Science, Math & STEM English 6 Interactive Flashcard Decks
Complete UK A-Level & International A-Level Study Guide: Biology, Chemistry, Physics, Psychology & Economics
Achieve A/A* in UK A-Levels and International A-Levels (AQA, OCR, Edexcel, CIE) with 192 high-yield flashcards covering A-Level Biology, Organic & Physical Chemistry Mechanisms, Physics Fields & Mechanics, AQA Psychology Studies, and Economics.
Interactive Premade Curricula
UK A-Level (AQA, OCR, Edexcel): Biology, Chemistry, Physics, Psychology & Economics Flashcard Decks (6)
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Foundational EN → EN
A-Level Biology I: Biological Molecules, Nucleic Acids, Cell Membranes, Immunity & Exchange Surfaces
Foundational EN → EN
A-Level Chemistry I (Physical & Inorganic): Amount of Substance, Bonding, Energetics, Kinetics, Equilibria & Periodicity
Intermediate EN → EN
A-Level Biology II: Photosynthesis, Cellular Respiration, Nervous/Endocrine Coordination, Homeostasis, Genetics & Gene Expression
Intermediate EN → EN
A-Level Chemistry II (Organic & Analysis): Reaction Mechanisms, Carbonyls, Aromatic Chemistry, Amines, Polymers & NMR/IR Spectroscopy
Achieve A/A* in UK A-Levels and International A-Levels (AQA, OCR, Edexcel, CIE) with 192 high-yield flashcards covering A-Level Biology, Organic & Physical Chemistry Mechanisms, Physics Fields & Mechanics, AQA Psychology Studies, and Economics.
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1. Complete 6-Deck Progressive Curriculum Matrix (192 Flashcards)
2. Module-by-Module Syllabus, Quick-Reference Tables & Sample Flashcards
Module 1 (BASIC): A-Level Biology I: Biological Molecules, Nucleic Acids, Cell Membranes, Immunity & Exchange Surfaces
Module Scope & Exam Blueprint: 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.
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| High-Yield Exam Concept / Question | Verified Answer & Core Rule |
|---|---|
| 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. |
| 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. |
| 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. |
| 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. |
| 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. |
| 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. |
| 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. |
| # | Front (Active Recall Prompt) | Back (Verified Exam Explanation) |
|---|---|---|
| 1 | 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). |
| 2 | 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. |
| 3 | 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. |
| 4 | 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). |
| 5 | 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). |
| 6 | 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. |
Module 2 (BASIC): A-Level Chemistry I (Physical & Inorganic): Amount of Substance, Bonding, Energetics, Kinetics, Equilibria & Periodicity
Module Scope & Exam Blueprint: Core AS & A-Level Physical and Inorganic Chemistry across AQA, OCR, and Edexcel: s/p/d electron configurations, ionisation energies, VSEPR shapes, intermolecular forces, Hess's Law, Born-Haber cycles, Kp/Kc, pH/Ka/buffers, electrode potentials, and Group 2/7 trends.
