How do ringing (girdling) experiments and radioactive ¹⁴CO2 tracer experiments provide evidence for phloem translocation?

In a ringing experiment, removing a ring of bark/phloem around a woody stem causes a bulge of sugars and tissue swelling ABOVE the ring (while tissue below withers), showing downward sugar transport occurs in the phloem. Supplying leaves with radioactive ¹⁴CO2 produces ¹⁴C-labelled sucrose that can be tracked along the phloem sieve tubes over time using autoradiography.

Compare Cell-Mediated Immunity (T-lymphocytes) and Humoral Immunity (B-lymphocytes).

Cell-mediated: Helper T-cells (TH) bind to specific antigens on APCs, undergo clonal expansion, and release cytokines that activate cytotoxic T-cells (TC, which secrete perforin to destroy virus-infected/cancerous body cells), phagocytes, and B-cells. Humoral: B-cells with complementary antibodies bind antigens, are activated by TH cytokines (clonal selection/expansion), and differentiate into plasma cells (secrete monoclonal antibodies) and memory B-cells.

Define the Bohr Effect (Bohr Shift) on the oxyhaemoglobin dissociation curve and explain its physiological importance.

High partial pressure of CO2 (pCO2) in actively respiring tissues lowers blood pH (via carbonic anhydrase forming H⁺ and HCO3⁻). H⁺ ions bind to haemoglobin, altering its quaternary/tertiary structure and lowering its affinity for oxygen. This shifts the oxyhaemoglobin dissociation curve to the RIGHT, causing haemoglobin to unload more O2 to respiring muscles.

Outline the stages of phagocytosis by a macrophage or neutrophil.

1) Phagocyte is attracted to the pathogen by chemical products (chemotaxis) and binds to pathogen antigens (aided by opsonins); 2) Phagocyte engulfs the pathogen by endocytosis to form a phagosome vesicle; 3) Lysosomes fuse with the phagosome (forming a phagolysosome) and release hydrolytic lysozymes that digest the pathogen; 4) Antigens are presented on MHC proteins on the cell surface (acting as an Antigen-Presenting Cell, APC).

State the formula for Simpson's Index of Diversity (D) and explain how species richness differs from index of diversity.

D = 1 − [Σ n(n − 1)] / [N(N − 1)] (or D = N(N − 1) / Σ n(n − 1) in AQA), where N is the total number of organisms of all species and n is the total number of organisms of each individual species. Species richness only counts the number of different species present, whereas an index of diversity takes into account both the number of species and the relative abundance (evenness) of each species.

Explain the Cohesion-Tension Theory of water transport in plant xylem vessels.

1) Water evaporates from mesophyll cell walls and transpires out through open stomata, lowering water potential in leaf cells; 2) This creates a negative hydrostatic pressure (transpiration pull / tension) at the top of the xylem; 3) Hydrogen bonds cause water molecules to stick together (cohesion) and to hydrophilic xylem walls (adhesion), pulling a continuous, unbroken column of water up from the roots.

Why is ATP described as the 'universal energy currency' of living cells, and which enzymes catalyse its hydrolysis and synthesis?

ATP (adenosine triphosphate) is hydrolysed by ATP hydrolase into ADP + Pi (inorganic phosphate), releasing a small, manageable packet of energy (30.5 kJ mol⁻¹) in a single-step reaction that can be coupled directly to metabolic processes, active transport, or muscle contraction, and phosphorylated biological molecules. ATP synthase resynthesises ATP from ADP + Pi during condensation.

How do antibodies lead to the destruction of pathogens via agglutination and opsonisation, and how does an ELISA test work?

Each antibody has two identical variable antigen-binding sites (complementary tertiary structure), allowing it to cross-link multiple pathogens into clumps (agglutination) for rapid phagocytosis, and acting as an opsonin marker. Direct/Indirect ELISA uses immobilised antigens/antibodies and a secondary antibody conjugated to an enzyme that catalyses a colour change when substrate is added.

Describe the process of semi-conservative DNA replication and the specific roles of DNA helicase and DNA polymerase.

