2023 Edexcel A-Level Biology B (9BI0) 模擬試題連答案詳解

(a) Activation energy is the minimum kinetic energy required for reactant molecules to collide successfully and react. Enzymes bind to substrates to form an enzyme-substrate complex, where they stabilize the unstable high-energy transition state (often by straining bonds or aligning substrates). This reduces the energy input needed, allowing more reactant molecules to have sufficient energy to react at physiological temperatures.

(b)(i) Change in product concentration over the first 20 seconds = \(3.4 - 0.0 = 3.4 \text{ mmol dm}^{-3}\). Rate = \(3.4 \text{ mmol dm}^{-3} / 20 \text{ s} = 0.17 \text{ mmol dm}^{-3} \text{ s}^{-1}\).

(b)(ii) As the reaction proceeds, substrate molecules are converted into products, causing substrate concentration to fall. This reduces the frequency of successful collisions between substrate molecules and enzyme active sites, leading to fewer enzyme-substrate complexes forming per unit time.

(c) Competitive inhibitors bind to the active site and increase the apparent \(K_m\) (affinity decreases), but do not affect \(V_{\text{max}}\) because high substrate concentrations can outcompete the inhibitor. Non-competitive inhibitors bind to an allosteric site and decrease \(V_{\text{max}}\) (overall rate capacity drops), but do not affect the affinity (\(K_m\)) of the unaffected active sites.

(a)
1. Definition of activation energy: minimum energy required to start a reaction (1)
2. Reference to formation of enzyme-substrate complex and stabilization of the transition state / weakening of substrate bonds (1)
3. Increases rate because more molecules have energy exceeding this lower activation threshold at standard physiological temperatures (1)

(b)(i)
1. Correct working showing change in concentration divided by time (e.g. \(3.4 / 20\)) (1)
2. Correct numerical value: \(0.17\) (1)
3. Correct units: \(\text{mmol dm}^{-3}\text{ s}^{-1}\) or equivalent (1)

(b)(ii)
1. Substrate concentration decreases / substrate is used up (1)
2. Fewer successful collisions per unit time / fewer enzyme-substrate complexes (ESCs) formed per second (1)

(c)
1. Competitive: increases \(K_m\) AND does not affect \(V_{\text{max}}\) (1)
2. Non-competitive: decreases \(V_{\text{max}}\) AND does not affect \(K_m\) (1)

(a) Pathogens are engulfed by antigen-presenting cells (APCs) such as macrophages, which present foreign antigens on their surface held in MHC class II proteins. T helper cells with complementary CD4 receptors bind to these presented antigens. This binding activates the T helper cells, causing them to divide by mitosis and release cytokines (e.g., interleukins). These cytokines bind to specific receptors on sensitized B cells (which have already bound the same antigen), stimulating the B cells to undergo clonal expansion and differentiate into antibody-producing plasma cells and memory B cells.

(b) Antibodies are Y-shaped quaternary proteins consisting of two identical heavy chains and two identical light chains held together by disulfide bridges. The variable regions at the tips of the 'Y' form highly specific antigen-binding sites that are complementary in shape and charge to a specific antigen. Because antibodies are bivalent (possess two antigen-binding sites), they can bind to antigens on two different pathogens simultaneously, linking them together into large aggregates (agglutination) to prevent spread. The constant region (Fc region) does not vary between antibodies of the same class and acts as a ligand that binds to specific Fc receptors on phagocytic cells, marking the pathogen for destruction (opsonisation).

(c) Active artificial immunity is achieved by vaccination, where an antigen is introduced to stimulate the host's own immune system to produce antibodies and memory cells, providing long-term protection. Passive artificial immunity involves the direct injection of pre-formed antibodies (such as antivenom or monoclonal antibodies), which provides immediate but temporary protection as no memory cells are formed and the antibodies are eventually broken down.

(a)
1. Antigens are presented on MHC class II molecules by antigen-presenting cells (APCs) (1)
2. T helper cells with complementary receptors (CD4) bind to the antigen-MHC complex (1)
3. Activated T helper cells secrete cytokines / interleukins (1)
4. Cytokines stimulate clonal expansion / mitosis of sensitized B cells into plasma cells / memory cells (1)

(b)
1. Four polypeptide chains (two heavy, two light) joined by disulfide bonds, creating a stable Y-shaped protein (1)
2. Variable regions provide highly specific antigen-binding sites complementary to a particular antigen (1)
3. Bivalent nature (two binding sites) allows binding to two different pathogens at once, causing agglutination (1)
4. Constant region / Fc region binds to receptors on phagocytes, acting as an opsonin to enhance phagocytosis (1)

(c)
1. Active artificial involves vaccination / introduction of antigens to stimulate the body's own antibody and memory cell production (1)
2. Passive artificial involves injecting pre-formed antibodies (e.g. anti-venom) which gives immediate but short-lived protection without memory cell production (1)

(a) Light energy absorbed by photosystem II (PSII) excites electrons, which are passed to an electron transport chain. Light energy also drives the photolysis of water, splitting water molecules into protons (\(\text{H}^+\)), electrons, and oxygen. The electrons from photolysis replace those lost by PSII. As electrons flow down the electron transport chain, energy is released and used to pump protons into the thylakoid space, establishing a proton gradient. Protons diffuse back through ATP synthase to produce ATP (photophosphorylation). At photosystem I (PSI), electrons are excited again and, along with protons, are used to reduce NADP to NADPH.

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

(b)(ii) Glycerate 3-phosphate (GP) is reduced to glyceraldehyde 3-phosphate (GALP). This reduction reaction requires hydrogen and electrons supplied by reduced NADP (NADPH) and energy supplied by the hydrolysis of ATP (both molecules being products of the light-dependent stage). Of the GALP produced, five out of every six molecules are recycled to regenerate ribulose bisphosphate (RuBP) using ATP, while one out of six is used to synthesize organic molecules such as glucose, lipids, and amino acids.

(c) A deficiency in magnesium ions reduces the rate of photosynthesis. Magnesium is a vital constituent of chlorophyll molecules. Without magnesium, plants suffer from chlorosis (yellowing of leaves) because they cannot synthesize chlorophyll. This reduces light absorption, decreasing the rate of photolysis and photophosphorylation, which in turn limits the ATP and reduced NADP available for the light-independent stage.

(a)
1. Light excites electrons in PSII which move down an electron transport chain (1)
2. Photolysis uses light energy to split water into \(\text{H}^+\), electrons, and \(\text{O}_2\), replacing electrons lost from PSII (1)
3. Proton gradient drives ATP synthesis (photophosphorylation) via ATP synthase (1)
4. Electrons and protons reduce NADP to NADPH at PSI (1)

(b)(i)
1. Stroma (1)

(b)(ii)
1. GP is reduced to GALP using reduced NADP and energy from ATP (1)
2. 5/6 of GALP is used to regenerate RuBP (1)
3. 1/6 of GALP is used to synthesize biological molecules like glucose / lipids / proteins (1)

(c)
1. Decreases rate of photosynthesis because magnesium is a component of chlorophyll / essential for chlorophyll synthesis (1)
2. Less chlorophyll means less light is absorbed, lowering the rate of the light-dependent stage / producing less ATP and NADPH (1)

(a) Directional selection occurs when a change in environmental conditions favors individuals expressing an extreme phenotype, shifting the population mean over time. In contrast, stabilizing selection occurs in a stable environment and selects against extreme phenotypes, favoring the intermediate phenotype and keeping the population mean constant. Directional selection occurs when a population of mice adapts to a newly formed dark lava environment.

