Name the location of osmoreceptors.

• 1. Adrenal cortex
• 2. Hypothalamus
• 3. Posterior pituitary

State the stimulus that is detected by osmoreceptors.

• 1. Changes in blood pressure.
• 2. Changes in blood glucose concentration.
• 3. Changes in the water potential of the blood (or blood solute concentration).

Name the structure that secretes ADH into the blood.

• 1. Anterior pituitary (gland).
• 2. Hypothalamus.
• 3. Posterior pituitary (gland).

Data shows that at an ADH concentration of 1 arbitrary unit, the rate of urine production is 13 cm³ min⁻¹. At an ADH concentration of 3 arbitrary units, the rate is 5 cm³ min⁻¹. Calculate the percentage decrease in the rate of urine production between these two ADH concentrations. Show your working and give your answer to one decimal place.

• 1. Result rounded to one decimal place: 61.5%
• 2. Calculation: Percentage decrease = (Decrease / Original rate) × 100% = (8 / 13) × 100%
• 3. Decrease in rate = 13 cm³ min⁻¹ – 5 cm³ min⁻¹ = 8 cm³ min⁻¹
• 4. Calculation: Percentage decrease = (Decrease / Final rate) × 100% = (8 / 5) × 100%

Explain the relationship where increasing ADH concentration decreases the rate of urine production.

• 1. Mechanism: ADH increases the permeability of the collecting duct walls to water.
• 2. Outcome: A smaller volume of more concentrated urine is produced.
• 3. Mechanism: ADH binds to receptors on collecting duct cells, triggering insertion of aquaporins into membranes.
• 4. Mechanism: ADH causes more salt to be reabsorbed, making the urine less concentrated.
• 5. Mechanism: More water is reabsorbed from the collecting duct filtrate into the blood by osmosis.

Describe the main factors that cause phenotypic variation within a population.

• 1. Factor: Environmental conditions experienced by individuals (e.g., diet, climate).
• 2. Factor: Genetic differences (genotype/alleles) inherited by individuals.
• 3. Factor: Acquired characteristics during an organism’s lifetime that are then inherited.
• 4. Interaction: Phenotype often results from genotype-environment interactions.
• 5. Sources of Genetic Variation: Mutation, meiosis (crossing over, independent assortment), random fertilisation.

Outline the theory of evolution.

• 1. It involves changes in the heritable characteristics of populations over time.
• 2. Evolution proposes that species are fixed and do not change over time.
• 3. Key mechanisms include natural selection and genetic drift.
• 4. It is the process by which new species arise from pre-existing species (common ancestry).
• 5. It leads to changes in the genetic makeup (allele frequencies) of populations.

Explain how DNA sequence data is used to show evolutionary relationships between species.

• 1. Bioinformatics tools align sequences and construct phylogenetic trees.
• 2. The more similar the DNA sequences, the more closely related the species are considered (more recent common ancestor).
• 3. Mutations accumulate over time; fewer differences imply more recent divergence.
• 4. Comparing DNA sequences reveals identical gene functions across all species.
• 5. Specific DNA sequences (e.g., mitochondrial DNA, nuclear genes) are compared between species.

The gene L controls colour pattern in tortoise beetles and has four known alleles: Lᵀ, Lᴿ, Lⱽ, and Lʳ. Explain why this is described as involving multiple alleles.

• 1. It involves four distinct genes controlling one trait.
• 2. A single gene (L) exists in the population in more than two different forms (alleles: Lᵀ, Lᴿ, Lⱽ, Lʳ).
• 3. Each beetle possesses four alleles for this gene.

A tortoise beetle with genotype LᵀLᴿ was crossed with another tortoise beetle with genotype LᵀLᴿ. Construct a genetic diagram to show the results of this cross, including the ratio of offspring phenotypes, given that LᵀLᵀ results in the ‘dfm-a’ phenotype, LᴿLᴿ results in the ‘red’ phenotype, and LᵀLᴿ results in the ‘dfm-b’ phenotype.

• 1. Offspring Phenotypes: dfm-a, dfm-b, red
• 2. Gametes produced by each parent: Lᵀ and Lᴿ
• 3. Ratio of Offspring Phenotypes: 1 dfm-a : 2 dfm-b : 1 red
• 4. Parental Genotypes: LᵀLᴿ × LᵀLᴿ
• 5. Offspring Genotypes possible: LᵀLᵀ, LᵀLᴿ, LᴿLᴿ
• 6. Ratio of Offspring Genotypes: 1 LᵀLᵀ : 1 LᵀLᴿ : 1 LᴿLᴿ

Given the following genotypes and phenotypes for gene L: LᵀLᵀ = dfm-a; LᴿLᴿ = red; LᵀLᴿ = dfm-b; LⱽLⱽ = veraguensis; LʳLʳ = metallic; LᵀLⱽ = dfm-a; LᵀLʳ = dfm-a; LⱽLʳ = veraguensis. Identify the codominant alleles and list the dominance hierarchy.

• 1. Dominance hierarchy: Lʳ > Lⱽ > Lᴿ > Lᵀ
• 2. Codominant alleles: Lᵀ and Lᴿ (because LᵀLᴿ has a distinct phenotype ‘dfm-b’).
• 3. Dominance hierarchy: (Lᵀ, Lᴿ) > Lⱽ > Lʳ (Lᵀ and Lᴿ are dominant over Lⱽ and Lʳ; Lⱽ is dominant over Lʳ).