1) DNA helicase breaks hydrogen bonds between complementary base pairs (A=T, G≡C), unwinding the double helix; 2) Both original strands act as templates as free activated DNA nucleotides align via complementary base pairing; 3) DNA polymerase catalyses condensation reactions forming phosphodiester bonds between adjacent nucleotides in the 5' to 3' direction.

How did Meselson and Stahl's ¹⁵N / ¹⁴N density-gradient centrifugation experiment prove semi-conservative DNA replication?

Bacteria grown in heavy nitrogen (¹⁵N) had a single low, dense DNA band. After 1 generation in light ¹⁴N medium, all DNA formed a single intermediate hybrid band (¹⁵N/¹⁴N), ruling out conservative replication. After 2 generations in ¹⁴N, two equal bands appeared (50% intermediate ¹⁵N/¹⁴N and 50% light ¹⁴N/¹⁴N), confirming semi-conservative replication.

Compare the structure and function of triglycerides and phospholipids.

Triglycerides have 1 glycerol bonded via 3 ester bonds to 3 fatty acid tails (hydrophobic, compact energy store, metabolic water source). Phospholipids replace one fatty acid with a hydrophilic phosphate head, making them amphipathic (hydrophilic head faces aqueous exterior/cytoplasm; two hydrophobic tails face inwards to form the cell membrane bilayer).

How do terrestrial insects balance efficient gas exchange through their tracheal system with minimising water loss?

Air enters via spiracles (muscular valves that close periodically to prevent water loss) into chitin-lined tracheae and branches into microscopic tracheoles whose ends penetrate directly into respiring muscle fibres (filled with tracheal fluid that withdraws during exercise via osmosis as lactate lowers muscle Ψ). A waterproof waxy exoskeleton cuticle also reduces evaporation.

Explain the Mass Flow Hypothesis for translocation of sucrose from source to sink in the phloem.

1) At the source (photosynthesising leaf), H⁺ ions are actively pumped out of companion cells (using ATP) and co-transport sucrose back in, which diffuses via plasmodesmata into sieve tube elements; 2) This lowers water potential (Ψ) in the sieve tube, so water enters from the adjacent xylem by osmosis, creating high hydrostatic pressure; 3) At the sink (respiring root/meristem), sucrose is actively unloaded and converted to starch/respired, raising Ψ so water leaves back to the xylem, lowering hydrostatic pressure and driving bulk mass flow from source to sink.

State Fick's Law of Diffusion and explain how fish gills achieve maximum gas exchange efficiency through counter-current flow.

Fick's Law: Rate of diffusion ∝ (Surface Area × Concentration Gradient) ÷ Diffusion Pathway Thickness. Fish gills have gill filaments covered in numerous lamellae (large SA) with a single-layer epithelium (short distance). In counter-current flow, blood flows through lamellae in the opposite direction to water, maintaining a steep O2 concentration gradient across the entire length of the gill lamella.

Compare the oxyhaemoglobin dissociation curves of fetal haemoglobin, lugworms, and llamas (high altitude) with adult human haemoglobin.

Curves shifted to the LEFT (fetal haemoglobin, lugworms, high-altitude llamas) indicate a HIGHER affinity for O2, allowing haemoglobin to load O2 efficiently in low-pO2 environments (e.g., across the placenta from maternal blood). Curves shifted to the RIGHT (mice, birds with high metabolic rates) indicate a LOWER affinity for O2, allowing rapid unloading of O2 for intense cellular respiration.

Explain how tissue fluid is formed at the arterial end of a capillary bed and reabsorbed at the venous end.

At the arterial end, high hydrostatic pressure (from left ventricular contraction) exceeds the opposing oncotic/water potential pull of plasma proteins, forcing water, glucose, amino acids, and O2 out through capillary fenestrations (ultrafiltration; large plasma proteins and RBCs remain in blood). At the venous end, hydrostatic pressure has dropped while blood water potential is very negative (due to retained plasma proteins), so ~90% of water returns by osmosis; the remaining ~10% drains into the lymphatic system.