(b) A random, spontaneous mutation occurs in a gene (such as Mc1r) regulating melanin production, producing a dark-furred allele. Within the pocket mouse population, variation in fur color (light vs. dark) now exists. On the dark basalt lava flows, predatory birds and mammals hunt visually and act as selection pressures. Light-colored mice are highly visible against the dark substrate and are easily preyed upon. Dark-colored mice are camouflaged, survive longer, and are more likely to reach reproductive age. The surviving dark-furred mice reproduce and pass on their advantageous allele to their offspring, causing the frequency of the dark-fur allele to increase in the population over generations.

(c) The founder effect occurs when a small number of colonizing individuals establish a new isolated population. This founder group represents only a small, non-representative sample of the parent population's gene pool, which greatly reduces genetic diversity. Due to chance, certain alleles may be lost completely, and other alleles may become much more common, increasing the rate of homozygosity and the risk of inbreeding depression.

(a)
1. Directional selection selects for one phenotypic extreme, changing the population mean (1)
2. Stabilizing selection selects for intermediate phenotypes, maintaining the mean (1)
3. Directional selection occurs in the pocket mouse population on dark lava (1)

(b)
1. Mutation occurs in a gene controlling pigment, creating a dark fur allele (1)
2. Variation in fur phenotype exists in the population (1)
3. Selection pressure: visual predators (owls, hawks) (1)
4. Differential survival and reproduction: dark mice are camouflaged, survive, and reproduce, whereas light mice are preyed upon (1)
5. Over generations, the frequency of the dark fur allele increases in the gene pool (1)

(c)
1. Founder effect leads to a significant reduction in genetic diversity / smaller gene pool (1)
2. Allele frequencies change by chance / some alleles are lost completely from the new population (1)

(a) Inner mitochondrial membrane (cristae).

(b) Reduced NAD and reduced FAD are oxidized, releasing high-energy electrons and protons (\(\text{H}^+\)) into the matrix. The electrons are passed along a series of electron carrier proteins in the electron transport chain (ETC) via redox reactions. As electrons flow down the chain, they lose energy, which is actively used by the carriers to pump protons from the matrix across the inner mitochondrial membrane into the intermembrane space, building up a high concentration of protons there (an electrochemical proton gradient).

(c) Protons diffuse down their concentration gradient from the intermembrane space back to the matrix through the channel of ATP synthase (chemiosmosis). This flow of protons provides the proton motive force that drives the rotation of ATP synthase, enabling the phosphorylation of ADP and inorganic phosphate (\(\text{P}_i\)) to form ATP. Oxygen acts as the terminal electron acceptor, combining with electrons leaving the ETC and protons in the matrix to form water (\(\text{H}_2\text{O}\)), which keeps the ETC running and prevents electron back-up.

(d) Oxygen consumption will decrease or stop. Because oligomycin blocks the proton channel of ATP synthase, protons cannot flow back into the matrix. The proton gradient becomes extremely steep, preventing further active pumping of protons by the electron transport chain because the energy required exceeds that released by electron flow. As a result, electron transport stops, and oxygen can no longer accept electrons at the end of the chain, preventing oxygen consumption.

(a)
1. Inner mitochondrial membrane / cristae (1)

(b)
1. Reduced NAD/FAD are oxidized, releasing protons and high-energy electrons (1)
2. Electrons are passed along a series of carrier proteins in the electron transport chain (1)
3. Energy is released as electrons pass down the chain (1)
4. This energy is used to actively pump protons from the matrix into the intermembrane space (1)

(c)
1. Protons diffuse back down their gradient into the matrix through ATP synthase (chemiosmosis) (1)
2. This flow provides energy for ATP synthase to phosphorylate ADP and Pi to ATP (1)
3. Oxygen acts as the terminal electron acceptor, combining with electrons and protons to form water, keeping the ETC flowing (1)

(d)
1. Oxygen consumption stops / decreases significantly (1)
2. Blocking proton flow halts the ETC because the proton gradient becomes too steep for further pumping; thus, no electrons reach oxygen (1)

(a) Transcription factors bind to specific promoter or enhancer regions of DNA. Activators assist transcription by helping RNA polymerase bind to the promoter, facilitating the assembly of the transcription initiation complex. Repressors inhibit transcription by physically blocking RNA polymerase from binding to the promoter or preventing its movement along the DNA template.

(b) Both DNA methylation and histone acetylation are epigenetic mechanisms that regulate gene expression by modifying chromatin architecture without altering the underlying nucleotide sequence. DNA methylation involves adding methyl (\(-\text{CH}_3\)) groups to cytosine bases (often in CpG islands) of DNA, which directly blocks transcription factors or recruits methyl-binding proteins to condense chromatin, suppressing transcription. In contrast, histone acetylation involves adding acetyl (\(-\text{COCH}_3\)) groups to lysine residues on histone tails, neutralizing their positive charge. This weakens the attraction between histones and negatively charged DNA, opening up the chromatin into euchromatin, which makes the DNA accessible to RNA polymerase and transcription factors, promoting transcription.

(c) Hypermethylation of the promoter region of the retinoblastoma (RB) gene would lead to chromatin condensation (heterochromatin) and silence the gene. Consequently, the tumor suppressor protein RB is not transcribed or translated. Without this regulatory protein, cell cycle checkpoints are lost, allowing cells to divide uncontrollably by mitosis, which leads to tumor formation.

(a)
1. Transcription factors bind to specific DNA sequences / promoter / enhancer regions (1)
2. Activators recruit or assist RNA polymerase binding to initiate transcription (1)
3. Repressors block RNA polymerase binding or movement to inhibit transcription (1)

(b)
1. Both alter chromatin structure / accessibility of DNA to transcription machinery without changing the base sequence (1)
2. DNA methylation involves adding methyl groups to cytosine bases, whereas histone acetylation adds acetyl groups to lysine on histones (1)
3. Methylation leads to gene silencing / closed chromatin, whereas acetylation leads to gene activation / open chromatin (1)
4. DNA methylation physically blocks transcription factor binding or recruits deacetylases (1)
5. Histone acetylation neutralizes positive charges on histones, reducing attraction to DNA to create euchromatin (1)

(c)
1. Hypermethylation of the promoter silences / represses the tumor suppressor gene (1)
2. Loss of the tumor suppressor protein removes checkpoints, leading to uncontrolled mitosis / cell division and tumor growth (1)

(a) The serial dilution process comprises four successive 1 in 10 dilutions:
- Dilution A: \(10^{-1}\)
- Dilution B: \(10^{-2}\)
- Dilution C: \(10^{-3}\)
- Dilution D: \(10^{-4}\)
Since 0.1 cm³ of dilution D is plated, the total volume of original stock represented on the plate is:
\(10^{-4} \times 0.1 \text{ cm}^3 = 10^{-5} \text{ cm}^3\).
Because 142 colonies grew, the concentration in the original culture is:
\(142 / 10^{-5} = 1.42 \times 10^7 \text{ viable cells cm}^{-3}\).

(b) Aseptic techniques are essential to prevent contamination of the bacterial cultures with unwanted microbes from the air or surfaces, ensuring results are valid, and to prevent the escape and exposure of laboratory personnel to potentially pathogenic bacteria.
Precautions:
1. Flame the necks of all culture tubes and bottles immediately before and after transferring liquid, creating a convection current that carries microbes away from the opening.
2. Work close to a lit Bunsen burner to utilize the upward convective air current, which keeps the immediate air zone free of falling microbes.
3. Hold the agar plate lid at an angle over the base when inoculating, rather than removing it completely, to shield the agar from airborne contaminants.

(c) A haemocytometer count measures all cells present (both living and dead cells) directly under a microscope, whereas serial dilution and plating is a viable count method that only counts living cells capable of dividing to form a visible colony. Additionally, a haemocytometer count provides immediate results, whereas the plating method requires incubation (typically 24 hours or more) to yield results.