Researchers carried out crosses involving veraguensis (genotypes LⱽLⱽ or LⱽLʳ) and metallic (genotype LʳLʳ) tortoise beetles. Deduce the genotypes of the parents used in cross 1 (female veraguensis × male metallic) given the offspring ratio observed (Cross 1 approx 1 veraguensis : 1 metallic).

• 1. Male Genotype: LʳLʳ (phenotype is metallic)
• 2. Female Genotype: LⱽLʳ (must carry Lʳ to produce LʳLʳ offspring)
• 3. Female Genotype: LⱽLⱽ (pure-breeding veraguensis)

Researchers carried out crosses involving veraguensis (genotypes LⱽLⱽ or LⱽLʳ) and metallic (genotype LʳLʳ) tortoise beetles. Deduce the genotypes of the parents used in cross 2 (female veraguensis × male veraguensis), given the offspring ratio observed (Cross 2 approx 3 veraguensis : 1 metallic).

• 1. Offspring Ratio Suggests: Both parents are heterozygous.
• 2. Male Genotype: LⱽLʳ
• 3. Female Genotype: LⱽLʳ
• 4. Male Genotype: LⱽLⱽ

In cross 2 (LⱽLʳ × LⱽLʳ), the offspring ratio was approximately 3 veraguensis : 1 metallic, with roughly equal numbers of males and females observed in each phenotypic group. Explain how this evidence supports the conclusion that gene L follows autosomal inheritance (is not located on the X chromosome), assuming XX females and XY males.

• 1. Evidence: Equal phenotypic ratios in males and females support autosomal inheritance.
• 2. If X-linked: Expected ratios often differ between sexes (e.g., recessive traits appear more in males).
• 3. If X-linked: Metallic females (XʳXʳ) could only be produced if the father was metallic (XʳY).
• 4. Evidence: The 3:1 ratio itself proves it cannot be X-linked.

Name a statistical test that can be used to determine whether an observed ratio of phenotypes is significantly different from an expected ratio.

• 1. Standard deviation calculation
• 2. Chi-squared (χ²) test
• 3. T-test

Haemophilia is an X-linked recessive condition where blood fails to clot properly due to a deficiency in clotting factor VIII. Explain the relationship between the gene involved (F8), the protein (factor VIII), and the phenotype (haemophilia).

• 1. Gene: The F8 gene on the X chromosome codes for factor VIII protein.
• 2. Allele: The recessive allele (Xᶠ) results from a mutation in the F8 gene.
• 3. Phenotype: Males (XᶠY) are affected; females are usually carriers (XᶠXᶠ) unless homozygous recessive (XᶠXᶠ).
• 4. Protein: Functional factor VIII is essential for the blood clotting cascade.
• 5. Mechanism: The dominant allele (Xᶠ) produces non-functional factor VIII, leading to haemophilia.
• 6. Mechanism: The recessive allele leads to non-functional or insufficient factor VIII, disrupting clotting and causing the haemophilia phenotype (prolonged bleeding).

In pea plants, the genotype lele results in dwarf plants. The Le allele codes for an enzyme (GA 3β-hydroxylase) involved in the final step of synthesising active gibberellin (GA₁), while the le allele codes for a non-functional version of this enzyme. Explain how the lele genotype results in dwarf plants.

• 1. Genotype lele means only non-functional GA 3β-hydroxylase is produced.
• 2. Active gibberellin (GA₁) stimulates stem elongation.
• 3. Genotype lele leads to overproduction of active gibberellin (GA₁).
• 4. Lack of functional enzyme blocks synthesis of active GA₁.
• 5. Insufficient active GA₁ reduces stem cell elongation, causing dwarfism.

DELLA proteins are repressors that inhibit plant growth. Gibberellins promote growth partly by causing the breakdown of DELLA proteins. In the absence of sufficient gibberellin, DELLA proteins accumulate and bind to transcription factors called PIFs, preventing them from activating growth-related genes. Explain how DELLA proteins act as repressors in this context.

• 1. DELLA binding prevents PIF transcription factors from binding to DNA target sequences.
• 2. DELLA binding activates RNA polymerase, increasing transcription of growth genes.
• 3. Lack of PIF binding blocks recruitment of RNA polymerase to growth gene promoters.
• 4. Transcription of growth-related genes is reduced or blocked.
• 5. Proteins required for growth are not synthesised.

The BRCA1 gene codes for a tumour suppressor protein involved in DNA repair. Certain mutations in BRCA1 increase the risk of breast cancer. State one reason why a doctor might recommend genetic testing for a BRCA1 mutation, and explain why such a mutation increases cancer risk.

• 1. Reason for Testing: Significant family history of breast/ovarian cancer (especially at young ages).
• 2. Mechanism: Mutated BRCA1 protein cannot repair damaged DNA effectively.
• 3. Mechanism: Cells accumulate mutations at a higher rate.
• 4. Mechanism: Non-functional BRCA1 directly transforms proto-oncogenes into oncogenes.
• 5. Outcome: Accumulation of mutations in cell cycle control genes can lead to uncontrolled cell division (cancer).
• 6. Protein Change: Mutation alters BRCA1 protein’s 3D shape and function.