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| High-Yield Exam Concept / Question | Verified Answer & Core Rule |
|---|---|
| State the Ideal Gas Equation with its strict SI units, and show how to calculate the Mr of a volatile liquid. | pV = nRT, where p = pressure in Pascals (Pa; 1 kPa = 1,000 Pa), V = volume in cubic metres (m³; 1 cm³ = 10⁻⁶ m³, 1 dm³ = 10⁻³ m³), n = moles (mol), R = 8.31 J K⁻¹ mol⁻¹, and T = temperature in Kelvin (K = °C + 273). Since n = m / Mr, substituting gives Mr = mRT / (pV). |
| Define Standard Enthalpy of Formation (ΔfH°) and Standard Enthalpy of Combustion (ΔcH°), and state their Hess's Law cycle equations. | ΔfH°: enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (100 kPa, 298 K). ΔcH°: enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions. Using Formation data: ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants). Using Combustion data: ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products). |
| How do you construct a Born–Haber Cycle for an ionic solid (e.g., NaCl), and why do experimental and theoretical lattice enthalpies sometimes differ? | ΔfH°(NaCl) = ΔatH°(Na) + IE1(Na) + ΔatH°(Cl) + EA1(Cl) + ΔLEH°(formation of NaCl). Theoretical lattice enthalpy assumes a purely 100% ionic model of spherical ions. If experimental lattice enthalpy is more exothermic (larger magnitude) than the theoretical value (e.g., in AgCl or MgI2), the compound has significant covalent character caused by a small, highly charged cation polarising a large anion. |
| State the Gibbs Free Energy equation (ΔG = ΔH − TΔS) and explain the four combinations of ΔH and ΔS that determine reaction feasibility. | A reaction is thermodynamically feasible (spontaneous) when ΔG ≤ 0 (where ΔS must be divided by 1,000 to convert J K⁻¹ mol⁻¹ into kJ K⁻¹ mol⁻¹). 1) ΔH < 0, ΔS > 0: feasible at ALL temperatures; 2) ΔH > 0, ΔS < 0: NEVER feasible at any temperature; 3) ΔH < 0, ΔS < 0: feasible only at LOW temperatures (T < ΔH/ΔS); 4) ΔH > 0, ΔS > 0: feasible only at HIGH temperatures (T > ΔH/ΔS). |
| Sketch and explain the Maxwell–Boltzmann distribution of molecular energies at two different temperatures (T2 > T1) and with a catalyst. | The curve starts at the origin (0,0), peaks at the most probable energy (Emp), has mean energy slightly to the right of Emp, and approaches the energy axis asymptotically; the total area under the curve equals the total number of molecules. At higher temperature T2, the peak shifts to the right and lowers, exponentially increasing the area past Ea. A catalyst shifts the Ea line to the left without changing the shape of the curve. |
| How do you determine the order of reaction (0, 1, or 2) with respect to a reactant from concentration–time and rate–concentration graphs, and state the Arrhenius equation? | Zero order: [A] vs t is a straight downward line (gradient = −k); Rate vs [A] is horizontal. First order: [A] vs t is exponential decay with constant half-life (t₁/₂ = ln 2 / k); Rate vs [A] is a straight line through origin. Second order: Rate vs [A] is an upward parabola (Rate vs [A]² is linear). Arrhenius equation: k = A e^(−Ea/RT) or ln k = −(Ea/R)(1/T) + ln A (gradient of ln k vs 1/T is −Ea/R). |
| How does the rate equation relate to the Rate-Determining Step (RDS) of a multi-step reaction mechanism? | Only the species involved in the slowest step (the Rate-Determining Step, RDS) or in fast equilibrium steps prior to the RDS appear in the rate equation, and the order with respect to each reactant equals the stoichiometric number of moles of that species involved up to and including the RDS. |
| Write the expressions for Kc and Kp for the general gas reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), and state what factor changes Kc or Kp. | Kc = ([C]^c [D]^d) / ([A]^a [B]^b). Kp = (p(C)^c × p(D)^d) / (p(A)^a × p(B)^b), where partial pressure p(X) = mole fraction χ(X) × total pressure P_total. ONLY temperature changes the numerical value of Kc or Kp (concentration, pressure, and catalysts have zero effect on Kc or Kp). |