(a)
1. Identifies dilution factor of dilution D as \(10^{-4}\) (1)
2. Multiplies by plated volume (0.1 cm³) to find total dilution of \(10^{-5}\) (or calculates \(142 / 0.1 = 1420\) cells per cm³ of dilution D) (1)
3. Correct calculation: \(1.42 \times 10^7\) (1)
4. Correct units: \(\text{viable cells cm}^{-3}\) or \(\text{CFU cm}^{-3}\) (or per cm³) (1)

(b)
1. Explains need: prevent contamination of culture with foreign microbes AND prevent contamination of environment/operator (1)
2. Precaution 1: Flame neck of tubes/bottles before/after opening (1)
3. Precaution 2: Work in the sterile zone of a lit Bunsen burner (1)
4. Precaution 3: Tilt agar plate lid over base / do not remove completely (1)
(Accept other valid precautions e.g. sterilize loop in flame, wipe bench with disinfectant)

(c)
1. Haemocytometer counts both living and dead cells, while plating counts only living (viable) cells (1)
2. Haemocytometer gives immediate results, while plating requires incubation (1)

(a) Non-overlapping means that each nucleotide base in a gene is part of only one triplet codon; the ribosome reads codons sequentially as discrete triplets without sharing any bases with neighboring codons. Degenerate means that more than one codon can code for the same amino acid, as there are 64 possible codons and only 20 amino acids. This degeneracy provides protection against mutation, because a single base substitution (point mutation) may result in a different codon that still codes for the exact same amino acid (a silent mutation), preserving protein structure.

(b) Similarities:
- Both processes require the unwinding of the DNA double helix by breaking hydrogen bonds between complementary bases.
- Both processes use a DNA strand as a template to build a complementary polynucleotide chain using complementary base-pairing rules.
- Both form phosphodiester bonds catalyzed by polymerase enzymes (DNA polymerase vs RNA polymerase).
Differences:
- DNA replication copies the entire genome, whereas transcription only copies a specific gene/segment of DNA.
- Replication produces a double-stranded DNA molecule, whereas transcription produces a single-stranded RNA molecule (mRNA, tRNA, or rRNA).
- Replication uses DNA nucleotides (containing deoxyribose and thymine), while transcription uses RNA nucleotides (containing ribose and uracil).

(c) Transfer RNA (tRNA) acts as an adapter molecule. Each tRNA molecule has a specific three-base sequence at one end called an anticodon, which is complementary to a specific codon on mRNA. At its 3' end, tRNA carries the specific amino acid that corresponds to its anticodon. During translation inside the ribosome, the tRNA anticodon binds via hydrogen bonding to the complementary codon on mRNA, ensuring that amino acids are brought to the ribosome and aligned in the correct sequence to build the polypeptide chain.

(a)
1. Non-overlapping: each base is read once / is part of only one triplet codon (1)
2. Degenerate: more than one codon can code for the same amino acid (1)
3. Advantage: point mutations may be silent / not change the amino acid sequence (1)

(b)
1. Similarity: both use DNA templates / complimentary base pairing / form phosphodiester bonds (1)
2. Similarity: both require DNA unwinding / breaking of hydrogen bonds (1)
3. Difference: replication uses DNA polymerase / DNA nucleotides, whereas transcription uses RNA polymerase / RNA nucleotides (1)
4. Difference: replication copies whole genome / double-stranded DNA, whereas transcription copies a gene / single-stranded RNA (1)
(Award max 2 marks for similarities and max 2 marks for differences)

(c)
1. tRNA contains a specific anticodon complementary to an mRNA codon (1)
2. tRNA carries a specific amino acid at its binding site (1)
3. tRNA aligns amino acids in the correct sequence by complementary base pairing with mRNA during translation (1)

### Solution

**(a)**
Oxygen acts as the terminal electron acceptor in the electron transport chain (ETC). It combines with electrons and protons (\(\text{H}^+\)) to form water (\(\text{H}_2\text{O}\)). Without oxygen, the electron transport chain cannot function because electron carriers remain in their reduced state and cannot accept further electrons. Consequently, the pumping of protons from the matrix into the intermembrane space stops, preventing the generation of the proton gradient necessary to power ATP synthase.

**(b)**
Compound Z inhibits the \(\text{F}_0\) subunit of ATP synthase, blocking the channel through which protons diffuse down their electrochemical gradient into the mitochondrial matrix. Because protons cannot pass through ATP synthase, ADP cannot be phosphorylated to ATP, causing ATP synthesis to decrease. As a result, the proton concentration in the intermembrane space remains extremely high, making the proton gradient too steep for the ETC complexes to pump more protons against it. This back-pressure stalls the ETC, slowing down electron transfer and leading to a significant decrease in oxygen consumption.

**(c)**
Adding DNP at 10 minutes will increase the rate of oxygen consumption back to a high level, but ATP synthesis will remain extremely low or zero. DNP acts as an uncoupler, transporting protons directly across the inner mitochondrial membrane into the matrix, bypassing ATP synthase. This dissipates the proton gradient, allowing the electron transport chain to resume pumping protons and consuming oxygen at a high rate. However, because the proton gradient is constantly dissipated and ATP synthase is still inhibited by Compound Z, there is no proton flow through ATP synthase to drive the phosphorylation of ADP, so ATP synthesis does not recover.

### Marking Scheme

**(a) [Maximum 2 marks]**
* **MP1:** Oxygen acts as the terminal electron acceptor / combines with protons and electrons to form water. (1)
* **MP2:** In the absence of oxygen, the electron transport chain (ETC) stops / electron carriers remain reduced, so no proton gradient is established to power ATP synthase. (1)
* *Accept: No proton motive force is created.*
* *Reject: Anaerobic respiration occurs (irrelevant to the specific mechanism of ATP synthase shutdown here).*

**(b) [Maximum 4 marks]**
* **MP1:** Compound Z blocks the flow of protons (\(\text{H}^+\)) through ATP synthase into the matrix. (1)
* **MP2:** This prevents the phosphorylation of ADP to ATP, decreasing the rate of ATP synthesis. (1)
* **MP3:** The accumulation of protons in the intermembrane space causes the electrochemical gradient / proton motive force to become extremely high. (1)
* **MP4:** The electron transport chain cannot pump protons against this steep gradient, halting/slowing electron transport and decreasing oxygen consumption. (1)
* *Accept: Electron carriers remain reduced because protons cannot be pumped.*

**(c) [Maximum 4 marks]**
* **MP1 (Prediction):** Rate of oxygen consumption increases / returns to a high rate, whereas the rate of ATP synthesis remains very low / zero. (1)
* **MP2 (Mechanism - Oxygen):** DNP allows protons to cross the inner mitochondrial membrane directly (into the matrix) bypassing ATP synthase. (1)
* **MP3 (Mechanism - Oxygen):** This dissipates/reduces the proton gradient, allowing the ETC to resume pumping protons and consuming oxygen. (1)
* **MP4 (Mechanism - ATP):** No ATP is synthesised because ATP synthase is still blocked / the proton gradient is bypassed, so there is no proton flow through ATP synthase to drive ADP phosphorylation. (1)

(a) 1. Rate of water uptake is equal to the rate of transpiration. 2. The plant is healthy and does not use a significant volume of water for photosynthesis or turgor. (b) 1. Xylem walls are thickened with lignin, providing mechanical strength to prevent collapse under tension. 2. They are dead, hollow tubes without end walls, forming continuous columns to reduce resistance to water flow. (c) 1. Radius: r = 0.4 mm. 2. Volume of cylinder: V = \pi \times r^2 \times h = \pi \times 0.4^2 \times 45 = 22.619 mm³. 3. Rate of uptake: 22.619 / 5 = 4.5238 mm³ min⁻¹. 4. To 2 significant figures: 4.5 mm³ min⁻¹.