Compare competitive and non-competitive enzyme inhibition and their effects on Vmax as substrate concentration increases.

Competitive inhibitors have a molecular shape similar to the substrate and bind reversibly to the active site; increasing substrate concentration outcompetes the inhibitor, so Vmax is still reached. Non-competitive inhibitors bind to an allosteric site (away from the active site), altering the enzyme's tertiary structure and active site shape; increasing substrate concentration cannot overcome this, so Vmax is permanently lowered.

State five biologically crucial properties of water and explain how each arises from hydrogen bonding between polar H2O molecules.

1) High specific heat capacity: H-bonds absorb heat energy, buffering aquatic habitats and intracellular temperatures; 2) High latent heat of vaporisation: evaporation provides an efficient cooling mechanism (sweating/transpiration); 3) Cohesion & surface tension: continuous xylem water columns; 4) Universal polar solvent: medium for metabolic reactions and mass transport; 5) Metabolite in hydrolysis and condensation reactions.

Define the primary, secondary, tertiary, and quaternary structures of a protein, specifying the exact chemical bonds involved.

Primary: linear sequence of amino acids joined by covalent peptide bonds. Secondary: folding into α-helices or β-pleated sheets stabilised by hydrogen bonds between -NH and -C=O groups of the polypeptide backbone. Tertiary: specific 3D folding stabilised by R-group interactions (ionic bonds, disulfide bridges between cysteine residues, hydrogen bonds, hydrophobic interactions). Quaternary: ≥2 polypeptide chains (and/or prosthetic groups, e.g., haem in haemoglobin).

How do crossing over (Prophase I) and independent assortment (Metaphase I) generate genetic variation during meiosis?

1) Crossing over: In Prophase I, homologous chromosomes pair up to form bivalents and non-sister chromatids exchange alleles at chiasmata, creating new combinations of linked alleles on recombinant chromatids. 2) Independent segregation/assortment: In Metaphase I, homologous pairs align randomly along the equator (2ⁿ possible combinations, where n is the haploid number), so maternal and paternal chromosomes are assorted independently into gametes.

How do alpha-glucose (α-glucose) and beta-glucose (β-glucose) isomers differ structurally, and which polymers do they form?

On carbon-1 (C1), α-glucose has the hydroxyl (-OH) group pointing BELOW the ring plane (forming starch [amylose + amylopectin] and glycogen via α-1,4 and α-1,6 glycosidic bonds), whereas β-glucose has the -OH group on C1 pointing ABOVE the ring plane (forming cellulose via β-1,4 glycosidic bonds).

Trace the myogenic electrical excitation of the mammalian heart through the SAN, AVN, Bundle of His, and Purkyne fibres.

1) Sinoatrial node (SAN, pacemaker in right atrium) initiates a wave of electrical excitation causing both atria to contract simultaneously (atrial systole); 2) Non-conducting atrioventricular septum stops impulse passing directly to ventricles; 3) Atrioventricular node (AVN) delays the impulse (~0.1 s) so atria finish emptying; 4) Impulse travels down the Bundle of His and Purkyne fibres to the apex, contracting ventricles from the bottom upwards.

Contrast directional selection (e.g., antibiotic resistance) and stabilising selection (e.g., human birth mass).

Directional selection occurs when the environment changes, favouring individuals with an extreme phenotype at one end of the distribution and shifting the population mean toward that extreme. Stabilising selection occurs in a stable environment, selecting against both extremes and favouring the intermediate mean phenotype, reducing standard deviation/phenotypic variation.

Relate the molecular structure of β-glucose chains in Cellulose to the high tensile strength of plant cell walls.

Adjacent β-glucose monomers are inverted by 180° to form straight, unbranched β-1,4 chains. Parallel chains are cross-linked by thousands of intermolecular hydrogen bonds, bundling into microfibrils and macrofibrils that provide high tensile strength to resist osmotic turgor pressure.

Contrast the 'induced-fit' model of enzyme action with the older 'lock-and-key' model.