| # | Front (Active Recall Prompt) | Back (Verified Exam Explanation) |
|---|---|---|
| 1 | describe the four stages of Time-of-Flight (ToF) Mass Spectrometry: Ionisation, Acceleration, Ion Drift, and Detection. | 1) Ionisation: Electrospray ionisation (sample dissolved in volatile solvent, pushed through high-voltage needle to gain H⁺ → MH⁺(g), no fragmentation) OR Electron impact (vaporised sample bombarded with high-energy electrons: X(g) + e⁻ → X⁺(g) + 2e⁻). 2) Acceleration: positive ions accelerated by an electric field so all ions have the same kinetic energy (Ek = ½mv²). 3) Ion Drift: lighter ions (lower m/z) travel faster through the flight tube (t = d√(m/2Ek)). 4) Detection: ions hit negative detector plate, gaining electrons and generating a current proportional to abundance. |
| 2 | Write the full electron configurations of Chromium (Z=24) and Copper (Z=29) and explain why they are exceptions to the Aufbau rule. | Chromium (Z=24): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹ (NOT 3d⁴ 4s²). Copper (Z=29): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (NOT 3d⁹ 4s²). Half-filled (3d⁵) and completely filled (3d¹⁰) d-subshells provide extra thermodynamic stability due to symmetrical electron distribution and exchange energy. When transition metals form positive ions, 4s electrons are ALWAYS lost before 3d electrons. |
| 3 | Explain why there is a slight drop in First Ionisation Energy between Magnesium and Aluminium, and between Phosphorus and Sulfur, across Period 3. | Mg (3s²) to Al (3s² 3p¹): Al's outer electron is in a higher-energy 3p subshell, which is slightly further from the nucleus and shielded by the 3s² electrons, so it requires less energy to remove. P (3p³) to S (3p⁴): in P, each of the three 3p orbitals is singly occupied, whereas in S, one 3p orbital contains a spin-paired pair of electrons; mutual repulsion between the paired electrons makes one easier to remove. |
| 4 | Using VSEPR theory, state the shapes and bond angles of molecules/ions with 2, 3, 4, 5, and 6 bonding pairs and 0 lone pairs. | 2 bonding pairs (e.g., BeCl2, CO2): Linear, 180°. 3 bp (e.g., BF3): Trigonal planar, 120°. 4 bp (e.g., CH4, NH4⁺): Tetrahedral, 109.5°. 5 bp (e.g., PCl5): Trigonal bipyramidal, 90° and 120°. 6 bp (e.g., SF6): Octahedral, 90°. |
| 5 | How do lone pairs of electrons affect bond angles in VSEPR theory? State the shapes and bond angles of NH3, H2O, SF4, and XeF4. | Lone pair–lone pair repulsion > lone pair–bonding pair repulsion > bonding pair–bonding pair repulsion, reducing bond angles by ~2.5° per lone pair. NH3 (3 bp, 1 lp): Trigonal pyramidal, 107°. H2O (2 bp, 2 lp): Bent / non-linear, 104.5°. SF4 (4 bp, 1 lp): Seesaw, 87° and 102°. XeF4 (4 bp, 2 lp): Square planar, 90°. |
| 6 | Compare the three types of intermolecular forces (London dispersion forces, permanent dipole–dipole, and hydrogen bonding) and explain why ice is less dense than liquid water. | 1) Induced dipole–dipole (London dispersion/van der Waals): present in all molecules due to temporary fluctuating electron clouds; increases with number of electrons/surface area. 2) Permanent dipole–dipole: between polar molecules with unsymmetrical electronegativity. 3) Hydrogen bonding: strongest IMF, occurs between a δ+ H bonded to N, O, or F and a lone pair on N, O, or F of another molecule. In ice, each H2O forms 4 tetrahedral H-bonds, holding molecules in an open lattice with large spaces. |
Module 3 (INTERMEDIATE): A-Level Biology II: Photosynthesis, Cellular Respiration, Nervous/Endocrine Coordination, Homeostasis, Genetics & Gene Expression
Module Scope & Exam Blueprint: A2 / Year 13 A-Level Biology mastery: Light-Dependent & Calvin Cycle reactions, Glycolysis/Krebs/Oxidative Phosphorylation, action potentials & cholinergic synapses, sliding filament muscle contraction, nephron osmoregulation, Hardy-Weinberg, epistasis, epigenetics, and PCR/CRISPR.