Part (a): (Max 2 marks) 1 mark for stating that water uptake is assumed to equal transpiration rate. 1 mark for stating that water used in metabolic processes/photosynthesis is assumed to be negligible. Part (b): (Max 4 marks) 1 mark for mentioning lignin deposition in walls. 1 mark for stating lignin provides structural support to resist negative pressure/tension. 1 mark for mentioning the absence of end walls/continuous hollow tube. 1 mark for describing how cohesion-tension pulls water without vessel collapse. Part (c): (Max 4 marks) 1 mark for calculating correct radius of 0.4 mm. 1 mark for using the cylinder volume formula correctly (\pi \times 0.4^2 \times 45 = 22.62 mm³). 1 mark for dividing the volume by 5 minutes. 1 mark for the correct final answer of 4.5 (mm³ min⁻¹) to 2 significant figures.

(a) Bacteriostatic antibiotics prevent the reproduction and growth of bacteria, whereas bactericidal antibiotics directly kill the bacterial cells. (b) Streptomycin binds to the 30S ribosomal subunit, causing conformational changes that lead to the misreading of mRNA codons during translation. This results in the insertion of incorrect amino acids and the production of non-functional or toxic proteins. These abnormal proteins disrupt cell membrane integrity and metabolic pathways, causing cell death. (c) Penicillin prevents the cross-linking of peptidoglycan in the cell wall. Active growth requires cell wall synthesis, so penicillin weakens the newly formed wall, causing lysis due to osmotic pressure. Dormant bacteria do not synthesise new cell walls, making them unaffected. Streptomycin targets protein synthesis, which continues to be necessary for basic cellular maintenance and metabolism in both growth phases.

Part (a): (Max 2 marks) 1 mark for bacteriostatic inhibiting growth/reproduction. 1 mark for bactericidal killing bacteria. Part (b): (Max 4 marks) 1 mark for binding to 30S subunit. 1 mark for causing misreading of mRNA codons. 1 mark for production of faulty/non-functional proteins. 1 mark for disruption of cell membrane/essential processes leading to cell death. Part (c): (Max 4 marks) 1 mark for identifying that penicillin inhibits peptidoglycan cross-linking. 1 mark for linking cell wall synthesis in growing bacteria to osmotic lysis. 1 mark for stating dormant bacteria do not synthesise new cell walls. 1 mark for explaining that protein synthesis is essential for basic metabolism in both active and dormant states.

(a) T helper cells are activated when their receptors bind to antigens presented on MHC Class II molecules of antigen-presenting cells. They then release cytokines (such as interleukins). These cytokines bind to receptors on clonal-selected B cells, triggering clonal expansion (mitosis) and differentiation into plasma cells and memory cells. (b) IgG is a Y-shaped glycoprotein with two heavy and two light polypeptide chains held by disulfide bridges. It has specific variable regions (Fab) that bind to a specific antigen, and a constant region (Fc). It is bivalent (two binding sites). For agglutination: the two binding sites allow it to bind to two pathogens simultaneously, clumping them. For opsonisation: the constant Fc region binds to receptors on phagocytes, marking the pathogen for phagocytosis. (c) The primary response has a longer lag phase and produces a lower concentration of antibodies. The secondary response is much faster and produces a higher concentration of antibodies due to the presence of memory cells.

Part (a): (Max 3 marks) 1 mark for T helper cell binding to antigen-MHC Class II complex. 1 mark for release of cytokines/interleukins. 1 mark for cytokines stimulating B cell clonal expansion/differentiation. Part (b): (Max 5 marks) 1 mark for describing the Y-shaped structure with 4 chains and disulfide bonds. 1 mark for identifying the variable region as complementary to a specific antigen. 1 mark for explaining bivalency (two antigen-binding sites). 1 mark for linking agglutination to cross-linking multiple pathogens. 1 mark for linking opsonisation to the constant region binding to phagocyte receptors. Part (c): (Max 2 marks) 1 mark for secondary response being faster/having a shorter lag phase. 1 mark for secondary response achieving a higher peak antibody concentration.

(a) Accessory pigments (e.g., chlorophyll b, carotenoids) absorb wavelengths of light that chlorophyll a cannot absorb. They transfer this light energy to the reaction centre (chlorophyll a), increasing the efficiency of light harvesting. (b) Photophosphorylation is the synthesis of ATP from ADP and Pi using light energy. In non-cyclic, light excites electrons in PSII, which pass through an electron transport chain (ETC) to PSI, generating a proton gradient to produce ATP via ATP synthase. Electrons from PSI are excited and transferred to NADP+ to form NADPH. Water photolysis resupplies PSII. In cyclic, electrons from PSI are recycled back through the ETC, generating a proton gradient for ATP synthesis, but no NADPH or photolysis occurs. (c) At high light intensities, light is no longer limiting. Limiting factors include CO2 concentration (limiting carbon fixation by RuBisCO), temperature (affecting enzyme kinetics), and the availability of enzymes/substrates (e.g., RuBP).

Part (a): (Max 2 marks) 1 mark for absorbing alternative wavelengths of light. 1 mark for transferring energy to the reaction centre / chlorophyll a. Part (b): (Max 5 marks) 1 mark for electron excitation by light. 1 mark for ETC driving H+ transport into the thylakoid space to create a proton gradient. 1 mark for H+ flow through ATP synthase producing ATP. 1 mark for non-cyclic involving PSII and PSI, photolysis of water, and production of ATP and NADPH. 1 mark for cyclic involving PSI only, returning electrons to the ETC, and producing only ATP. Part (c): (Max 3 marks) 1 mark each for any three of: carbon dioxide concentration, temperature, RuBisCO concentration/activity, or water availability.

(a) Genetic variation exists in the Agrostis capillaris population due to mutation, resulting in some plants having alleles for copper tolerance. On contaminated soil, copper acts as a selection pressure; non-tolerant plants die, while tolerant plants survive, reproduce, and pass on their tolerant alleles. Over time, the allele frequency increases. On uncontaminated pastures, there is no copper selection pressure. Non-tolerant plants have a competitive advantage because maintaining tolerance mechanisms carries a metabolic/energy cost, so tolerant plants are outcompeted. (b) To compare distributions: 1. Set up a line transect from the mine boundary into the pasture. 2. Place quadrats at regular intervals (e.g., every 5 metres) along the transect line. 3. Estimate percentage cover of each phenotype within each quadrat. 4. Control variables such as quadrat size and sampling season. 5. Record environmental variables (e.g., soil copper levels) and repeat the transect multiple times to calculate means.

Part (a): (Max 5 marks) 1 mark for identifying that genetic variation arises from mutation. 1 mark for copper acting as a selection pressure. 1 mark for survival and reproduction of tolerant individuals passing on the allele. 1 mark for increase in the frequency of the copper-tolerant allele. 1 mark for explaining that non-tolerant plants outcompete tolerant ones in clean soil due to the metabolic cost of tolerance. Part (b): (Max 5 marks) 1 mark for setting up a line/belt transect away from the mine. 1 mark for placing quadrats at systematic/regular intervals. 1 mark for estimating percentage cover or frequency. 1 mark for standardising quadrat size / experimental conditions. 1 mark for repeating the transect to ensure reliability / measuring soil copper levels to correlate with abundance.