In the induced-fit model, the enzyme's active site is not initially a rigid, exact fit for the substrate. As the substrate binds, it induces a conformational change in the tertiary structure of the active site, moulding tightly around the substrate and placing strain on substrate bonds to lower the activation energy of the transition state.

Explain the mechanism of sodium-glucose co-transport across the epithelial cells of the mammalian ileum.

1) Na⁺/K⁺ ATPase pumps actively transport Na⁺ ions out of the epithelial cell into the blood (using ATP), creating a low intracellular Na⁺ concentration; 2) Na⁺ ions diffuse down their concentration gradient from the ileum lumen into the epithelial cell via a sodium-glucose co-transporter protein, carrying glucose molecules against their concentration gradient; 3) Glucose moves into the blood via facilitated diffusion through channel/carrier proteins.

In the Fluid Mosaic Model of cell membranes, what is the exact role of cholesterol, glycoproteins, and glycolipids?

Cholesterol binds to hydrophobic fatty acid tails, restricting phospholipid movement at high temperatures (reducing fluidity/permeability) and preventing close packing/crystallisation at low temperatures. Glycoproteins and glycolipids act as cell-surface receptors for hormones/neurotransmitters, cell-to-cell recognition markers (antigens), and cell adhesion sites.

Relate the molecular structure of Starch (amylose and amylopectin) and Glycogen to their functions as energy storage molecules.

1) Coiled/helical α-1,4 chains make them compact, storing lots of glucose in a small volume; 2) Insoluble in water, so they do not lower the water potential (Ψ) of cells or cause osmotic lysis; 3) Branched α-1,6 glycosidic bonds (highly branched in glycogen) provide many terminal ends for rapid enzymatic hydrolysis back to α-glucose.

Why is the genetic code described as 'degenerate', 'non-overlapping', and 'universal'?

Degenerate: Most amino acids are coded for by more than one base triplet/codon (64 possible codons for 20 amino acids), which means some base substitution mutations are silent. Non-overlapping: Each nucleotide base is read only once as part of a single consecutive triplet. Universal: The same triplet codons specify the same amino acids in almost all known living organisms.

Describe the mechanism of human inspiration (inhaling) and expiration (exhaling) in terms of muscle contraction, volume, and pressure.

Inspiration (active): External intercostal muscles contract (ribs move up and out) and diaphragm muscles contract (diaphragm flattens) → thoracic volume increases → pulmonary pressure drops below atmospheric pressure → air flows into lungs. Normal expiration (passive): External intercostals and diaphragm relax (internal intercostals contract only in forced expiration) → thoracic volume decreases → pressure rises above atmospheric.

Describe the two-stage process of protein synthesis: Transcription in the nucleus and Translation at the ribosomes.

Transcription: RNA polymerase unwinds DNA and joins free RNA nucleotides complementary to the DNA template strand (with U replacing T) to form pre-mRNA; introns (non-coding regions) are spliced out to form mature mRNA (in eukaryotes). Translation: mRNA binds to a ribosome; tRNA molecules with complementary anticodons bring specific amino acids according to the mRNA codon sequence; peptidyl transferase forms peptide bonds using ATP.

Write the A-Level Biology water potential equation and explain what happens to an animal cell vs a plant cell in a hypotonic (high Ψ) solution.

Ψ (Water Potential) = Ψs (Solute Potential, always negative) + Ψp (Pressure Potential). Pure water has the maximum Ψ = 0 kPa. In a hypotonic solution (higher Ψ outside), water enters by osmosis: an animal red blood cell swells and bursts (haemolysis) because it lacks a cell wall; a plant cell swells until the inelastic cellulose cell wall exerts positive turgor pressure (Ψp), becoming turgid.

A-Level Biology I: Biological Molecules, Nucleic Acids, Cell Membranes, Immunity & Exchange Surfaces

AS / Year 12 A-Level Biology foundations across AQA, OCR, and Edexcel: alpha/beta glucose polysaccharides, protein structure, semi-conservative DNA replication, fluid mosaic membrane transport, humoral/cell-mediated immunity, gas exchange, and cardiac mass transport.