Suggest why the polymerase chain reaction (PCR) is used prior to DNA sequencing when testing for mutations in the BRCA1 gene.

• 1. PCR purifies the DNA sample by removing contaminants.
• 2. PCR makes many copies (amplifies) the specific target DNA region (BRCA1 gene).
• 3. PCR generates enough DNA material for accurate sequencing analysis from a potentially small initial sample.
• 4. PCR introduces specific mutations to make sequencing easier.

In the UK, approximately 1 in 400 females have a mutated BRCA1 allele, and 70% of these individuals will develop breast cancer by age 80. In 2022, the UK female population was 3.5 × 10⁷. Calculate the estimated number of these females who will develop breast cancer by age 80 due to a BRCA1 mutation. Give your answer in standard form.

• 1. Number developing cancer = 0.70 × 87,500 = 61,250
• 2. Number with BRCA1 mutation = (1/400) × (3.5 × 10⁷) = 87,500
• 3. Number developing cancer = 70 / 87,500
• 4. Final answer in standard form = 6.125 × 10⁴

A study compared survival probabilities for women with breast cancer treated with a DNA-damaging drug. One group had a BRCA1 mutation, the other did not. The results showed both groups started with survival probability near 1.0. Over 180 months, probability decreased for both, but the decrease was less steep for the BRCA1 mutation group, resulting in a consistently higher survival probability (e.g., at 180 months, approx 0.7 vs 0.47). Discuss whether these results support the hypothesis that the drug is more effective in women with a BRCA1 mutation.

• 1. Conclusion: The results support the hypothesis.
• 2. Evidence: The BRCA1 mutation group shows consistently higher survival probability over time.
• 3. Evidence: The rate of decrease in survival is slower for the BRCA1 group.
• 4. Conclusion: The results show the drug is equally effective in both groups.
• 5. Evidence: Specific data points (e.g., at 180 months, ~0.7 vs ~0.47 survival) show a clear difference favouring the mutated group.

Describe the pathway oxygen takes to diffuse from the cytoplasm, across the mitochondrial membranes, and into the mitochondrial matrix.

• 1. Oxygen diffuses across the outer mitochondrial membrane.
• 2. Oxygen diffuses through the intermembrane space.
• 3. Oxygen diffuses across the inner mitochondrial membrane.
• 4. Oxygen binds to haemoglobin within the matrix for transport.
• 5. Oxygen diffuses into the mitochondrial matrix.

Within the inner mitochondrial membrane are electron transport chain proteins (A, B, C) and a final protein complex (D) responsible for ATP synthesis. Identify protein D and the general name for proteins A, B, and C, and describe their roles in oxidative phosphorylation.

• 1. Identity: D = ATP synthase; A, B, C = Electron transport chain (ETC) proteins / electron carriers.
• 2. Role (A, B, C): Pump protons (H⁺) from the matrix to the intermembrane space using energy from electron transport.
• 3. Role (A, B, C): Accept electrons from reduced coenzymes (NADH, FADH₂) and pass them sequentially along the chain.
• 4. Role (D): Allows protons to flow back into the matrix, using the energy to synthesize ATP (chemiosmosis).
• 5. Role (A, B, C): Synthesize ATP directly as electrons pass through them.

Suggest three types of functional molecules that could be coded for by these mitochondrial genes, essential for mitochondrial function.

• 1. Some protein subunits of ATP synthase.
• 2. Glycolysis enzymes.
• 3. Some protein subunits of the electron transport chain complexes.
• 4. tRNAs and rRNAs required for mitochondrial protein synthesis.
• 5. Enzymes involved in the Krebs cycle.

The results at low light intensity showed that the rate of oxygen production remained low and relatively constant (around 0.5 mm³ h⁻¹) across the temperature range 10°C to 50°C. Describe and explain this result.

• 1. Description: Rate is low and relatively constant across the temperature range.
• 2. Explanation: Light intensity is the limiting factor.
• 3. Explanation: Enzymes are working at their maximum rate regardless of temperature.
• 4. Explanation: Limited light restricts the production of ATP and reduced NADP from light-dependent reactions, limiting the overall rate.
• 5. Explanation: Increasing temperature has little effect when light is limiting.

At high light intensity, the rate of oxygen production peaked at 4.0 mm³ h⁻¹ around 30°C, then decreased sharply above this temperature (to ~2.5 mm³ h⁻¹ at 40°C and ~0.65 mm³ h⁻¹ at 50°C). Describe and explain this decrease above 30°C.

• 1. Explanation: High temperatures cause denaturation of crucial photosynthetic enzymes (e.g., RuBisCO, ATP synthase).
• 2. Description: Rate decreases significantly above the optimum temperature (~30°C).
• 3. Explanation: Denaturation changes enzyme tertiary structure and active site shape, reducing activity.
• 4. Explanation: Increased photorespiration at high temperatures consumes oxygen.
• 5. Outcome: Reduced enzyme activity leads to a sharp decline in the overall rate of photosynthesis.

Describe the process that produces oxygen during non-cyclic photophosphorylation.