(a) Cristae increase the surface area of the inner mitochondrial membrane, allowing for a greater number of electron transport chain carrier proteins and ATP synthase complexes, which increases the rate of ATP synthesis. (b) 1. NADH and FADH2 donate high-energy electrons to the ETC. 2. As electrons move through carrier proteins, energy is released. 3. This energy is used to pump protons (H+) from the matrix into the intermembrane space, creating an electrochemical proton gradient. 4. Protons diffuse down this gradient back into the matrix through ATP synthase (chemiosmosis). 5. This proton flow drives the phosphorylation of ADP and Pi to ATP. (c) DNP dissipates the proton gradient by allowing protons to bypass ATP synthase. Consequently, ATP production decreases dramatically, and the unused potential energy of the gradient is released as heat, increasing body temperature.

Part (a): (Max 2 marks) 1 mark for stating that cristae increase surface area. 1 mark for explaining that this accommodates more electron carriers / ATP synthase. Part (b): (Max 6 marks) 1 mark for NADH/FADH2 oxidation donating electrons. 1 mark for electrons moving through redox carriers in the ETC. 1 mark for energy from electron flow pumping protons. 1 mark for protons entering the intermembrane space. 1 mark for establishing a proton/electrochemical gradient. 1 mark for protons diffusing through ATP synthase to phosphorylate ADP. Part (c): (Max 2 marks) 1 mark for predicting decreased ATP production due to a lost/weakened proton gradient. 1 mark for predicting increased body temperature because energy is dissipated as heat.

(a) DNA methylation adds methyl groups to cytosine bases in DNA (CpG islands). This blocks transcription factors from binding to promoters and recruits proteins that condense chromatin, silencing transcription. Histone acetylation adds acetyl groups to lysine residues on histone tails. This neutralises their positive charge, weakening their interaction with negatively charged DNA. The chromatin loosens (forming euchromatin), allowing RNA polymerase and transcription factors access to initiate transcription. (b) Normal MECP2 binds to methylated CpG sites and recruits HDACs, which remove acetyl groups from histones, condensing chromatin and silencing genes. A non-functional MECP2 protein cannot recruit HDACs. Consequently, histones remain acetylated, chromatin remains open (euchromatin), and target genes are continuously transcribed/over-expressed instead of being properly silenced.

Part (a): (Max 6 marks) 1 mark for methylation adding methyl groups to cytosine. 1 mark for methylation blocking transcription factor/RNA polymerase binding. 1 mark for DNA methylation leading to gene silencing. 1 mark for acetylation adding acetyl groups to histones. 1 mark for acetylation reducing positive charges, weakening histone-DNA attraction. 1 mark for chromatin loosening/opening (euchromatin) to promote transcription. Part (b): (Max 4 marks) 1 mark for normal MECP2 recruiting HDACs to deacetylate/condense chromatin. 1 mark for explaining that non-functional MECP2 fails to recruit HDACs. 1 mark for stating that histones remain acetylated and chromatin remains open. 1 mark for describing the continuous transcription / failure of target gene silencing.

(a) A double circulatory system separates pulmonary and systemic circuits. It allows blood to be pumped at low pressure to the lungs (preventing damage to delicate capillaries) and re-pressurised by the heart to high pressure for systemic circulation. High pressure ensures rapid oxygen/nutrient delivery to active tissues. (b) During ventricular systole, the contracting left ventricle generates pressure higher than the left atrium, forcing the atrioventricular (bicuspid) valve to close. As ventricular pressure rises further and exceeds aortic pressure, the semi-lunar (aortic) valve is forced open, allowing blood to enter the aorta. (c) At the arterial end: Net filtration pressure = 4.3 - 3.2 = +1.1 kPa. The positive pressure drives net fluid movement OUT of the capillary (filtration). At the venous end: Net filtration pressure = 1.6 - 3.2 = -1.6 kPa. The negative pressure drives net fluid movement INTO the capillary (absorption).

Part (a): (Max 3 marks) 1 mark for separation of pulmonary and systemic circuits. 1 mark for low pulmonary pressure preventing damage to alveoli/capillaries. 1 mark for high systemic pressure ensuring rapid delivery to metabolising tissues. Part (b): (Max 4 marks) 1 mark for stating ventricular pressure is higher than atrial pressure. 1 mark for stating this closes the AV valve. 1 mark for stating ventricular pressure is higher than aortic pressure. 1 mark for stating this opens the semi-lunar valve. Part (c): (Max 3 marks) 1 mark for arterial end net pressure calculation (+1.1 kPa) and outward movement. 1 mark for venous end net pressure calculation (-1.6 kPa) and inward movement. 1 mark for showing correct working (subtracting oncotic from hydrostatic).

Part (a):
- Carbon dioxide reacts with water to form carbonic acid, which dissociates into hydrogen ions, lowering the pH of the blood.
- This change is sensed by chemoreceptors located in the carotid bodies, aortic bodies, and the medulla oblongata.
- These receptors send impulses to the cardiovascular control centre in the medulla oblongata, which responds by sending more frequent impulses down the sympathetic (accelerator) nerve to the SAN to increase cardiac output and clear \(CO_2\).

Part (b):
- High arterial blood pressure stretches the walls of the aorta and carotid arteries, activating stretch-sensitive baroreceptors.
- This increases the frequency of action potentials transmitted to the cardiovascular control centre via the glossopharyngeal and vagus sensory pathways.
- The medulla oblongata coordinates a response that increases vagal (parasympathetic) output and reduces sympathetic output.
- The release of acetylcholine at the SAN lowers the heart rate, reducing cardiac output and bringing blood pressure back down to homeostatic levels.

Part (c):
- The key difference lies in the neurotransmitters released and their opposing effects on the membrane potential of the SAN pacemaker cells.
- Noradrenaline (sympathetic) increases the slope of the pacemaker potential (depolarisation) by increasing inward \(Na^+\) and \(Ca^{2+}\) currents, shortening the interval between action potentials.
- Acetylcholine (parasympathetic) increases \(K^+\) permeability, hyperpolarising the SAN cell membrane and reducing the slope of the pacemaker potential, lengthening the interval between action potentials.

Part (a) [Maximum 3 marks]:
1. (Increase in carbon dioxide lowers blood pH / forms carbonic acid) which is detected by chemoreceptors in the carotid artery / aorta / medulla oblongata. (1)
2. Nerve impulses are sent at a higher frequency along sensory neurones to the cardiovascular control centre (in the medulla oblongata). (1)
3. (The cardiovascular centre sends) more impulses along sympathetic / accelerator nerves to the sinoatrial node / SAN. (1)

Part (b) [Maximum 4 marks]:
1. Elevated blood pressure is detected by baroreceptors / stretch receptors (in the wall of the carotid sinus / aortic arch). (1)
2. (This leads to) an increase in the frequency of nerve impulses sent along sensory neurones to the cardiovascular control centre (in the medulla). (1)
3. The cardiovascular control centre increases impulses along the parasympathetic / vagus nerve AND decreases impulses along the sympathetic nerve. (1)
4. Acetylcholine is released (at the SAN) to slow down the rate of depolarisation / increase the time between heartbeats. (1)

Part (c) [Maximum 3 marks]:
1. Sympathetic nerves release noradrenaline WHEREAS parasympathetic nerves release acetylcholine. (1)
2. Noradrenaline increases the rate of depolarisation / SAN firing / decreases threshold potential time WHEREAS acetylcholine hyperpolarises the SAN membrane / decreases the rate of depolarisation / slows SAN firing. (1)
3. Sympathetic stimulation increases calcium / sodium ion influx (into SAN cells) WHEREAS parasympathetic stimulation increases potassium ion efflux / decreases calcium influx. (1)

[Total: 10 Marks]

(a) Incubating at \(30^\circ\text{C}\) prevents the growth of human pathogens (which grow best at human body temperature, \(37^\circ\text{C}\)) and is safer for school/college laboratories.
(b) Dilution of Tube 4 is \(10^{-4}\). Volume plated is \(0.1\text{ cm}^3\). Concentration of cells in Tube 4 is \(142 / 0.1 = 1420\text{ cells cm}^{-3}\). Therefore, original concentration is \(1420 \times 10^4 = 1.42 \times 10^7\text{ CFU cm}^{-3}\) (or \(1.42 \times 10^7\text{ cells cm}^{-3}\).
(c) Without serial dilution, the bacterial concentration would be too high, resulting in overlapping colonies (a lawn of growth) on the agar plate. This makes it impossible to count individual colonies and calculate an accurate concentration.
(d) Use a colorimeter to measure the absorbance/turbidity of the bacterial culture at regular intervals over 24 hours. Calibrate with sterile broth as a blank. Plot a graph of absorbance against time. A limitation of turbidimetry is that it measures both living (viable) and dead bacterial cells, whereas the viable plate count only measures living cells.