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| High-Yield Exam Concept / Question | Verified Answer & Core Rule |
|---|---|
| How is a resting potential (−70 mV) maintained across a neurone axon membrane, and how is an action potential (+40 mV) generated and repolarised? | Resting potential (−70 mV): Na⁺/K⁺ ATPase pumps 3 Na⁺ out and 2 K⁺ in, while the membrane is much more permeable to K⁺ diffusing out than Na⁺ diffusing in. Depolarisation (+40 mV): Stimulus opens voltage-gated Na⁺ channels; Na⁺ rushes in down its electrochemical gradient (positive feedback once threshold −55 mV is reached). Repolarisation & Hyperpolarisation: At +40 mV, Na⁺ channels close and voltage-gated K⁺ channels open, so K⁺ diffuses out; K⁺ channels close slowly, causing temporary hyperpolarisation (−80 mV) and the refractory period. |
| Why is saltatory conduction along a myelinated neurone much faster than conduction along an unmyelinated neurone, and what are the roles of the refractory period? | The lipid myelin sheath (formed by Schwann cells) acts as an electrical insulator preventing ion movement across the membrane except at the Nodes of Ranvier. Local currents therefore jump from node to node (saltatory conduction), vastly increasing conduction velocity. The refractory period ensures: 1) Action potentials are propagated unidirectionally; 2) Action potentials remain discrete (do not merge); 3) It limits the maximum frequency of impulses. |
| Describe the sequence of transmission across a cholinergic (acetylcholine, ACh) synapse and distinguish spatial vs temporal summation. | 1) Action potential arrives at the synaptic knob, opening voltage-gated Ca²⁺ channels so Ca²⁺ ions diffuse in; 2) Ca²⁺ influx causes synaptic vesicles containing ACh to fuse with the presynaptic membrane and release ACh into the synaptic cleft by exocytosis; 3) ACh diffuses across the cleft and binds to nicotinic Na⁺ receptor channels on the postsynaptic membrane, opening them so Na⁺ enters (generating an EPSP); 4) Acetylcholinesterase (AChE) hydrolyses ACh into choline + ethanoic acid (recycled using ATP). Spatial summation = multiple presynaptic neurones; Temporal summation = high-frequency impulses from one neurone. |
| Explain the Sliding Filament Theory of skeletal muscle contraction, specifying the roles of Ca²⁺, Tropomyosin, Troponin, and ATP. | 1) Action potential travels down T-tubules, causing the sarcoplasmic reticulum (SR) to release Ca²⁺ ions into the sarcoplasm; 2) Ca²⁺ binds to troponin, causing it to change shape and pull tropomyosin away from the actin-myosin binding sites on the actin filament; 3) Myosin heads (with ADP+Pi attached) bind to actin to form actomyosin cross-bridges, nod forward in a 'power stroke' (releasing ADP+Pi), and pull actin filaments inwards; 4) A new ATP molecule binds to the myosin head to break the cross-bridge, and ATPase hydrolyses ATP → ADP + Pi to recock the myosin head. |
| What happens to the lengths of the Sarcomere, I-band, A-band, and H-zone when a myofibril contracts? | Remember the mnemonic 'HIA' (only A stays constant): As actin and myosin filaments slide past each other without shortening themselves, the Z-lines move closer together (sarcomere shortens), the light I-band (actin only) becomes NARROWER, the H-zone (myosin only) becomes NARROWER, while the dark A-band (full length of myosin filaments) stays CONSTANT in width. |
| Explain the Second Messenger Model of adrenaline and glucagon action on liver cells (hepatocytes). | 1) Adrenaline or glucagon (first messenger, peptide/catecholamine hormone unable to cross the phospholipid bilayer) binds to a specific complementary transmembrane receptor on the liver cell surface; 2) This activates a G-protein, which activates the membrane-bound enzyme adenylyl cyclase; 3) Adenylyl cyclase converts ATP into cyclic AMP (cAMP, the second messenger); 4) cAMP activates protein kinase A enzymes via an intracellular cascade that catalyses glycogenolysis (glycogen → glucose). |
| Explain how ultrafiltration in the Renal Bowman's Capsule and selective reabsorption in the Proximal Convoluted Tubule (PCT) work. | Ultrafiltration: The afferent arteriole is wider than the efferent arteriole, creating high hydrostatic pressure in the glomerulus that forces water, glucose, amino acids, urea, and ions through the capillary endothelium, basement membrane (the main molecular sieve stopping RBCs and large plasma proteins), and podocyte foot processes into the Bowman's capsule. Selective reabsorption in PCT: 100% of glucose and amino acids and ~85% of water/Na⁺ are reabsorbed into peritubular capillaries via Na⁺ co-transport (driven by basolateral Na⁺/K⁺ ATPase pumps and microvilli brush border). |
