A diagram of a Zika virus shows an outer protein structure enclosing the genetic material. Name this protein coat.

• 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 = (10 mm * 1,000,000 nm/mm) / 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 viewed.
• 2. The ability to distinguish between two separate points that are close together.
• 3. The amount by which the image is enlarged.

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 transmitted by mosquito vectors.
• 2. Difference: Zika is transmitted by Aedes mosquitoes, while malaria is transmitted by Anopheles mosquitoes.
• 3. Similarity: Both are transmitted via contaminated water sources.
• 4. Similarity: Both are transmitted when an infected mosquito feeds on human blood.
• 5. Similarity: The pathogen (virus or parasite) is taken up by the mosquito from an infected person and passed to another person during a subsequent blood meal by the infected mosquito.

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. Vaccine proteins act as non-self antigens, recognised by the immune system and stimulating a primary immune response.
• 2. As part of this response, long-lived memory B-cells and memory T-cells specific to Zika antigens are produced and persist.
• 3. The vaccine directly provides pre-formed antibodies against Zika virus that circulate and last for the person’s entire lifetime.
• 4. During the primary response, selected B-lymphocytes that recognize the antigens differentiate into antibody-producing plasma cells and memory B-cells.
• 5. Upon subsequent natural infection with Zika virus, these memory cells initiate a rapid, strong, and effective secondary immune response, eliminating the virus before disease develops.
• 6. Specific T-lymphocytes and B-lymphocytes recognize the vaccine antigens, undergo clonal selection and expansion (proliferation), and differentiate 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 by introducing weakened virus.
• 2. It aims to establish herd immunity, which occurs when a sufficiently large proportion of the population becomes immune.
• 3. Herd immunity leads to reduced transmission because there are fewer susceptible hosts available for the virus to infect and replicate in, hindering its spread (e.g., fewer infected people for mosquitoes to pick up the virus from).
• 4. By breaking the chain of transmission, herd immunity indirectly protects individuals who are not vaccinated or cannot be vaccinated (e.g., infants, immunocompromised).

Describe the structure of a collagen molecule (tropocollagen).

• 1. It consists of three polypeptide chains wound together into a right-handed triple helix.
• 2. Each individual polypeptide chain is folded into a typical right-handed alpha-helix structure before forming the triple helix.
• 3. Each polypeptide chain often exhibits 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 chains.
• 5. In the Gly-X-Y sequence, X and Y are frequently the amino acids proline and hydroxyproline, which help stabilize the helical structure of the individual chains.

Describe the structure of a collagen fibre.

• 1. Many collagen fibrils associate side-by-side and end-to-end to form larger collagen fibres.
• 2. Collagen molecules (tropocollagen) align in a parallel, overlapping fashion to form collagen fibrils. Adjacent molecules are typically staggered by about one-quarter of their length.
• 3. Strong covalent cross-links form between lysine/hydroxylysine residues on adjacent collagen molecules within a fibril, greatly increasing tensile strength.
• 4. Hydrogen bonds are the only significant forces holding the multiple collagen molecules together within a fibril and fibre.
• 5. The regular, staggered arrangement of collagen molecules within fibrils often creates a characteristic banded or striated appearance when viewed with an electron microscope.

In collagen, a disaccharide made of two hexose monosaccharides, D and E, can be bonded to the amino acid 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.
• 2. It is a chain attached to the alpha carbon, which in hydroxylysine is -CH₂-CH₂-CH(OH)-CH₂-NH₂.
• 3. It has an amino (-NH₂) group on the terminal (epsilon) carbon atom of the side chain.
• 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 the third carbon atom counting away from the alpha carbon (the delta 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 stronger and denser due to compensatory mechanisms.
• 2. Named Structure: Bone. Effect: Becomes brittle, weak, and prone to fractures easily.
• 3. Named Structure: Tendons/Ligaments. Effect: May be weaker and more lax, leading to joint hypermobility or instability.
• 4. Named Structure: Skin. Effect: May be thinner, more fragile, and bruise easily.

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

• 1. Endothelium: Present (✓) in Arteries, Capillaries, and Veins (forms the innermost lining of all blood vessels).
• 2. Smooth muscle: Present (✓) in Arteries and Veins; Generally Absent (X) in Capillaries.
• 3. Tunica media: Present (✓) only in Arteries; Absent (X) in Capillaries and Veins.
• 4. Tunica media: Present (✓) in Arteries and Veins (though thickness varies); Absent (X) 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) 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 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 oxygen partial pressure low near the alveoli.
• 2. Mechanism: Continuous flow of blood through the pulmonary capillaries constantly brings deoxygenated blood (low O₂) to the alveoli and carries away oxygenated blood (high O₂).
• 3. Mechanism: Slow, shallow breathing allows maximum time for oxygen to diffuse across the alveolar-capillary barrier before the air is exhaled.
• 4. Mechanism: Continuous ventilation (breathing) constantly replenishes the air in the alveoli, maintaining a high partial pressure of oxygen.
• 5. Mechanism: The extremely short diffusion path due to the very thin alveolar and capillary walls facilitates rapid oxygen movement down its gradient.

