuBP.
- 1 mark: Existing GP continues to be converted into TP, so GP levels drop.
- 1 mark: RuBP continues to be regenerated from existing TP.
- 1 mark: Because RuBP cannot combine with CO2, it accumulates.

Part (b) [2.7 marks]:
- 1 mark: Correct absolute decrease of 335 a.u. calculated.
- 1 mark: Correct percentage calculation process: \(\frac{335}{420} \times 100\).
- 0.7 marks: Correct rounding to 3 significant figures (79.8%). Accept 79.8%.

Part (c) [4 marks]:
- 1 mark: Less GP is available to be reduced to TP.
- 1 mark: Mention that this reduction requires ATP and reduced NADP.
- 1 mark: The rate of TP production decreases.
- 1 mark: Since TP is needed for RuBP regeneration, the rate of RuBP regeneration will also decrease.

(a) First, calculate the radius (r) of the capillary tube: \(r = \frac{0.8 \text{ mm}}{2} = 0.4 \text{ mm}\).
Next, calculate the volume of oxygen consumed (volume of cylinder):
\(\text{Volume} = \pi r^2 h = 3.142 \times (0.4)^2 \times 24 = 3.142 \times 0.16 \times 24 = 12.06528 \text{ mm}^3\).
Then, divide by the mass of the seeds and the duration of the experiment:
\(\text{Rate} = \frac{12.06528 \text{ mm}^3}{10.0 \text{ g} \times 20 \text{ min}} = \frac{12.06528}{200} = 0.0603264 \text{ mm}^3\,\text{g}^{-1}\,\text{min}^{-1}\).
In standard form to three significant figures, this is \(6.03 \times 10^{-2} \text{ mm}^3\,\text{g}^{-1}\,\text{min}^{-1}\).

(b) The control tube ensures that any movement of the fluid is solely due to oxygen uptake by respiration and not due to abiotic factors such as changes in ambient temperature or atmospheric pressure. Boiled, sterile seeds cannot respire, and the glass beads ensure that the volume of air inside the control tube remains equal to the experimental tube.

(c) The germinating seeds actively carry out aerobic respiration, during which they absorb oxygen from the air. The carbon dioxide gas they release is chemically absorbed by the KOH solution. Consequently, as oxygen is consumed and carbon dioxide is absorbed, the total volume and pressure of gas inside the tube decrease. The external atmospheric pressure is now greater, pushing the fluid along the capillary tube towards the experimental tube.

Part (a) [4.7 marks]:
- 1 mark: Correctly calculates radius (0.4 mm).
- 1 mark: Correctly calculates volume of cylinder (\(12.07 \text{ mm}^3\) or \(12.1 \text{ mm}^3\)).
- 1 mark: Divides by both mass (10 g) and time (20 min).
- 1 mark: Obtains the correct decimal rate (\(0.0603\)).
- 0.7 marks: Correctly expresses in standard form to 3 s.f. (\(6.03 \times 10^{-2}\)).

Part (b) [3 marks]:
- 1 mark: Controls for changes in temperature and/or atmospheric pressure.
- 1 mark: Sterile, boiled seeds do not respire (no oxygen consumption).
- 1 mark: Glass beads equalize the air volume in both tubes.

Part (c) [3 marks]:
- 1 mark: Oxygen is removed from the air by the respiring seeds.
- 1 mark: Carbon dioxide produced is absorbed by the KOH.
- 1 mark: This creates a vacuum/drop in gas volume/pressure inside the tube, pulling the fluid in.

(a) Step 1: Calculate Net Secondary Productivity (NSP) of the primary consumers.
\(\text{NSP} = \text{Energy consumed} - (\text{Fecal waste} + \text{Respiration})\)
\(\text{NSP} = 3.50 \times 10^3 - (1.40 \times 10^3 + 1.60 \times 10^3)\)
\(\text{NSP} = 3500 - (1400 + 1600) = 3500 - 3000 = 500 \text{ kJ m}^{-2} \text{ year}^{-1}\).

Step 2: Calculate the efficiency of transfer from NPP.
\(\text{Efficiency} = \frac{\text{NSP}}{\text{NPP}} \times 100\)
\(\text{Efficiency} = \frac{500}{1.20 \times 10^4} \times 100 = \frac{500}{12000} \times 100 = 4.1667\%\).
Rounded to three significant figures, the efficiency is \(4.17\%\).

(b) NPP represents the chemical energy store remaining after respiratory losses in plants are accounted for. The relationship is \(\text{NPP} = \text{GPP} - \text{Respiration}\). Trees use a significant portion of the organic molecules synthesized during photosynthesis (GPP) for cellular respiration to produce ATP for metabolic processes. Some energy is lost as heat, meaning NPP is always less than GPP.