| How does the Loop of Henle act as a counter-current multiplier to concentrate urine in the kidney medulla? | 1) In the ascending limb (impermeable to water), Na⁺ and Cl⁻ ions are actively pumped out into the medulla interstitial fluid, lowering the water potential (Ψ) of the medulla; 2) In the descending limb (permeable to water, impermeable to ions), water moves out by osmosis into the hypertonic medulla and is carried away by the vasa recta capillaries, so filtrate reaches maximum concentration at the hairpin apex; 3) As filtrate flows past the hypertonic medulla down the collecting duct, water is reabsorbed by osmosis through aquaporins controlled by ADH. |
| # | Front (Active Recall Prompt) | Back (Verified Exam Explanation) |
|---|---|---|
| 1 | Describe the Light-Dependent Reaction (non-cyclic photophosphorylation) of photosynthesis across the thylakoid membrane. | 1) Photoionisation: Light excites electrons in chlorophyll at Photosystem II (PSII) and PSI; 2) Photolysis of water at PSII (2H2O → 4H⁺ + 4e⁻ + O2) replaces lost electrons; 3) Excited electrons pass along an electron transport chain (ETC), releasing energy to pump H⁺ protons from the stroma into the thylakoid lumen; 4) Chemiosmosis: H⁺ diffuses back into the stroma through ATP synthase, synthesising ATP from ADP + Pi; 5) Electrons from PSI + H⁺ reduce NADP to NADPH (catalysed by NADP reductase). |
| 2 | Describe the three stages of the Light-Independent Reaction (Calvin Cycle) in the chloroplast stroma. | 1) Carbon Fixation: CO2 combines with 5-carbon ribulose bisphosphate (RuBP), catalysed by the enzyme Rubisco, to form two molecules of 3-carbon glycerate 3-phosphate (GP); 2) Reduction: GP is reduced into two molecules of triose phosphate (TP / GALP) using energy from ATP hydrolysis and H atoms from reduced NADP (NADPH); 3) Regeneration: 5 out of every 6 TP molecules are used to regenerate RuBP (using ATP), while 1 TP is used to synthesise hexose sugars, glycerol, fatty acids, or amino acids. |
| 3 | Trace the four stages of aerobic cellular respiration, stating the exact cellular location and net products of Glycolysis and the Link Reaction. | 1) Glycolysis (cytoplasm, anaerobic): Glucose (6C) is phosphorylated by 2 ATP to hexose bisphosphate, splits into 2× triose phosphate (TP), and is oxidised to 2× Pyruvate (3C), yielding a net gain of 2 ATP (substrate-level phosphorylation) and 2 reduced NAD (NADH). 2) Link Reaction (mitochondrial matrix): Pyruvate (3C) is decarboxylated (−CO2) and oxidised (forming NADH) and combines with Coenzyme A to form Acetyl Coenzyme A (2C). |
| 4 | Describe the Krebs Cycle (in the mitochondrial matrix) and Oxidative Phosphorylation (across the inner mitochondrial cristae). | Krebs Cycle: Acetyl CoA (2C) combines with a 4C acceptor (oxaloacetate) to form a 6C molecule (citrate), which undergoes serial decarboxylation and dehydrogenation to regenerate the 4C acceptor; per cycle it yields 2 CO2, 1 ATP (substrate-level), 3 reduced NAD, and 1 reduced FAD. Oxidative Phosphorylation: NADH and FADH2 donate H atoms (splitting into H⁺ + e⁻); electrons pass along the cristae ETC, pumping H⁺ into the intermembrane space; H⁺ flows back through ATP synthase (chemiosmosis) to make ATP, and O2 acts as the final electron acceptor (½O2 + 2H⁺ + 2e⁻ → H2O). |
| 5 | How do you calculate NPP (Net Primary Production), Net Production of Consumers (N), and Respiratory Quotient (RQ)? | NPP = GPP (Gross Primary Production) − R (respiratory heat losses by plants). Consumer Net Production: N = I − (F + R), where I = chemical energy in ingested food, F = energy lost in faeces/urine, and R = respiratory energy loss. Respiratory Quotient: RQ = Volume of CO2 produced ÷ Volume of O2 consumed (Carbohydrates = 1.0, Proteins ≈ 0.9, Lipids ≈ 0.7; RQ > 1.0 indicates anaerobic respiration is occurring). |
| 6 | Trace the stages of the Nitrogen Cycle, naming the specific bacterial processes involved. | 1) Nitrogen fixation: Nitrogen-fixing bacteria (free-living Azotobacter or mutualistic Rhizobium in leguminous root nodules) reduce atmospheric N2 gas into ammonium ions (NH4⁺); 2) Ammonification: Saprobionts (decomposers) secrete extracellular enzymes to hydrolyse proteins/urea/DNA in dead organic matter into NH4⁺; 3) Nitrification (aerobic): Nitrifying bacteria oxidise NH4⁺ → Nitrites (NO2⁻) → Nitrates (NO3⁻); 4) Denitrification (anaerobic waterlogged soil): Denitrifying bacteria convert NO3⁻ back into N2 gas. |
Module 4 (INTERMEDIATE): A-Level Chemistry II (Organic & Analysis): Reaction Mechanisms, Carbonyls, Aromatic Chemistry, Amines, Polymers & NMR/IR Spectroscopy
Module Scope & Exam Blueprint: High-yield A-Level Organic Chemistry and Spectroscopy across AQA, OCR, and Edexcel: E/Z & optical isomerism, free-radical substitution, electrophilic/nucleophilic mechanisms, Friedel-Crafts acylation, condensation polymers, and 1H/13C NMR & IR interpretation.