• 1. Process: Photolysis of water, catalysed by an enzyme complex associated with Photosystem II.
• 2. Location: Occurs in the stroma of the chloroplast.
• 3. Trigger: Replaces electrons lost from Photosystem II after photoactivation by light energy.
• 4. Products: Water (H₂O) is split into electrons (e⁻), protons (H⁺), and oxygen (O₂) molecules.
• 5. Fate of Products: Electrons go to PSII, protons contribute to the gradient, oxygen is released.

Name the structure at the beginning of the sensory pathway that detects a stimulus.

• 1. Axon terminal
• 2. Receptor (e.g., sensory receptor cell, nerve ending)
• 3. Synapse

Name a type of cell that might form a synapse with the axon terminal of a sensory neurone.

• 1. Intermediate neurone (relay neurone)
• 2. Motor neurone
• 3. Schwann cell
• 4. Sensory receptor cell

Name the cells that form the myelin sheath around axons in the peripheral nervous system.

• 1. Oligodendrocytes
• 2. Astrocytes
• 3. Schwann cells

State and explain the differences in the speed and mechanism of nerve impulse transmission between myelinated and unmyelinated neurones.

• 1. Speed: Faster in myelinated neurones.
• 2. Mechanism – Myelinated: Saltatory conduction (impulse ‘jumps’ between nodes of Ranvier).
• 3. Mechanism – Unmyelinated: Continuous wave of depolarisation along the entire axon.
• 4. Ion Channels: Concentrated at nodes of Ranvier in myelinated; distributed along the axon in unmyelinated.
• 5. Insulation: Myelin sheath acts as an insulator in unmyelinated neurones, speeding up conduction.
• 6. Local Circuits: Longer local circuits between nodes in myelinated conduction contribute to speed.

In a cholinergic synapse, voltage-gated calcium channels (A) are on the presynaptic membrane, while ligand-gated sodium channels (B) are on the postsynaptic membrane, activated by acetylcholine. Describe the differences between these two types of ion channels.

• 1. Gating: Channel A opens due to membrane potential change (voltage-gated); Channel B opens due to acetylcholine binding (ligand-gated).
• 2. Location: Channel A is presynaptic; Channel B is postsynaptic.
• 3. Ion Selectivity: Channel A is selective for Ca²⁺ ions; Channel B is selective for acetylcholine molecules.
• 4. Ion Selectivity: Channel A is selective for Ca²⁺ ions; Channel B is primarily selective for Na⁺ ions.

Morphine binds to opioid receptors on the presynaptic membrane, preventing the opening of voltage-gated calcium channels (Channel A). Suggest and describe the effects of this action on the functioning of the cholinergic synapse.

• 1. Effect: Stops Ca²⁺ influx into the presynaptic terminal.
• 2. Effect: Prevents fusion of synaptic vesicles containing acetylcholine with the presynaptic membrane.
• 3. Effect: Blocks release of acetylcholine into the synaptic cleft.
• 4. Effect: Causes continuous opening of postsynaptic sodium channels (Channel B).
• 5. Effect: Prevents depolarisation of the postsynaptic membrane and transmission of the impulse.

Morphine also binds to opioid receptors on the postsynaptic membrane, causing associated potassium channels (Channel C) to open, allowing K⁺ to diffuse out. Suggest how this would affect the excitability of the postsynaptic neurone.

• 1. Effect: Outward K⁺ flow causes hyperpolarisation (membrane potential becomes more negative).
• 2. Effect: Moves the membrane potential further away from the threshold potential.
• 3. Effect: Makes it harder to depolarise the postsynaptic neurone to threshold.
• 4. Effect: Increases the excitability of the postsynaptic neurone.
• 5. Outcome: Inhibits the postsynaptic neurone, making it less likely to fire an action potential.

The Asian common toad, an invasive alien species introduced to Madagascar, breeds fast, spreads rapidly, and has toxic skin poisonous to predators. Suggest four ways this toad may negatively affect the ecosystem of eastern Madagascar.

• 1. Competition: Competes with native amphibians for food resources (insects) and breeding sites.
• 2. Predation: Consumes large numbers of native insects, potentially impacting insect populations and species that rely on them. It might also prey on small native vertebrates.
• 3. Disease Introduction: May carry or introduce novel pathogens or parasites that can infect and harm native wildlife populations (amphibians, reptiles).
• 4. Habitat Improvement: Enhances local soil quality and aeration through its burrowing activities, benefiting certain soil organisms.
• 5. Poisoning Predators: Native predators (snakes, birds, mammals) that attempt to eat the toxic toad may be poisoned and killed, disrupting natural food webs.
• 6. Biodiversity Reduction: Through competition, predation, poisoning, and disease spread, the invasive toad can lead to declines or local extinctions of native species, reducing the overall biodiversity of the ecosystem.

Outline the main steps involved in the process of in vitro fertilisation (IVF) as used in assisted reproduction for mammals like the endangered eastern black rhino.

• 1. Ovarian Stimulation: Female is treated with hormones (like FSH) to induce the development and maturation of multiple ova (superovulation).
• 2. Ovum Collection (Egg Retrieval): Mature ova are collected from the female’s ovaries, often via ultrasound-guided aspiration.
• 3. Fertilisation: Occurs naturally inside the female’s uterus or oviduct after hormone treatment and timed mating/insemination.
• 4. Sperm Preparation: Semen is collected from the male and processed in the lab to select motile, healthy sperm.
• 5. Fertilisation (In Vitro): Collected ova and prepared sperm are mixed together in a laboratory dish (“in vitro”) to allow fertilisation to occur, or Intracytoplasmic Sperm Injection (ICSI) may be performed.
• 6. Embryo Culture: Fertilized eggs (zygotes) are cultured in a special medium in the lab for several days, allowing them to develop into embryos (e.g., to blastocyst stage).
• 7. Embryo Transfer: One or more selected embryos are carefully transferred into the uterus of the recipient female (original donor or a surrogate mother).