(a) 1 mark: To prevent/reduce growth of potential human pathogens / organisms adapted to human body temperature.
1 mark: For safety reasons in a school/college lab environment.

(b) 1 mark: Identifies dilution factor of Tube 4 as \(10^{-4}\) (or 1 in 10000).
1 mark: Correctly divides number of colonies by plated volume (\(142 / 0.1 = 1420\)).
1 mark: Correct final answer in standard form with correct units: \(1.42 \times 10^7\text{ CFU cm}^{-3}\) or \(1.42 \times 10^7\text{ cells cm}^{-3}\) (accept \(1.4 \times 10^7\)).

(c) 1 mark: Prevents colonies from merging/overlapping / avoids a bacterial lawn.
1 mark: Allows individual, distinct colonies to be seen and counted accurately.

(d) 1 mark: Use a colorimeter / spectrophotometer to measure light transmission / absorbance.
1 mark: Take measurements at regular time intervals (e.g., every 2 hours) and use sterile nutrient broth to calibrate the colorimeter (blank).
1 mark: Plot a growth curve of absorbance against time.
1 mark: Limitation: Turbidimetry counts both dead and living cells (or counts non-bacterial debris), leading to an overestimation of viable cells.

(a) \(Q_{10} = \frac{\text{Rate at } (T + 10)^\circ\text{C}}{\text{Rate at } T^\circ\text{C}} = \frac{3.6}{1.8} = 2.0\).
(b) Between \(25^\circ\text{C}\) and \(35^\circ\text{C}\), increasing temperature increases the kinetic energy of enzyme and substrate molecules. They move faster, resulting in more frequent successful collisions and the formation of more enzyme-substrate complexes. Between \(35^\circ\text{C}\) and \(55^\circ\text{C}\), the rate decreases rapidly because high temperatures break hydrogen and ionic bonds holding the tertiary structure of the enzyme together. This alters the shape of the active site, denaturing the enzyme so the substrate no longer fits.
(c) 1. Substrate (starch) concentration: controlled by using the same volume and concentration of stock starch solution for each trial. 2. pH: controlled by adding a specific buffer solution (e.g., pH 7.0) to the reaction mixture.

(a) 1 mark: Correct formula or substitution shown (e.g., \(3.6 / 1.8\)).
1 mark: Correct final answer: \(2.0\) (or 2).

(b) 1 mark: (Between 25 and 35) Increased kinetic energy leads to faster movement of molecules.
1 mark: More frequent successful collisions / more enzyme-substrate complexes formed.
1 mark: (Between 35 and 55) Heat breaks hydrogen / ionic bonds within the enzyme's tertiary structure.
1 mark: Active site changes shape / becomes denatured.
1 mark: Substrate is no longer complementary / cannot bind to active site.

(c) 1 mark: Identify variable 1 (e.g., pH) and 1 mark for control method (e.g., use a buffer solution).
1 mark: Identify variable 2 (e.g., enzyme concentration / substrate concentration / volume of mixture) and 1 mark for control method (e.g., use the same volume/concentration from the same stock source).

(a) \(R_f = \frac{\text{Distance moved by spot}}{\text{Distance moved by solvent}} = \frac{11.7}{12.4} = 0.9435... \approx 0.94\).
(b) Spot D is Carotene (or Carotenoids). It has a high \(R_f\) because it is highly soluble in the mobile phase (non-polar solvent) and has low affinity for/adsorption to the stationary phase (polar chromatography paper).
(c) 1. Draw the origin line in pencil (not ink) so it does not dissolve and run. 2. Ensure the solvent level is below the pencil line so the pigments do not wash off into the solvent. 3. Avoid touching the paper with bare hands to prevent transfer of skin lipids.
(d) Extract pigments from equal masses of leaves of both species, run chromatography for both under identical conditions, and compare the presence, intensity, and relative abundance of different pigments (e.g. Chlorophyll a vs b ratio, carotenoids).

(a) 1 mark: Correct substitution: \(11.7 / 12.4\).
1 mark: Correct answer rounded to 2 decimal places: \(0.94\).

(b) 1 mark: Identifies Carotene / Carotenoid.
1 mark: Explains that it is highly soluble in the mobile phase / solvent.
1 mark: Explains that it has low adsorption/affinity to the stationary phase / paper.

(c) Award 1 mark per valid precaution (max 3):
- Draw start line in pencil (not ink).
- Ensure solvent level is below the start line.
- Hold paper only by the edges / wear gloves to avoid contamination with finger oils.
- Ensure the paper is vertical / does not touch the sides of the tube/beaker.
- Cover the container with a lid to saturate the atmosphere with solvent vapor.

(d) 1 mark: Extract pigments from both types of plants using the same solvent and volume.
1 mark: Run chromatograms simultaneously under the same conditions (same solvent, temperature).
1 mark: Compare the \(R_f\) values and the size/intensity of the spots to assess the ratio of chlorophyll b (absorbs shade wavelengths) to chlorophyll a / carotenoids.

(a) Diameter = \(0.80\text{ mm}\), so radius \(r = 0.40\text{ mm}\).
Distance \(h = 45\text{ mm}\).
Volume \(V = \pi \times 0.40^2 \times 45 = 22.6195\text{ mm}^3\).
Time = \(15\text{ minutes} = 0.25\text{ hours}\).
Rate = \(22.6195 / 0.25 = 90.478... \approx 90.5\text{ mm}^3\text{ h}^{-1}\).
(b) Not all water absorbed is transpired. Some water is used in photosynthesis, and some is used to maintain cell turgidity.
(c) Use a fan placed at fixed distances from the plant to vary the wind speed. Use an anemometer to measure and verify the wind speed at each distance. Keep other factors constant: use a lamp for constant light intensity, perform in the same room for constant temperature and humidity. Allow the shoot to acclimate at each wind speed before measuring bubble movement.
(d) The air spaces in the spongy mesophyll are saturated with water vapor (high water potential), whereas the air outside the leaf typically has a lower water vapor concentration (lower water potential), creating a water vapor potential gradient out of the leaf.

(a) 1 mark: Correct radius \(r = 0.40\text{ mm}\) used in volume calculation: \(\pi \times 0.4^2 \times 45 = 22.62\text{ mm}^3\).
1 mark: Correctly scales time from 15 mins to 1 hour (multiplies by 4).
1 mark: Correct final answer to 3 significant figures: \(90.5\text{ mm}^3\text{ h}^{-1}\).

(b) 1 mark: Water is used in photosynthesis / metabolic reactions.
1 mark: Water is retained in cells to maintain turgor pressure / plant structure.

(c) 1 mark: Use an electric fan to generate wind and vary speed by altering the distance of the fan (or using different fan settings).
1 mark: Measure wind speed using an anemometer.
1 mark: Control other environmental factors (e.g., constant light using a lamp, constant room temperature).
1 mark: Allow time for the plant to adjust/acclimate to each new wind speed before recording.