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| High-Yield Exam Concept / Question | Verified Answer & Core Rule |
|---|---|
| How do you oxidise a primary alcohol to an aldehyde vs a carboxylic acid, and what happens with secondary and tertiary alcohols? | Reagent: acidified potassium dichromate(VI) (K2Cr2O7 / H2SO4, colour changes from orange Cr2O7²⁻ to green Cr³⁺). 1) Primary alcohol + distillation: distil off the aldehyde product immediately before further oxidation. 2) Primary alcohol + heating under reflux with excess oxidising agent: oxidises fully to a carboxylic acid. 3) Secondary alcohol + reflux: oxidises to a ketone. 4) Tertiary alcohol: resists oxidation (stays orange). |
| State the chemical tests to distinguish between an Aldehyde and a Ketone using Tollens' Reagent and Fehling's Solution. | Aldehydes are easily oxidised to carboxylic acids (or carboxylate ions), whereas ketones cannot be oxidised easily: 1) Tollens' reagent (ammoniacal silver nitrate, [Ag(NH3)2]⁺): warm gently in a water bath → aldehydes reduce Ag⁺ to metallic silver(0), forming a 'silver mirror' on the test tube wall; ketones give no change. 2) Fehling's / Benedict's solution (blue aqueous Cu²⁺ complex): warm → aldehydes reduce blue Cu²⁺ to a brick-red precipitate of copper(I) oxide (Cu2O); ketones remain blue. |
| Describe the reagents and Nucleophilic Addition mechanism for reducing aldehydes/ketones to alcohols and for adding HCN to form hydroxynitriles. | Reduction: aqueous sodium tetrahydridoborate(III) (NaBH4), which generates the hydride nucleophile (:H⁻) that attacks the δ+ carbonyl carbon, followed by protonation from water to form a primary alcohol (from aldehyde) or secondary alcohol (from ketone). Hydroxynitrile formation: KCN followed by dilute acid (HCl/H2SO4)—never HCN gas alone for safety; :CN⁻ attacks the δ+ C=O carbon, and the intermediate alkoxide ion O⁻ is protonated by H⁺/HCN, extending the carbon chain by 1 carbon. |
| Describe the Nucleophilic Addition–Elimination mechanism of Acyl Chlorides (RCOCl) and Acid Anhydrides with water, alcohols, ammonia, and primary amines. | The lone pair on the nucleophile (:OH2, :OHR', :NH3, :NH2R') attacks the strongly δ+ carbonyl carbon of the acyl chloride; the π-bond breaks to form a tetrahedral intermediate, then the lone pair on oxygen reforms the C=O double bond and expels the Cl⁻ leaving group, followed by loss of H⁺. Products with RCOCl: 1) + H2O → carboxylic acid + misty fumes of HCl; 2) + R'OH → ester + HCl; 3) + 2NH3 → primary amide + NH4Cl; 4) + 2R'NH2 → N-substituted secondary amide + R'NH3⁺Cl⁻. |
| Why is ethanoic anhydride preferred to ethanoyl chloride in the industrial manufacture of Aspirin? | Ethanoyl chloride is more reactive, but ethanoic anhydride is used industrially because: 1) It is cheaper to manufacture; 2) It is less corrosive and less vulnerable to hydrolysis by moisture; 3) Its by-product is weak ethanoic acid (CH3COOH) rather than toxic, corrosive fumes of hydrogen chloride gas (HCl). |