(c) 1. Restricting movement of livestock (e.g., penning animals in small spaces) or keeping them indoors in warm environments. This minimizes energy lost via muscular contraction and heat generation to maintain body temperature, meaning more consumed energy is converted into animal biomass.
2. Using selective herbicides to kill weeds. This eliminates competition for light and minerals, ensuring crop plants capture more solar energy and convert it into biomass available for human consumption.

Part (a) [4.7 marks]:
- 2 marks: Correct calculation of NSP (500 kJ m-2 year-1). (1 mark for working, 1 mark for value).
- 1 mark: Divides NSP by NPP (500 / 12000).
- 1 mark: Multiplies by 100 to get a percentage.
- 0.7 marks: Correctly rounds to 3 s.f. (4.17%).

Part (b) [3 marks]:
- 1 mark: Explicitly states the formula NPP = GPP - R.
- 1 mark: Explains that plants use GPP products for cellular respiration.
- 1 mark: Explains that respiration results in energy lost as heat (reducing chemical energy stored in biomass).

Part (c) [3 marks]:
- 1 mark: Describes one valid practice (e.g., intensive farming/indoor housing, use of herbicides/pesticides, high-digestibility food).
- 1 mark: Explains how it reduces energy loss (e.g., less respiratory loss, less energy lost in feces, reduced competition).
- 1 mark: Mentions that this results in more energy channeled into human food/biomass.

(a) Let \(p\) be the frequency of the dominant allele and \(q\) be the frequency of the recessive allele.
The frequency of homozygous recessive individuals (those with chondrodysplasia) is represented by \(q^2\).
\(q^2 = \frac{34}{850} = 0.04\).
To find \(q\), take the square root of \(q^2\):
\(q = \sqrt{0.04} = 0.2\).
Since \(p + q = 1\), we can find \(p\):
\(p = 1 - 0.2 = 0.8\).
The frequency of the heterozygous carrier genotype is represented by \(2pq\):
\(2pq = 2 \times 0.8 \times 0.2 = 0.32\).
Expressed to three decimal places, the carrier frequency is \(0.320\).

(b) For the Hardy-Weinberg principle to hold, the following assumptions must be met: 1) The population is extremely large; 2) Mating is entirely random; 3) There is no mutation; 4) There is no immigration or emigration (no gene flow); and 5) There is no selection pressure favoring one genotype over another.

(c) Although the homozygous recessive sheep (genotype \(qq\)) are removed, the recessive allele remains hidden in the gene pool within heterozygous carrier sheep (genotype \(pq\)). These carriers are phenotypically normal, so they are not selected against or removed by the breeder. They will continue to breed and pass the recessive allele on to their offspring, maintaining it in the population.

Part (a) [4.7 marks]:
- 1 mark: Identifies \(q^2 = 0.04\).
- 1 mark: Calculates recessive allele frequency \(q = 0.2\).
- 1 mark: Calculates dominant allele frequency \(p = 0.8\).
- 1 mark: Uses \(2pq\) to calculate carrier frequency (0.32).
- 0.7 marks: Correctly expresses answer to 3 decimal places (0.320).

Part (b) [3 marks]:
- 1 mark each for any three valid assumptions: large population, random mating, no mutation, no migration, no selection.

Part (c) [3 marks]:
- 1 mark: Recessive alleles are present in heterozygous carriers.
- 1 mark: Heterozygotes do not display symptoms / have a normal phenotype.
- 1 mark: Therefore, carriers are not identified or removed, and can pass on the recessive allele.

(a) Denitrification is the chemical reduction of nitrate ions (\(\text{NO}_3^-\)) into nitrogen gas (\(\text{N}_2\)). This process is carried out by anaerobic denitrifying bacteria. Waterlogging displaces oxygen from the soil pores, creating an anaerobic environment. Because denitrifying bacteria only respire anaerobically, they thrive under these conditions and use nitrates as alternative electron acceptors, dramatically increasing the rate of denitrification.

(b) Nitrate concentration decreased from 28.1 to 2.1 mg dm-3 because:
1) Anaerobic conditions trigger denitrifying bacteria to convert existing nitrates into gaseous nitrogen, which escapes into the atmosphere.
2) Nitrification, which converts ammonium into nitrites and then nitrates, is inhibited because nitrifying bacteria are aerobic and require oxygen, which is absent in waterlogged soils.
Ammonium concentration increased from 12.4 to 18.2 mg dm-3 because:
1) Saprobiontic microorganisms continue to perform ammonification (breaking down dead organic matter, proteins, and urea into ammonium ions), as many decomposers can function anaerobically.
2) The generated ammonium ions accumulate in the soil because they cannot be converted into nitrites or nitrates due to the inhibition of aerobic nitrifying bacteria.

Part (a) [4.7 marks]:
- 1 mark: Denitrification is the conversion of nitrate ions to nitrogen gas.
- 1 mark: It is performed by anaerobic denitrifying bacteria.
- 1 mark: Waterlogging creates anaerobic conditions.
- 1 mark: Denitrifying bacteria utilize nitrate for anaerobic respiration.
- 0.7 marks: Correctly explains that lack of oxygen is what directly drives this process.