Name the mechanism, involving detection of deviation from a set point and triggering corrective responses, that maintains blood glucose concentration relatively constant.

• 1. Positive feedback
• 2. Homeostasis
• 3. Negative feedback

Graph shows blood glucose drops during exercise (15-40 min) and rises during recovery (40-70 min). Explain these changes.

• 1. Decrease (Exercise): Increased rate of glucose uptake from the blood and utilisation (respiration) by active muscles lowers blood glucose levels.
• 2. Increase (Recovery): Secretion of glucagon is stimulated (by low glucose), promoting the liver to break down stored glycogen (glycogenolysis) and release glucose into the blood.
• 3. Decrease (Exercise): Insulin secretion increases significantly during exercise to facilitate glucose uptake by muscles.
• 4. Increase (Recovery): The liver releases stored glucose into the bloodstream to counteract the exercise-induced drop and restore normal blood glucose levels (homeostasis).

Urine test strips for glucose involve: (1) glucose + O₂ –(glucose oxidase)–> A + H₂O₂; (2) H₂O₂ + B –(peroxidase)–> oxidised B + H₂O. B is colourless, oxidised B is coloured. Identify A and the type of substrate B.

• 1. A = Gluconic acid (or its lactone form, gluconolactone)
• 2. B = An enzyme cofactor required for the peroxidase reaction.
• 3. B = Chromogen (a substance that develops colour upon oxidation in this reaction).
• 4. A = Carbon dioxide

Suggest two advantages of using a blood glucose biosensor compared to urine test strips for monitoring diabetes.

• 1. Advantage: Biosensors measure the actual blood glucose concentration at that moment, providing real-time information, whereas urine glucose reflects blood levels only after they have exceeded the renal threshold sometime previously.
• 2. Advantage: Biosensors typically provide a precise quantitative (numerical) result (e.g., mmol/L or mg/dL), allowing for accurate tracking and adjustment of treatment.
• 3. Advantage: Urine strips are generally better and more reliable for detecting low blood glucose levels (hypoglycaemia).
• 4. Advantage: Biosensors are able to detect hypoglycaemia (low blood sugar), which urine tests cannot do as glucose only appears in urine when blood levels are high.
• 5. Advantage: Many modern biosensors allow for data storage, trend analysis, and connectivity with other devices (e.g., insulin pumps, smartphones), facilitating better diabetes management.

Name a structure found in viruses but absent in bacteria.

• 1. Cell wall
• 2. Capsid
• 3. Protein coat

The genetic material of the Zika virus consists of a single-stranded nucleic acid molecule. Suggest the name of this nucleic acid.

• 1. Deoxyribonucleic acid (DNA)
• 2. Ribonucleic acid (RNA)
• 3. Protein

A transmission electron micrograph shows a Zika virus with a measured diameter of 10 mm on the image. The magnification of the micrograph is x200 000. Calculate the actual diameter of the virus in nanometres (nm).

• 1. Working: Actual size = (Image size in nm) / Magnification = (10 mm × 1,000,000 nm/mm) / 200,000 = 10,000,000 nm / 200,000
• 2. Result: 500 nm
• 3. Formula: Actual size (A) = Image size (I) / Magnification (M)
• 4. Result: 50 nm

Describe what is meant by the resolution of a microscope.

• 1. The maximum size of an object that can be clearly viewed through the microscope.
• 2. The ability of the microscope to distinguish between two separate points or objects that are very close together; the minimum distance between two distinguishable points.
• 3. The amount by which the microscope lens system enlarges the image of the object (magnification).

The vector for Zika virus is the Aedes aegypti mosquito, while the vector for malaria is the Anopheles mosquito. Describe the similarities and differences between the transmission of Zika virus disease and the transmission of malaria.

• 1. Similarity: Both diseases are primarily transmitted between humans through the bite of an infected female mosquito vector.
• 2. Difference: The specific genus of mosquito responsible differs (Aedes for Zika, Anopheles for malaria).
• 3. Similarity: Both diseases are typically transmitted via contaminated drinking water sources in tropical regions.
• 4. Similarity: In both cases, transmission occurs when an infected mosquito takes a blood meal from a human, injecting the pathogen (virus or parasite) into the bloodstream.
• 5. Similarity: For the cycle to continue, a non-infected mosquito must bite an infected human to pick up the pathogen, which then develops within the mosquito before it can transmit it to another human.

A potential Zika virus vaccine contains small proteins derived from the virus. Explain how giving this vaccine to a person can lead to the development of long-term immunity against Zika virus disease.