| Present the three pieces of evidence (bond lengths, enthalpy of hydrogenation, and resistance to bromination) that disprove Kekulé's alternating C=C/C–C model of Benzene. | 1) X-ray diffraction shows all six C–C bond lengths in benzene are identical (0.139 nm), intermediate between a C–C single bond (0.154 nm) and a C=C double bond (0.134 nm). 2) Thermochemistry: ΔH(hydrogenation) of cyclohexene is −120 kJ mol⁻¹, so Kekulé's cyclohexa-1,3,5-triene would be −360 kJ mol⁻¹, whereas actual benzene is −208 kJ mol⁻¹ (152 kJ mol⁻¹ more stable due to delocalisation energy). 3) Reactivity: Benzene does not decolourise bromine water (undergoes electrophilic substitution rather than addition to preserve the delocalised π-ring). |
| Write the equations and Electrophilic Substitution mechanism for the Nitration of Benzene. | Reagents: concentrated nitric acid (HNO3) + concentrated sulfuric acid (H2SO4 catalyst) at 50°C. Generation of the nitronium ion electrophile: HNO3 + 2H2SO4 → NO2⁺ + 2HSO4⁻ + H3O⁺ (or HNO3 + H2SO4 → NO2⁺ + HSO4⁻ + H2O). Mechanism: curly arrow from benzene's delocalised π-electron ring to N of NO2⁺ forming a horseshoe Wheland intermediate (covering 5 carbons with a + charge), then curly arrow from C–H bond back into the ring to restore aromaticity and release H⁺ (which regenerates the H2SO4 catalyst: H⁺ + HSO4⁻ → H2SO4). |
| Write the equations and Electrophilic Substitution mechanism for Friedel–Crafts Acylation of benzene with ethanoyl chloride and AlCl3. | Reagents: acyl chloride (e.g., CH3COCl) + anhydrous aluminium chloride catalyst (AlCl3) under reflux in dry ether. Generation of acylium electrophile: CH3COCl + AlCl3 → CH3C⁺=O + AlCl4⁻. Mechanism: curly arrow from benzene's delocalised π-ring to the positively charged C⁺ of CH3CO⁺, forming the horseshoe intermediate, then curly arrow from the C–H bond back into the ring to form phenylethanone (a ketone) + H⁺, which reacts with AlCl4⁻ → AlCl3 + HCl. |
| # | Front (Active Recall Prompt) | Back (Verified Exam Explanation) |
|---|---|---|
| 1 | Contrast Structural Isomerism (chain, position, functional group) with Stereoisomerism (E/Z and optical isomerism). | Structural isomers have the same molecular formula but different structural formulae (chain branching, functional group position, or different functional groups such as aldehydes vs ketones). Stereoisomers have the same structural connectivity but a different 3D spatial arrangement of atoms: E/Z isomers arise from restricted rotation around a C=C double bond (using Cahn–Ingold–Prelog atomic number priority rules: Z = 'zusammen' on same side, E = 'entgegen' on opposite sides); Optical isomers (enantiomers) are non-superimposable mirror images around a chiral carbon atom bonded to 4 different groups. |
| 2 | What is a racemic mixture (racemate), why is it optically inactive, and why does nucleophilic addition to an unsymmetrical planar carbonyl (RCHO) produce a racemate? | A racemic mixture contains an equimolar (50:50) mixture of two enantiomers; it is optically inactive because the two enantiomers rotate plane-polarised light by equal angles in opposite directions, cancelling out. In nucleophilic addition of HCN (CN⁻) to an unsymmetrical aldehyde (e.g., ethanal), the C=O group is trigonal planar, so the CN⁻ nucleophile attacks with equal probability from above or below the plane, producing a 50:50 racemate. |
| 3 | Write the three stages (Initiation, Propagation, Termination) of the free-radical substitution of methane (CH4) with chlorine (Cl2). | 1) Initiation (UV light homolytic fission): Cl2 → 2Cl•. 2) Propagation (chain reaction): Cl• + CH4 → •CH3 + HCl, followed by •CH3 + Cl2 → CH3Cl + Cl• (overall: CH4 + Cl2 → CH3Cl + HCl). 3) Termination (two radicals collide): 2Cl• → Cl2, 2•CH3 → C2H6, or •CH3 + Cl• → CH3Cl. Further substitution produces CH2Cl2, CHCl3, and CCl4. |