Level: basic
Languages: en → en
Cards: 32
Markdown: A-Level Biology I: Biological Molecules, Nucleic Acids, Cell Membranes, Immunity & Exchange Surfaces (.md)

Flashcards (32)

#	Front (Term / Question)	Back (Definition / Answer)
1	How do ringing (girdling) experiments and radioactive ¹⁴CO2 tracer experiments provide evidence for phloem translocation?	In a ringing experiment, removing a ring of bark/phloem around a woody stem causes a bulge of sugars and tissue swelling ABOVE the ring (while tissue below withers), showing downward sugar transport occurs in the phloem. Supplying leaves with radioactive ¹⁴CO2 produces ¹⁴C-labelled sucrose that can be tracked along the phloem sieve tubes over time using autoradiography.
2	Compare Cell-Mediated Immunity (T-lymphocytes) and Humoral Immunity (B-lymphocytes).	Cell-mediated: Helper T-cells (TH) bind to specific antigens on APCs, undergo clonal expansion, and release cytokines that activate cytotoxic T-cells (TC, which secrete perforin to destroy virus-infected/cancerous body cells), phagocytes, and B-cells. Humoral: B-cells with complementary antibodies bind antigens, are activated by TH cytokines (clonal selection/expansion), and differentiate into plasma cells (secrete monoclonal antibodies) and memory B-cells.
3	Define the Bohr Effect (Bohr Shift) on the oxyhaemoglobin dissociation curve and explain its physiological importance.	High partial pressure of CO2 (pCO2) in actively respiring tissues lowers blood pH (via carbonic anhydrase forming H⁺ and HCO3⁻). H⁺ ions bind to haemoglobin, altering its quaternary/tertiary structure and lowering its affinity for oxygen. This shifts the oxyhaemoglobin dissociation curve to the RIGHT, causing haemoglobin to unload more O2 to respiring muscles.
4	Outline the stages of phagocytosis by a macrophage or neutrophil.	1) Phagocyte is attracted to the pathogen by chemical products (chemotaxis) and binds to pathogen antigens (aided by opsonins); 2) Phagocyte engulfs the pathogen by endocytosis to form a phagosome vesicle; 3) Lysosomes fuse with the phagosome (forming a phagolysosome) and release hydrolytic lysozymes that digest the pathogen; 4) Antigens are presented on MHC proteins on the cell surface (acting as an Antigen-Presenting Cell, APC).
5	State the formula for Simpson's Index of Diversity (D) and explain how species richness differs from index of diversity.	D = 1 − [Σ n(n − 1)] / [N(N − 1)] (or D = N(N − 1) / Σ n(n − 1) in AQA), where N is the total number of organisms of all species and n is the total number of organisms of each individual species. Species richness only counts the number of different species present, whereas an index of diversity takes into account both the number of species and the relative abundance (evenness) of each species.
6	Explain the Cohesion-Tension Theory of water transport in plant xylem vessels.	1) Water evaporates from mesophyll cell walls and transpires out through open stomata, lowering water potential in leaf cells; 2) This creates a negative hydrostatic pressure (transpiration pull / tension) at the top of the xylem; 3) Hydrogen bonds cause water molecules to stick together (cohesion) and to hydrophilic xylem walls (adhesion), pulling a continuous, unbroken column of water up from the roots.
7	Why is ATP described as the 'universal energy currency' of living cells, and which enzymes catalyse its hydrolysis and synthesis?	ATP (adenosine triphosphate) is hydrolysed by ATP hydrolase into ADP + Pi (inorganic phosphate), releasing a small, manageable packet of energy (30.5 kJ mol⁻¹) in a single-step reaction that can be coupled directly to metabolic processes, active transport, or muscle contraction, and phosphorylated biological molecules. ATP synthase resynthesises ATP from ADP + Pi during condensation.
8	How do antibodies lead to the destruction of pathogens via agglutination and opsonisation, and how does an ELISA test work?	Each antibody has two identical variable antigen-binding sites (complementary tertiary structure), allowing it to cross-link multiple pathogens into clumps (agglutination) for rapid phagocytosis, and acting as an opsonin marker. Direct/Indirect ELISA uses immobilised antigens/antibodies and a secondary antibody conjugated to an enzyme that catalyses a colour change when substrate is added.