• 1. The vaccine proteins act as non-self antigens, recognized by the recipient’s immune system, triggering a primary immune response.
• 2. As a result of this primary response, long-lived memory B-lymphocytes and memory T-lymphocytes specific to Zika virus antigens are generated and persist in the body.
• 3. The vaccine directly provides a supply of pre-formed antibodies against Zika virus proteins that circulate and provide protection for the person’s entire lifetime.
• 4. During the primary response, selected B-lymphocytes specific to the vaccine antigens proliferate and differentiate into antibody-producing plasma cells and memory B-cells.
• 5. If the vaccinated person is later exposed to the actual Zika virus, these memory cells initiate a much faster, stronger, and more effective secondary immune response, eliminating the virus before it can cause significant disease.
• 6. Specific T-lymphocytes (helper and cytotoxic) and B-lymphocytes recognize the presented vaccine antigens, leading to clonal expansion and differentiation into effector and memory cells.

Explain how a vaccination programme using an effective Zika vaccine may limit the spread of Zika virus disease through a population.

• 1. Vaccination increases the number of susceptible individuals in the population, making transmission easier.
• 2. It helps establish herd immunity (community immunity) when a sufficiently large proportion of the population becomes immune through vaccination.
• 3. Widespread vaccination reduces the overall transmission rate because there are fewer susceptible hosts available for the virus to infect (either directly or via mosquito bites).
• 4. Herd immunity effectively breaks or disrupts the chain of transmission, thereby indirectly protecting unvaccinated individuals (like newborns or immunocompromised people) because the virus circulates less readily.

Describe the structure of a collagen molecule (tropocollagen).

• 1. It consists of three polypeptide chains wound together to form a right-handed triple helix structure.
• 2. Each individual polypeptide chain is folded into a typical right-handed alpha-helix structure, similar to that found in globular proteins.
• 3. Each polypeptide chain typically features a repeating amino acid sequence pattern of Gly-X-Y, where Glycine (Gly) is essential for tight packing within the triple helix.
• 4. The overall triple helix structure is stabilized by numerous hydrogen bonds formed between the amino acid residues of adjacent polypeptide chains.
• 5. In the Gly-X-Y sequence, X and Y are frequently the amino acids proline and hydroxyproline, which contribute to the helical conformation.

Describe the structure of a collagen fibre.

• 1. Many collagen fibrils associate laterally and longitudinally to form a larger collagen fibre, visible with light microscopy.
• 2. Collagen molecules (tropocollagen) align in a parallel, staggered array to form microfibrils and then fibrils.
• 3. Covalent cross-links (e.g., involving lysine and hydroxylysine residues) form between adjacent collagen molecules within and between fibrils, providing high tensile strength.
• 4. Hydrogen bonds between water molecules trapped within the fibre are the primary forces responsible for holding the entire fibre structure together.
• 5. The staggered arrangement of collagen molecules within a fibril creates gaps (hole zones) and overlaps, leading to a characteristic banded or striated appearance in electron micrographs.

In collagen, a disaccharide made of two hexose monosaccharides, D and E, can be bonded to hydroxylysine. Monosaccharide D is commonly found in glycoproteins. Give the precise name of monosaccharide D.

• 1. Fructose
• 2. Ribose
• 3. α-glucose
• 4. Deoxyribose

Identify the type of bond that links monosaccharides D and E in the disaccharide described in part (b)(i).

• 1. Peptide bond
• 2. Hydrogen bond
• 3. Phosphodiester bond
• 4. Glycosidic bond

The amino acid hydroxylysine is derived from lysine. Describe the chemical structure of the R group (side chain) of hydroxylysine, based on its derivation from lysine (-CH₂-CH₂-CH₂-CH₂-NH₂) and the addition of a hydroxyl group.

• 1. It contains only carbon and hydrogen atoms, making it nonpolar.
• 2. It is based on a four-carbon chain attached to the alpha carbon atom of the amino acid backbone.
• 3. It possesses an amino (-NH₂) group at the end of the carbon chain (the epsilon carbon).
• 4. It has a hydroxyl (-OH) group attached to the carbon atom immediately adjacent to the alpha carbon (the beta carbon).
• 5. It has a hydroxyl (-OH) group attached to one of the internal carbon atoms of the chain (specifically, the delta carbon, the third from the alpha carbon).

Osteogenesis imperfecta (brittle bone disease) results from a deficiency in collagen. Suggest how a named tissue or structure containing collagen is affected in a person who has this disease.

• 1. Named Structure: Bone. Effect: Becomes significantly stronger, denser, and more resistant to fracture due to compensatory mechanisms.
• 2. Named Structure: Bone. Effect: Lacks sufficient tensile strength and proper mineralization framework, becoming brittle and prone to easy fractures.
• 3. Named Structure: Tendons/Ligaments. Effect: May exhibit weakness and increased laxity, leading to joint instability and dislocations.
• 4. Named Structure: Skin. Effect: May be thinner than normal, fragile, and bruise easily due to reduced structural integrity.

State whether smooth muscle, endothelium, and tunica media are typically present (✓) or absent (X) in arteries, capillaries, and veins.

• 1. Endothelium: Present as the innermost lining in Arteries, Capillaries, and Veins.
• 2. Smooth muscle: Present in the walls of Arteries and Veins; essentially Absent in Capillaries (which may have pericytes).
• 3. Tunica media: Present only in Arteries; completely Absent in Capillaries and Veins.
• 4. Tunica media: Present (and often thick) in Arteries, present (but typically thinner) in Veins; Absent in Capillaries.

Describe the direction of movement of oxygen and carbon dioxide between the alveolar air space and the capillary lumen during gas exchange in the lungs.