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): Lignin deposition in xylem cell walls provides significant structural reinforcement, preventing the vessels from collapsing under the negative pressure (tension) generated during transpiration.
• 2. Suberin Role: Suberin forms the waterproof Casparian strip within the cell walls of the root endodermis. This blocks the apoplast pathway (movement through cell walls), forcing water and minerals to enter the symplast (cross cell membranes) before reaching the xylem, allowing for selective uptake.
• 3. General Property: Both lignin and suberin are highly hydrophilic polymers, readily absorbing water and facilitating its movement through cell walls where they are present.
• 4. Lignin Role (Waterproofing): Lignification makes the xylem vessel walls largely impermeable to water, ensuring water stays within the vessel and doesn’t leak out laterally as it moves up the plant.
• 5. Suberin Role: In addition to the Casparian strip, suberin deposition in other tissues (like cork or later in the endodermis) can act as a general barrier preventing unwanted water loss or entry.

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 laccase enzyme possesses a rigid structure that is a perfect, pre-formed complementary shape to the monolignol substrates before binding occurs.
• 2. Improved Complementarity: As the monolignol substrate binds to the laccase active site, it induces a conformational change in the enzyme, leading to a more precise fit between the active site and the substrate.
• 3. Enzyme Reverts: After the reaction is complete and the products (e.g., lignin precursors) are released, the laccase active site returns to its original, less complementary conformation.
• 4. Lowering Activation Energy: The binding process and induced fit can strain bonds within the monolignol substrate or correctly orient catalytic groups in the active site, thus facilitating the chemical reaction by lowering its activation energy.
• 5. Conformational Change: The key feature is that substrate binding causes the enzyme’s active site to change shape.
• 6. Enzyme-Substrate Complex: The formation of the tightly fitting enzyme-substrate complex occurs as a result of the induced conformational change.

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

• 1. Reception: The process begins when a signalling molecule (ligand) binds specifically to its corresponding receptor protein, which is often located on the target cell’s surface membrane.
• 2. Response: The final stage where the signal transduction pathway culminates in a specific cellular activity, such as altered gene expression, enzyme activation/inactivation, changes in cell shape, or secretion.
• 3. Signal Transduction: Upon ligand binding, the receptor typically undergoes a conformational change, initiating a cascade of intracellular events. This pathway often involves secondary messengers and/or phosphorylation cascades that amplify and relay the signal within the target cell.
• 4. Ligand Synthesis: The target cell actively synthesizes the original signalling molecule (ligand) immediately after the ligand binds to its own receptor.
• 5. Secretion/Transport: (Occurs before reception) The signalling cell produces and secretes the ligand, which then travels through the extracellular environment (e.g., bloodstream, synaptic cleft) 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: The binding of the inhibitor to this separate (allosteric) site causes a conformational change in the CDK enzyme, likely altering the shape of the active site so that the normal substrate (e.g., cyclin) can no longer bind effectively.
• 2. Type: This mechanism represents competitive inhibition.
• 3. Type: This mechanism represents non-competitive inhibition (or a form of allosteric inhibition).
• 4. Location: The inhibitor binds to an allosteric site, which is any site on the enzyme other than 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 progression before S phase (DNA replication) occurs means the chromosomes will not be duplicated, hence they will consist of only a single chromatid.
• 2. Inhibitor: p21Cip1 (inhibits CDK2 regulating S phase).
• 3. Inhibitor: Palbociclib (inhibits CDK4 blocking G₁/S transition).
• 4. Explanation: Inhibition of CDK2, which acts during S phase, would likely arrest cells during or after DNA replication has begun, meaning chromosomes would likely have two chromatids (or be in the process of replicating).

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: A block at the G₂/M transition occurs after S phase (DNA replication is complete), so chromosomes will consist of two sister chromatids.
• 2. Inhibitor: Palbociclib (blocks G₁/S).
• 3. Explanation: Mitochondria replicate primarily during interphase (G₁, S, G₂). Blocking entry into mitosis (M phase) after G₂ allows the cell to continue growing and accumulating components like mitochondria without dividing, leading to a relatively high concentration.
• 4. Inhibitor: RO-3306 (blocks G₂/M).
• 5. Inhibitor: p21Cip1 (blocks S).

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

• 1. Cancer is fundamentally a disease characterized by uncontrolled cell division, often resulting from dysregulation of the cell cycle control system.
• 2. CDK inhibitors work by specifically enhancing the activity of Cyclin-Dependent Kinases (CDKs), thereby promoting more controlled and regulated cell growth.
• 3. By inhibiting the activity of specific CDKs that are essential for driving progression through different phases of the cell cycle (e.g., G₁/S or G₂/M transitions), these drugs can arrest the cell cycle in cancer cells.
• 4. Blocking cell cycle progression prevents the uncontrolled proliferation of cancer cells, which can slow down or stop tumour growth.

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 Component: mRNA contains the pentose sugar ribose in its nucleotide backbone, whereas DNA contains deoxyribose (which lacks an oxygen atom at the 2′ carbon).
• 2. Nitrogenous Bases: mRNA molecules utilize the 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 the DNA molecules comprising nuclear chromosomes are typically very long, complex polymers of roughly similar overall lengths within a cell.
• 4. Number of Strands: mRNA is typically synthesized and functions as a single-stranded molecule, whereas nuclear DNA exists as a double-stranded helix.
• 5. Base Pairing: In DNA, specific base pairing (A with T, C with G) occurs between the two complementary strands. While mRNA is single-stranded, it can fold upon itself forming localized regions of base pairing, but it doesn’t have a continuous complementary strand like DNA.

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 Adenine:Thymine (A:T) base pair in natural DNA is held together by only two hydrogen bonds.
• 2. Base Pair: A:T (Adenine:Thymine).
• 3. Reason: The Cytosine:Guanine (C:G) base pair in natural DNA is held together by three hydrogen bonds, which matches the number of hydrogen bonds formed by the synthetic pairs Z:P and S:B.
• 4. Base Pair: C:G (Cytosine:Guanine).