9	Describe the process of semi-conservative DNA replication and the specific roles of DNA helicase and DNA polymerase.	1) DNA helicase breaks hydrogen bonds between complementary base pairs (A=T, G≡C), unwinding the double helix; 2) Both original strands act as templates as free activated DNA nucleotides align via complementary base pairing; 3) DNA polymerase catalyses condensation reactions forming phosphodiester bonds between adjacent nucleotides in the 5' to 3' direction.
10	How did Meselson and Stahl's ¹⁵N / ¹⁴N density-gradient centrifugation experiment prove semi-conservative DNA replication?	Bacteria grown in heavy nitrogen (¹⁵N) had a single low, dense DNA band. After 1 generation in light ¹⁴N medium, all DNA formed a single intermediate hybrid band (¹⁵N/¹⁴N), ruling out conservative replication. After 2 generations in ¹⁴N, two equal bands appeared (50% intermediate ¹⁵N/¹⁴N and 50% light ¹⁴N/¹⁴N), confirming semi-conservative replication.
11	Compare the structure and function of triglycerides and phospholipids.	Triglycerides have 1 glycerol bonded via 3 ester bonds to 3 fatty acid tails (hydrophobic, compact energy store, metabolic water source). Phospholipids replace one fatty acid with a hydrophilic phosphate head, making them amphipathic (hydrophilic head faces aqueous exterior/cytoplasm; two hydrophobic tails face inwards to form the cell membrane bilayer).
12	How do terrestrial insects balance efficient gas exchange through their tracheal system with minimising water loss?	Air enters via spiracles (muscular valves that close periodically to prevent water loss) into chitin-lined tracheae and branches into microscopic tracheoles whose ends penetrate directly into respiring muscle fibres (filled with tracheal fluid that withdraws during exercise via osmosis as lactate lowers muscle Ψ). A waterproof waxy exoskeleton cuticle also reduces evaporation.
13	Explain the Mass Flow Hypothesis for translocation of sucrose from source to sink in the phloem.	1) At the source (photosynthesising leaf), H⁺ ions are actively pumped out of companion cells (using ATP) and co-transport sucrose back in, which diffuses via plasmodesmata into sieve tube elements; 2) This lowers water potential (Ψ) in the sieve tube, so water enters from the adjacent xylem by osmosis, creating high hydrostatic pressure; 3) At the sink (respiring root/meristem), sucrose is actively unloaded and converted to starch/respired, raising Ψ so water leaves back to the xylem, lowering hydrostatic pressure and driving bulk mass flow from source to sink.
14	State Fick's Law of Diffusion and explain how fish gills achieve maximum gas exchange efficiency through counter-current flow.	Fick's Law: Rate of diffusion ∝ (Surface Area × Concentration Gradient) ÷ Diffusion Pathway Thickness. Fish gills have gill filaments covered in numerous lamellae (large SA) with a single-layer epithelium (short distance). In counter-current flow, blood flows through lamellae in the opposite direction to water, maintaining a steep O2 concentration gradient across the entire length of the gill lamella.
15	Compare the oxyhaemoglobin dissociation curves of fetal haemoglobin, lugworms, and llamas (high altitude) with adult human haemoglobin.	Curves shifted to the LEFT (fetal haemoglobin, lugworms, high-altitude llamas) indicate a HIGHER affinity for O2, allowing haemoglobin to load O2 efficiently in low-pO2 environments (e.g., across the placenta from maternal blood). Curves shifted to the RIGHT (mice, birds with high metabolic rates) indicate a LOWER affinity for O2, allowing rapid unloading of O2 for intense cellular respiration.
16	Explain how tissue fluid is formed at the arterial end of a capillary bed and reabsorbed at the venous end.	At the arterial end, high hydrostatic pressure (from left ventricular contraction) exceeds the opposing oncotic/water potential pull of plasma proteins, forcing water, glucose, amino acids, and O2 out through capillary fenestrations (ultrafiltration; large plasma proteins and RBCs remain in blood). At the venous end, hydrostatic pressure has dropped while blood water potential is very negative (due to retained plasma proteins), so ~90% of water returns by osmosis; the remaining ~10% drains into the lymphatic system.