• 1. Oxygen: Moves from the capillary lumen (blood) across the respiratory membrane into the alveolar air space.
• 2. Carbon Dioxide: Moves from the capillary lumen (blood), across the alveolar and capillary epithelia, into the alveolar air space.
• 3. Oxygen: Moves from the alveolar air space, across the alveolar and capillary epithelia, into the capillary lumen (blood).
• 4. Carbon Dioxide: Moves from the alveolar air space, across the respiratory membrane, into the capillary lumen (blood).

Suggest and explain how a steep oxygen concentration gradient (partial pressure gradient) is maintained between the alveolar air space and the blood in the alveolar capillaries.

• 1. Mechanism: Rapid binding of oxygen to haemoglobin within red blood cells effectively removes dissolved oxygen from the plasma, keeping the plasma PO₂ low and favouring further diffusion from alveoli.
• 2. Mechanism: Continuous flow of blood through the alveolar capillaries constantly brings deoxygenated blood (low PO₂) to the lungs and carries away oxygenated blood (high PO₂ bound to Hb).
• 3. Mechanism: Slow, shallow breathing patterns maximize the time available for oxygen to diffuse across the respiratory membrane into the blood.
• 4. Mechanism: Ventilation (breathing) constantly replenishes the air in the alveoli, maintaining a high PO₂ in the alveolar air space compared to the incoming blood.
• 5. Mechanism: The very short diffusion path, consisting of thin alveolar and capillary endothelial cells and their fused basement membranes, facilitates rapid oxygen diffusion, helping to keep pace with removal by blood flow and Hb binding.

Complete the passage about blood vessels associated with the heart using the most appropriate scientific terms: “The pulmonary vein carries blood to the left atrium. After passing from the left atrium to the left ventricle, blood is pumped into the (1)______. The (1)______ is one of two large arteries that carry blood away from the ventricles of the heart. Blood that leaves the heart to enter these arteries must pass through the (2)______ valves. Oxygenated blood is supplied to the cardiac muscle cells through the (3)______ arteries.”

• 1. (1) = Aorta, (2) = Semilunar, (3) = Coronary
• 2. (1) = Vena Cava, (2) = Atrioventricular, (3) = Pulmonary
• 3. (1) = Aorta, (2) = Atrioventricular, (3) = Carotid
• 4. (1) = Pulmonary Artery, (2) = Semilunar, (3) = Coronary

Lignin and suberin are polymers found in plant tissues. Describe and explain the roles of lignin and suberin in the transport of water through the roots and stem of a plant.

• 1. Lignin Role (Strength): Provides mechanical strength and rigidity to xylem vessels, preventing them from collapsing under the negative pressure (tension) created during transpiration.
• 2. Suberin Role (Pathway Control): Forms the impermeable Casparian strip within the cell walls of the root endodermis, blocking the apoplast pathway and forcing water and minerals to enter the symplast (cross cell membranes) before reaching the xylem.
• 3. General Property: Both lignin and suberin are hydrophilic (water-attracting) polymers, which facilitates the rapid movement of water through the cell walls they impregnate.
• 4. Lignin Role (Waterproofing): Makes the walls of mature xylem vessels impermeable to water, ensuring water stays within the transport system and doesn’t leak out sideways excessively.
• 5. Suberin Role (Prevention of Backflow): The Casparian strip also helps prevent water and solutes from leaking back out of the xylem into the root cortex once they have entered the vascular cylinder.

Describe and explain the induced fit model of enzyme action, using laccase and monolignols as an example.

• 1. Initial Binding: The active site of the enzyme (laccase) possesses a rigid, pre-formed shape that is exactly complementary to the shape of the substrate (monolignols) before binding occurs.
• 2. Improved Complementarity: The binding of the substrate induces a subtle change in the conformation (shape) of the active site, leading to a more precise and tighter fit.
• 3. Enzyme Reverts: After the chemical reaction is complete and the product (e.g., lignin precursors) is released, the enzyme’s active site returns to its original, un-induced conformation.
• 4. Lowering Activation Energy: The induced fit helps to stabilize the transition state of the reaction, perhaps by straining bonds within the substrate or optimally aligning catalytic residues, thus lowering the activation energy.
• 5. Conformational Change: The interaction between the substrate (monolignol) and the enzyme (laccase) triggers a change in the three-dimensional shape of the active site region.
• 6. Enzyme-Substrate Complex: The formation of this precisely fitting enzyme-substrate complex is crucial for catalysis.

Outline the main stages in cell signalling that involve ligands binding to receptors on target cells and lead to specific cellular responses.

• 1. Reception: A signalling molecule (ligand) binds to a specific receptor protein located on the target cell’s surface or inside the cell.
• 2. Response: The transduced signal ultimately triggers a specific cellular activity, such as altered gene expression, activation or inhibition of an enzyme, changes in cell shape, or secretion.
• 3. Signal Transduction: The binding of the ligand causes a conformational change in the receptor, initiating a cascade of intracellular events (e.g., involving second messengers like cAMP, phosphorylation cascades by kinases) that relay and amplify the signal.
• 4. Ligand Synthesis: The target cell synthesizes the specific signalling ligand molecule immediately after the receptor binds it.
• 5. Secretion/Transport: (Occurs before reception) The signalling cell produces and secretes the ligand, which then travels (e.g., via diffusion or bloodstream) to reach the target cell.