(d) 1 mark: Evaporation of water from cell walls into the air spaces of spongy mesophyll creates high humidity/water vapor concentration inside the leaf.
1 mark: Lower water vapor concentration/relative humidity outside the leaf creates a concentration/diffusion gradient through the stomata.

(a) Total \(N = 18 + 12 + 5 + 25 = 60\).
For each species, calculate \(\left(\frac{n}{N}\right)^2\):
- Meadow buttercup: \((18/60)^2 = 0.0900\)
- Red clover: \((12/60)^2 = 0.0400\)
- Dandelion: \((5/60)^2 = 0.0069\)
- Ribwort plantain: \((25/60)^2 = 0.1736\)
\(\sum \left(\frac{n}{N}\right)^2 = 0.0900 + 0.0400 + 0.0069 + 0.1736 = 0.3106\).
\(d = 1 - 0.3106 = 0.6894 \approx 0.689\).
(b) Meadow A has a higher Simpson's Index of Diversity (0.689 vs 0.482), indicating higher biodiversity. High diversity means the ecosystem is more stable and resilient to environmental changes (such as disease or drought), as the loss of one species is less likely to disrupt the food web or community structure.
(c) 1. Set up a grid system over the meadow using two tape measures at right angles. 2. Use a random number generator to obtain coordinates. 3. Place a quadrat (e.g., \(0.5\text{ m} \times 0.5\text{ m}\)) at the generated coordinates. 4. Count the number of individuals of each species inside the quadrat. 5. Repeat this process many times (at least 10–20 times) to ensure a representative sample and calculate a reliable estimate.

(a) 1 mark: Correctly calculates \(N = 60\).
1 mark: Correct sum of \((n/N)^2 = 0.311\) (or individually calculated values shown).
1 mark: Correct final calculation of \(d\) to 3 decimal places: \(0.689\).

(b) 1 mark: Meadow A has higher biodiversity than Meadow B (0.689 vs 0.482).
1 mark: Meadow A is more stable / less susceptible to changes (e.g. disease, climate change).
1 mark: If one species is affected, others can survive and ecological niches remain filled.

(c) 1 mark: Establish a coordinate grid system using tape measures at right angles.
1 mark: Use a random number generator/table to select coordinates (prevents bias).
1 mark: Use a quadrat of standard size (e.g., \(0.25\text{ m}^2\) or \(1\text{ m}^2\)).
1 mark: Count individuals / estimate percentage cover of each species within each quadrat.
1 mark: Repeat with a large number of quadrats (minimum 10) to calculate a representative mean.

(a) Tidal volume is the volume of air inhaled or exhaled during a single normal breath. Minute ventilation rate = Tidal volume \(\times\) Breathing rate = \(0.55\text{ dm}^3 \times 12\text{ breaths min}^{-1} = 6.6\text{ dm}^3\text{ min}^{-1}\).
(b) The drop in baseline is \(0.35\text{ dm}^3\) in 1 minute. Converting to \(\text{cm}^3\): \(0.35 \times 1000 = 350\text{ cm}^3\text{ min}^{-1}\).
(c) The baseline falls because the subject inhales oxygen from the chamber, but the exhaled carbon dioxide is absorbed by the soda lime canister, reducing the total volume of gas in the chamber. The function of the soda lime is to absorb exhaled carbon dioxide to prevent its accumulation, which would be toxic/stimulate hyperventilation.
(d) 1. Use medical-grade oxygen to fill the chamber. 2. Disinfect the mouthpiece before and after use. 3. Ensure the subject does not have asthma or cardiovascular conditions (health screening).

(a) 1 mark: Definition of tidal volume: volume of air breathed in or out in a single breath at rest.
1 mark: Correct calculation: \(0.55 \times 12 = 6.6\).
1 mark: Correct units: \(\text{dm}^3\text{ min}^{-1}\) (or \(\text{L min}^{-1}\)).

(b) 1 mark: Identifies baseline fall of \(0.35\text{ dm}^3\) represents oxygen consumption in 1 min.
1 mark: Correct conversion to \(\text{cm}^3\text{ min}^{-1}\): \(350\text{ cm}^3\text{ min}^{-1}\).

(c) 1 mark: Oxygen is consumed during respiration and gas volume decreases.
1 mark: Carbon dioxide produced is absorbed (not returned to chamber).
1 mark: Soda lime absorbs carbon dioxide.

(d) Award 1 mark for each safety precaution (max 3):
- Disinfect/sterilise mouthpiece.
- Fill spirometer with fresh, medical-grade oxygen.
- Use a nose clip to ensure no gas leaks.
- Check subject has no underlying respiratory/cardiovascular diseases (asthma/heart issues).
- Ensure the water trap/seal is level.

(a) The student can perform simple dilutions using pond water to dilute the 1.0% stock solution. For example, to prepare \(10\text{ cm}^3\) of each concentration:
- 0.4%: \(4\text{ cm}^3\) of 1% stock + \(6\text{ cm}^3\) of pond water.
- 0.3%: \(3\text{ cm}^3\) of 1% stock + \(7\text{ cm}^3\) of pond water.
- 0.2%: \(2\text{ cm}^3\) of 1% stock + \(8\text{ cm}^3\) of pond water.
- 0.1%: \(1\text{ cm}^3\) of 1% stock + \(9\text{ cm}^3\) of pond water.
- 0.0%: \(10\text{ cm}^3\) of pond water (control).
(b) Place a single Daphnia on a cavity slide. Use cotton wool fibers to immobilize the Daphnia without crushing it. Add a few drops of the test caffeine solution. Do not place a coverslip, allowing oxygen diffusion. Place under low power of a light microscope. Allow 5 minutes for the Daphnia to acclimate. Use a tally counter or make dots on paper to count the heartbeats over a set time period (e.g., 15 seconds) and multiply by 4 to find beats per minute. Control temperature by using a heat shield (e.g., a water bath beaker between the light source and slide) or turn off the light between counts.
(c) Daphnia is suitable because its body transparent, allowing the heart to be viewed directly without invasive dissection. It is more ethical because it has a simpler nervous system with fewer pain receptors, making it less likely to experience suffering compared to mammals.

(a) 1 mark: Describes dilution using pond water (not distilled water, to prevent osmotic shock).
1 mark: Mentions calculating correct proportions of stock and water to make up fixed volumes.
1 mark: Shows correct volumes for at least two concentrations (e.g., for 0.1%: 1 part stock to 9 parts pond water).

(b) 1 mark: Immobilise Daphnia using cotton wool fibers on a cavity slide (accept using a small droplet of water/solution, no coverslip).
1 mark: Allow equilibration/acclimation time (e.g., 5 minutes) in the test solution before counting.
1 mark: Count heartbeats for a set time (e.g., 15 seconds or 30 seconds) using a tally counter / clicker, then calculate beats per minute.
1 mark: Control temperature by using a heat shield / cold light source / turning off microscope bulb between measurements.
1 mark: Repeat measurements with multiple Daphnia (e.g., at least 5) per concentration to calculate a mean.

(c) 1 mark: Transparency allows direct, non-invasive visualization of the heart.
1 mark: Simpler nervous system / lacks complex pain perception.
1 mark: No dissection needed / invertebrates are subject to fewer ethical restrictions than vertebrates.