17	Compare competitive and non-competitive enzyme inhibition and their effects on Vmax as substrate concentration increases.	Competitive inhibitors have a molecular shape similar to the substrate and bind reversibly to the active site; increasing substrate concentration outcompetes the inhibitor, so Vmax is still reached. Non-competitive inhibitors bind to an allosteric site (away from the active site), altering the enzyme's tertiary structure and active site shape; increasing substrate concentration cannot overcome this, so Vmax is permanently lowered.
18	State five biologically crucial properties of water and explain how each arises from hydrogen bonding between polar H2O molecules.	1) High specific heat capacity: H-bonds absorb heat energy, buffering aquatic habitats and intracellular temperatures; 2) High latent heat of vaporisation: evaporation provides an efficient cooling mechanism (sweating/transpiration); 3) Cohesion & surface tension: continuous xylem water columns; 4) Universal polar solvent: medium for metabolic reactions and mass transport; 5) Metabolite in hydrolysis and condensation reactions.
19	Define the primary, secondary, tertiary, and quaternary structures of a protein, specifying the exact chemical bonds involved.	Primary: linear sequence of amino acids joined by covalent peptide bonds. Secondary: folding into α-helices or β-pleated sheets stabilised by hydrogen bonds between -NH and -C=O groups of the polypeptide backbone. Tertiary: specific 3D folding stabilised by R-group interactions (ionic bonds, disulfide bridges between cysteine residues, hydrogen bonds, hydrophobic interactions). Quaternary: ≥2 polypeptide chains (and/or prosthetic groups, e.g., haem in haemoglobin).
20	How do crossing over (Prophase I) and independent assortment (Metaphase I) generate genetic variation during meiosis?	1) Crossing over: In Prophase I, homologous chromosomes pair up to form bivalents and non-sister chromatids exchange alleles at chiasmata, creating new combinations of linked alleles on recombinant chromatids. 2) Independent segregation/assortment: In Metaphase I, homologous pairs align randomly along the equator (2ⁿ possible combinations, where n is the haploid number), so maternal and paternal chromosomes are assorted independently into gametes.
21	How do alpha-glucose (α-glucose) and beta-glucose (β-glucose) isomers differ structurally, and which polymers do they form?	On carbon-1 (C1), α-glucose has the hydroxyl (-OH) group pointing BELOW the ring plane (forming starch [amylose + amylopectin] and glycogen via α-1,4 and α-1,6 glycosidic bonds), whereas β-glucose has the -OH group on C1 pointing ABOVE the ring plane (forming cellulose via β-1,4 glycosidic bonds).
22	Trace the myogenic electrical excitation of the mammalian heart through the SAN, AVN, Bundle of His, and Purkyne fibres.	1) Sinoatrial node (SAN, pacemaker in right atrium) initiates a wave of electrical excitation causing both atria to contract simultaneously (atrial systole); 2) Non-conducting atrioventricular septum stops impulse passing directly to ventricles; 3) Atrioventricular node (AVN) delays the impulse (~0.1 s) so atria finish emptying; 4) Impulse travels down the Bundle of His and Purkyne fibres to the apex, contracting ventricles from the bottom upwards.
23	Contrast directional selection (e.g., antibiotic resistance) and stabilising selection (e.g., human birth mass).	Directional selection occurs when the environment changes, favouring individuals with an extreme phenotype at one end of the distribution and shifting the population mean toward that extreme. Stabilising selection occurs in a stable environment, selecting against both extremes and favouring the intermediate mean phenotype, reducing standard deviation/phenotypic variation.
24	Relate the molecular structure of β-glucose chains in Cellulose to the high tensile strength of plant cell walls.	Adjacent β-glucose monomers are inverted by 180° to form straight, unbranched β-1,4 chains. Parallel chains are cross-linked by thousands of intermolecular hydrogen bonds, bundling into microfibrils and macrofibrils that provide high tensile strength to resist osmotic turgor pressure.