Part (b) [6 marks]:
- 1 mark: Nitrate decrease is due to conversion into nitrogen gas by denitrifying bacteria.
- 1 mark: Nitrate is also not replenished because nitrification is inhibited.
- 1 mark: Nitrifying bacteria are aerobic and require oxygen to function.
- 1 mark: Ammonium increase is because ammonification still occurs.
- 1 mark: Saprobionts break down nitrogenous organic compounds into ammonium.
- 1 mark: Ammonium is not converted to nitrate because of the lack of nitrification.

(a) First, calculate the absolute increase in yeast cells:
\(\text{Increase} = 4.80 \times 10^6 - 1.50 \times 10^3 = 4,800,000 - 1,500 = 4,798,500 \text{ cells mL}^{-1}\).
Next, divide this increase by the time duration (24 hours):
\(\text{Rate} = \frac{4,798,500 \text{ cells mL}^{-1}}{24 \text{ hours}} = 199,937.5 \text{ cells mL}^{-1} \text{ hour}^{-1}\).
Expressed in standard form to two decimal places, this is \(2.00 \times 10^5 \text{ cells mL}^{-1} \text{ hour}^{-1}\).

(b) Between 48 and 72 hours, the yeast population size levels off (stationary phase) because the rate of cell division (birth rate) is equal to the rate of cell death. This is due to abiotic factors becoming limiting: 1) depletion of glucose/nutrients in the broth, and 2) accumulation of toxic waste products (such as ethanol, which lowers pH and is toxic to yeast). Biotically, there is intense competition for space and remaining resources.

(c) Intraspecific competition is competition between organisms of the same species, while interspecific competition is between different species. Intraspecific competition regulates population size: if the population exceeds the carrying capacity, intense competition for resources leads to an increased death rate and decreased birth rate. This causes the population to fall. If the population drops below carrying capacity, competition decreases, allowing birth rates to exceed death rates, raising the population size back to carrying capacity (negative feedback).

Part (a) [3.7 marks]:
- 1 mark: Correctly calculates change in cell density (4,798,500).
- 1 mark: Divides by time (24 hours).
- 1 mark: Obtains 199,937.5.
- 0.7 marks: Correct standard form and rounding to two decimal places (\(2.00 \times 10^5\)).

Part (b) [4 marks]:
- 1 mark: Explains that stationary phase means birth rate equals death rate.
- 1 mark: Identifies nutrient depletion (e.g., glucose) as a limiting abiotic factor.
- 1 mark: Identifies accumulation of ethanol/toxic wastes or change in pH as a limiting abiotic factor.
- 1 mark: Identifies competition for resources/space as a factor.

Part (c) [3 marks]:
- 1 mark: Defines difference between intraspecific (same species) and interspecific (different species).
- 1 mark: Explains that exceeding carrying capacity increases competition, raising death rate/lowering birth rate.
- 1 mark: Explains that this forms a negative feedback loop to stabilize the population size.

(a) Geographical isolation prevents gene flow (interbreeding) between the split beetle populations. The environments on the different islands will feature different abiotic conditions and biotic selection pressures (e.g., different food sources, predators). Natural selection acts independently on each population, favoring different alleles that enhance survival in their respective environments. Additionally, random mutations arise separately in each group. Over generations, these differences in allele frequencies accumulate until the two populations develop reproductive isolation, meaning they can no longer interbreed to produce fertile offspring.

(b) A very small starting population contains a limited gene pool and is subject to the founder effect, where allele frequencies differ significantly from the parent population by chance. In small populations, genetic drift has a much more powerful effect. Chance events can cause alleles to become fixed or lost rapidly, accelerating the rate of genetic divergence compared to larger, more buffered populations.

(c) Biologists can bring together sexually mature individuals from both island populations under laboratory conditions and allow them to mate. They would observe whether they successfully produce offspring, and then test if these offspring are viable and fertile. If they cannot produce viable, fertile offspring, they are classified as two distinct species.

Part (a) [4.7 marks]:
- 1 mark: Geographical isolation prevents gene flow / interbreeding.
- 1 mark: Different environments have different selection pressures.
- 1 mark: Natural selection favors different beneficial alleles in each population.
- 1 mark: Random mutations occur independently in each population.
- 0.7 marks: Leads to reproductive isolation (inability to produce fertile offspring).

Part (b) [3 marks]:
- 1 mark: Small population has a smaller gene pool / lower genetic diversity.
- 1 mark: Genetic drift has a larger impact on small populations.
- 1 mark: Alleles can become fixed or lost much faster due to chance events.

Part (c) [3 marks]:
- 1 mark: Attempt cross-breeding of beetles from both populations.
- 1 mark: Determine if they produce offspring.
- 1 mark: Verify that the offspring are viable and fertile (if not, they are separate species).