A diagram shows a CDK inhibitor binding to a CDK molecule at a site distinct from the enzyme’s active site, preventing its activity. State and explain the type of inhibition shown.

• 1. Explanation: Binding of the inhibitor to this separate (allosteric) site causes a conformational change in the CDK molecule, altering the shape of the active site so that the substrate (e.g., a target protein) can no longer bind effectively.
• 2. Type: Competitive inhibition.
• 3. Type: Non-competitive inhibition (or allosteric inhibition).
• 4. Location: The inhibitor binds to the enzyme at an allosteric site, which is physically distinct from the active site where the substrate binds.

State which CDK inhibitor (palbociclib – inhibits CDK4 blocking G₁/S; or p21Cip1 – inhibits CDK2 regulating S phase) is likely to result in a cell containing chromosomes with only one chromatid each. Explain your answer.

• 1. Explanation: Blocking the cell cycle before S phase (the DNA synthesis phase) prevents DNA replication from occurring, meaning chromosomes remain in their unreplicated state (consisting of a single chromatid).
• 2. Inhibitor: p21Cip1 (inhibits CDK2, acting within or after S phase).
• 3. Inhibitor: Palbociclib (inhibits CDK4, blocking G₁ to S transition).
• 4. Explanation: Inhibition of CDK2 by p21Cip1 affects progression through S phase or later checkpoints; cells arrested by it would likely have already initiated or completed DNA replication, resulting in chromosomes with two sister chromatids.

Considering CDK inhibitors (palbociclib blocks G₁/S; p21Cip1 blocks S; RO-3306 blocks G₂/M), state which inhibitor is likely to result in a cell with both a relatively high concentration of mitochondria and chromosomes consisting of two chromatids. Explain your answer.

• 1. Explanation: The block must occur after S phase (so chromosomes have replicated into two chromatids) but before mitosis/cytokinesis (allowing organelles like mitochondria, which duplicate during interphase, to accumulate).
• 2. Inhibitor: Palbociclib (blocks G₁/S transition).
• 3. Explanation: Blocking the G₂/M transition (with RO-3306) arrests cells after DNA replication (S phase) and after G₂ growth (including mitochondrial replication), fitting the criteria.
• 4. Inhibitor: RO-3306 (blocks G₂/M transition).
• 5. Inhibitor: p21Cip1 (blocks S phase progression).

Explain why CDK inhibitors can be used to treat cancerous tumours.

• 1. Cancer is fundamentally characterized by uncontrolled cell division, often resulting from dysregulation of the cell cycle control system (including CDKs and checkpoints).
• 2. CDK inhibitors specifically enhance the activity of cyclin-dependent kinases (CDKs), thereby promoting more controlled and orderly cell growth in tumours.
• 3. By binding to and inhibiting the activity of essential CDKs (like CDK4/6 or CDK2), these drugs block the progression of cancer cells through critical checkpoints in the cell cycle.
• 4. Halting the cell cycle prevents the uncontrolled proliferation of cancer cells, thus slowing down or stopping tumour growth and potentially inducing cell death (apoptosis).

Describe three distinct ways in which the structure of a typical messenger RNA (mRNA) molecule differs from the structure of a typical DNA molecule found in a eukaryotic nucleus.

• 1. Sugar Composition: mRNA contains the five-carbon sugar ribose in its nucleotide backbone, whereas DNA contains deoxyribose (which lacks an oxygen atom on the 2′ carbon).
• 2. Nitrogenous Bases: mRNA uses the pyrimidine base Uracil (U) in place of Thymine (T), which is found in DNA. Both molecules contain Adenine (A), Guanine (G), and Cytosine (C).
• 3. Molecular Length: Both mRNA molecules (transcripts of genes) and nuclear DNA molecules (chromosomes) are typically very long, often of similar overall lengths within a cell.
• 4. Number of Strands: mRNA is typically a single-stranded polynucleotide chain, whereas nuclear DNA exists as a double-stranded helix.
• 5. Base Pairing: In DNA, specific complementary base pairing (A with T, C with G) occurs between the two strands; in single-stranded mRNA, bases are generally unpaired, although intramolecular base pairing can occur to form secondary structures (like hairpins).

Scientists synthesised four artificial bases: Z (nitro-substituted pyrimidine derivative), P (imidazole derivative), S (methylated pyrimidine derivative), and B (imidazole derivative). Identify the two synthetic bases listed that are purine derivatives (possessing a double-ring structure).

• 1. Z and S
• 2. P and B
• 3. Z and P
• 4. S and B

The synthetic base pairs Z:P and S:B each form three hydrogen bonds between them. State and explain which natural DNA base pair (A:T or C:G) is most similar in terms of hydrogen bonding to these synthetic pairs.

• 1. Reason: The natural Adenine:Thymine (A:T) base pair forms only two hydrogen bonds between the bases.
• 2. Base Pair Most Similar: A:T (Adenine:Thymine).
• 3. Reason: The natural Cytosine:Guanine (C:G) base pair forms three hydrogen bonds, which matches the number of hydrogen bonds formed by the synthetic pairs (Z:P and S:B).
• 4. Base Pair Most Similar: C:G (Cytosine:Guanine).