(a) KOH absorbs the carbon dioxide produced by the yeast during respiration, ensuring that any change in gas volume inside the tube is solely due to the uptake of oxygen.
(b) Since carbon dioxide is absorbed by the KOH, the volume of gas in the tube decreases as oxygen is consumed by the yeast. This decrease in pressure draws the bubble toward the chamber, directly representing the rate of oxygen consumption.
(c) The respiratory quotient (RQ) for glucose is 1.0, meaning the volume of carbon dioxide produced equals the volume of oxygen consumed (\(\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}\)). Without KOH, the carbon dioxide produced replaces the oxygen consumed, so there is no net change in gas volume or pressure, and the bubble does not move.
(d) Lactose is a disaccharide that yeast cannot directly use. Yeast cells lack the enzyme \(\beta\)-galactosidase (lactase) required to hydrolyze lactose into glucose and galactose. Furthermore, yeast lacks the specific transport proteins (permeases) in its cell membrane required to import lactose into the cell. Without transport and hydrolysis, yeast cannot enter glycolysis or aerobic respiration using lactose.

(a) 1 mark: To absorb carbon dioxide.
1 mark: Prevents carbon dioxide from affecting the pressure/gas volume inside the respirometer.

(b) 1 mark: Aerobic respiration consumes oxygen, reducing gas volume inside the tube.
1 mark: Since CO2 is absorbed, the pressure drops, drawing the bubble towards the chamber.

(c) 1 mark: RQ of glucose is 1.0.
1 mark: Volume of CO2 produced is equal to the volume of O2 consumed.
1 mark: Therefore, there is no net change in gas volume/pressure inside the tube, so the bubble remains stationary.

(d) 1 mark: Lactose is a disaccharide (glucose + galactose) which must be hydrolyzed first.
1 mark: Yeast lacks the gene for / cannot synthesize the enzyme lactase / \(\beta\)-galactosidase.
1 mark: Yeast lacks the membrane transport proteins / permeases needed to transport lactose into the cytoplasm.
1 mark: Therefore, lactose cannot undergo glycolysis / respiration.

(a) Aseptic techniques are used to prevent contamination of the agar plates by unwanted environmental microbes and to protect the operator from potential pathogens. Specific practices include flaming the forceps in a Bunsen burner flame before transferring discs, and working close to a Bunsen flame to create an upward current of warm air that prevents airborne contaminants from settling. (b) To perform a two-fold serial dilution: Add 5 cm³ of sterile water to three separate test tubes. Transfer 5 cm³ of the 100% stock solution to the first tube and mix thoroughly to produce a 50% solution. Transfer 5 cm³ of this 50% solution to the second tube to produce a 25% solution. Finally, transfer 5 cm³ of the 25% solution to the third tube to produce a 12.5% solution. (c) Mean diameter = \((14 + 16) / 2 = 15 \text{ mm}\), which gives a radius \(r = 7.5 \text{ mm}\). Using the formula \(\text{Area} = \pi r^2\): \(\text{Area} = \pi \times 7.5^2 = 176.71 \text{ mm}^2\). (d) Incubating at a maximum of 25°C prevents the rapid growth of potential human pathogens that thrive at human body temperature (37°C), maintaining a safer working environment in a school laboratory.

(a) 1 mark: To prevent contamination of the agar plate / escape of pathogens. 1 mark each (max 2 marks) for describing valid aseptic practices (e.g. flaming inoculating loops/forceps, working near Bunsen flame, opening Petri dish lid at an angle). (b) 1 mark: Adding 5 cm³ of sterile water to the tubes. 1 mark: Describing transfer of 5 cm³ of 100% stock to first tube to make 50% solution. 1 mark: Describing sequential transfer of 5 cm³ from 50% to make 25%, and from 25% to make 12.5%. (c) 1 mark: Correct mean diameter of 15 mm or radius of 7.5 mm. 1 mark: Correct formula stated (\(\text{Area} = \pi r^2\) or \(\text{Area} = \pi d^2 / 4\)). 1 mark: Correct calculation of area as 176.7 mm² or 177 mm² (allow 176.71 mm²). (d) 1 mark: Temperatures above 25°C / near body temperature (37°C) encourage the growth of microbes that are pathogenic to humans. 1 mark: 25°C is a safety compromise that allows bacterial growth while minimising risk.

(a) To extract pigments, grind fresh spinach leaves using a pestle and mortar with a small volume of solvent such as acetone. Draw a straight pencil line (the origin) approximately 2 cm from the bottom of the chromatography paper. Use a capillary tube to apply a concentrated spot of the pigment extract onto the line, allowing it to dry between applications to keep the spot small. Suspend the paper vertically in a boiling tube containing the running solvent, ensuring the solvent level is below the pencil line. Seal the tube with a bung to saturate the atmosphere. Remove the paper just before the solvent front reaches the top, and immediately mark the solvent front in pencil. (b) Rf is calculated as distance travelled by pigment divided by distance travelled by solvent front. Pigment A: \(3.4 / 8.5 = 0.40\). Pigment B: \(6.8 / 8.5 = 0.80\). (c) Pigment B has a higher Rf value because it is more soluble in the mobile phase (solvent) and/or has less affinity for the stationary phase (paper). Pigment A has a lower Rf value because it is less soluble in the mobile phase and/or adsorbs more strongly to the stationary phase. (d) Ink contains soluble dyes that would dissolve in the running chromatography solvent and separate up the paper along with the leaf pigments, obscuring the results. Pencil lead (graphite) is insoluble in the solvent and will not run.

(a) 1 mark: Grinding leaves with solvent (acetone/propanone). 1 mark: Drawing pencil line and applying extract as a small, concentrated spot. 1 mark: Suspending paper in solvent ensuring the solvent level is below the origin line. 1 mark: Using a sealed container to saturate the atmosphere with solvent vapour. 1 mark: Removing paper and marking the solvent front before it reaches the top. (b) 1 mark: Pigment A Rf = 0.40 (or 0.4). 1 mark: Pigment B Rf = 0.80 (or 0.8). (c) 1 mark: Pigment B is more soluble in the solvent / has less affinity for paper. 1 mark: Pigment A is less soluble in the solvent / has greater affinity for paper. (d) 1 mark: Ink is soluble in the solvent and would run/separate. 1 mark: Pencil graphite is insoluble and will not run/contaminate the chromatogram.

(a) Place a single Daphnia on a cavity slide using a pipette and remove excess pond water. Add a few strands of cotton wool to restrict its movement. Allow the Daphnia to acclimatise to the solution for 5 minutes. Position the slide under the light microscope and count the heartbeats for 15 seconds using a tally/clicker, then multiply by 4 to find beats per minute (bpm). To ensure ethical treatment, turn off the microscope light between viewings to prevent heating up the Daphnia, and return it to its original culture aquarium immediately after. (b) The standard deviation range for the 0.0% caffeine group is 168 to 192 bpm (180 - 12 to 180 + 12). The range for the 0.5% caffeine group is 212 to 268 bpm (240 - 28 to 240 + 28). Since these ranges do not overlap, the difference in mean heart rate is likely to be statistically significant. (c) Variable 1: Temperature. Controlled by using a water bath or turning off the microscope lamp when not counting to avoid heating. Variable 2: Size or developmental stage of Daphnia. Controlled by selecting Daphnia of similar length/age from the same laboratory culture stock.

(a) 1 mark: Use of cavity slide and cotton wool to restrict movement. 1 mark: Acclimatisation period in the solution before counting. 1 mark: Accurate method for counting (e.g., counting for 15 seconds and multiplying by 4, or using a video recording). 1 mark: Ethical measure described (e.g., turning off the microscope light between observations to prevent thermal stress, or returning Daphnia safely to culture). (b) 1 mark: Correctly calculating the SD range for 0.0% caffeine (168 - 192 bpm). 1 mark: Correctly calculating the SD range for 0.5% caffeine (212 - 268 bpm). 1 mark: Stating that because the standard deviation ranges do not overlap, there is a statistically significant difference. (c) 1 mark for each of two variables identified (e.g., temperature, Daphnia size/source, volume of caffeine solution) (max 2 marks). 1 mark for each corresponding control method linked to that variable (max 2 marks).