25	Contrast the 'induced-fit' model of enzyme action with the older 'lock-and-key' model.	In the induced-fit model, the enzyme's active site is not initially a rigid, exact fit for the substrate. As the substrate binds, it induces a conformational change in the tertiary structure of the active site, moulding tightly around the substrate and placing strain on substrate bonds to lower the activation energy of the transition state.
26	Explain the mechanism of sodium-glucose co-transport across the epithelial cells of the mammalian ileum.	1) Na⁺/K⁺ ATPase pumps actively transport Na⁺ ions out of the epithelial cell into the blood (using ATP), creating a low intracellular Na⁺ concentration; 2) Na⁺ ions diffuse down their concentration gradient from the ileum lumen into the epithelial cell via a sodium-glucose co-transporter protein, carrying glucose molecules against their concentration gradient; 3) Glucose moves into the blood via facilitated diffusion through channel/carrier proteins.
27	In the Fluid Mosaic Model of cell membranes, what is the exact role of cholesterol, glycoproteins, and glycolipids?	Cholesterol binds to hydrophobic fatty acid tails, restricting phospholipid movement at high temperatures (reducing fluidity/permeability) and preventing close packing/crystallisation at low temperatures. Glycoproteins and glycolipids act as cell-surface receptors for hormones/neurotransmitters, cell-to-cell recognition markers (antigens), and cell adhesion sites.
28	Relate the molecular structure of Starch (amylose and amylopectin) and Glycogen to their functions as energy storage molecules.	1) Coiled/helical α-1,4 chains make them compact, storing lots of glucose in a small volume; 2) Insoluble in water, so they do not lower the water potential (Ψ) of cells or cause osmotic lysis; 3) Branched α-1,6 glycosidic bonds (highly branched in glycogen) provide many terminal ends for rapid enzymatic hydrolysis back to α-glucose.
29	Why is the genetic code described as 'degenerate', 'non-overlapping', and 'universal'?	Degenerate: Most amino acids are coded for by more than one base triplet/codon (64 possible codons for 20 amino acids), which means some base substitution mutations are silent. Non-overlapping: Each nucleotide base is read only once as part of a single consecutive triplet. Universal: The same triplet codons specify the same amino acids in almost all known living organisms.
30	Describe the mechanism of human inspiration (inhaling) and expiration (exhaling) in terms of muscle contraction, volume, and pressure.	Inspiration (active): External intercostal muscles contract (ribs move up and out) and diaphragm muscles contract (diaphragm flattens) → thoracic volume increases → pulmonary pressure drops below atmospheric pressure → air flows into lungs. Normal expiration (passive): External intercostals and diaphragm relax (internal intercostals contract only in forced expiration) → thoracic volume decreases → pressure rises above atmospheric.
31	Describe the two-stage process of protein synthesis: Transcription in the nucleus and Translation at the ribosomes.	Transcription: RNA polymerase unwinds DNA and joins free RNA nucleotides complementary to the DNA template strand (with U replacing T) to form pre-mRNA; introns (non-coding regions) are spliced out to form mature mRNA (in eukaryotes). Translation: mRNA binds to a ribosome; tRNA molecules with complementary anticodons bring specific amino acids according to the mRNA codon sequence; peptidyl transferase forms peptide bonds using ATP.
32	Write the A-Level Biology water potential equation and explain what happens to an animal cell vs a plant cell in a hypotonic (high Ψ) solution.	Ψ (Water Potential) = Ψs (Solute Potential, always negative) + Ψp (Pressure Potential). Pure water has the maximum Ψ = 0 kPa. In a hypotonic solution (higher Ψ outside), water enters by osmosis: an animal red blood cell swells and bursts (haemolysis) because it lacks a cell wall; a plant cell swells until the inelastic cellulose cell wall exerts positive turgor pressure (Ψp), becoming turgid.