| 4 | Explain the mechanism of Electrophilic Addition of HBr to an unsymmetrical alkene (propene) and why Markovnikov's Rule favours the major product. | 1) The high electron density of the C=C π-bond attacks the δ+ H of H–Br, breaking the H–Br bond heterolytically; 2) In propene, protonation forms either a secondary carbocation intermediate (CH3–CH⁺–CH3) or a primary carbocation (CH3–CH2–CH2⁺); 3) Alkyl groups are electron-releasing (+I positive inductive effect), stabilising the positive charge, so the more stable secondary carbocation forms faster and is attacked by the :Br⁻ ion to give 2-bromopropane as the major product. |
| 5 | Compare the conditions and mechanisms for converting a haloalkane (e.g., 2-bromopropane) into an alcohol (Nucleophilic Substitution) vs an alkene (Elimination). | In both reactions the reagent is potassium/sodium hydroxide (KOH/NaOH), where OH⁻ acts in two different roles depending on the solvent: 1) Warm AQUEOUS NaOH: :OH⁻ acts as a NUCLEOPHILE, attacking the δ+ carbon to substitute Br⁻ and form an alcohol. 2) Hot ETHANOLIC (anhydrous) KOH: :OH⁻ acts as a BASE, removing a proton (H⁺) from an adjacent β-carbon to eliminate HBr and form an alkene. |
| 6 | Compare the rate of hydrolysis of chloroalkanes, bromoalkanes, and iodoalkanes with aqueous silver nitrate in ethanol. | Rate of nucleophilic substitution increases down Group 7: 1-iodobutane > 1-bromobutane > 1-chlorobutane (yellow AgI precipitate appears fastest, then cream AgBr, then white AgCl slowest). Although C–Cl is the most polar bond, bond enthalpy is the dominant factor: the C–I bond is the longest and weakest (lowest bond enthalpy, 238 kJ mol⁻¹ vs 346 kJ mol⁻¹ for C–Cl), so C–I breaks with the lowest activation energy. |
Module 5 (ADVANCED): A-Level Physics & Mathematics: Mechanics, Electric/Magnetic/Gravitational Fields, Nuclear/Particle Physics & Calculus Integration
Module Scope & Exam Blueprint: High-yield A-Level Physics and Mathematics across AQA, OCR, and Edexcel: particle physics (quarks/leptons), photoelectric effect, projectile motion, Young modulus, Simple Harmonic Motion, gravitational/electric/magnetic fields, capacitor RC decay, nuclear binding energy, and calculus integration.
- Direct Deck Link: A-Level Physics & Mathematics: Mechanics, Electric/Magnetic/Gravitational Fields, Nuclear/Particle Physics & Calculus Integration (32 Cards)
- Browse Level Sets: ADVANCED Flashcard Sets on Flashcards.world
| High-Yield Exam Concept / Question | Verified Answer & Core Rule |
|---|---|
| How do you solve 2D Projectile Motion problems using the SUVAT equations? | Resolve initial velocity u at angle θ above the horizontal into independent perpendicular components: Horizontal velocity u_x = u cos θ (constant velocity since horizontal air resistance = 0, so horizontal range x = (u cos θ)t). Vertical velocity u_y = u sin θ (experiences constant downward acceleration a_y = −g = −9.81 m s⁻²). At maximum height, v_y = 0, so v_y² = u_y² − 2gh_max gives h_max = (u² sin² θ) / (2g). |
| Define Tensile Stress (σ), Tensile Strain (ε), and the Young Modulus (E), and state what the area under a force–extension vs stress–strain graph represents. | Tensile Stress: σ = F / A (in Pa or N m⁻², where A = πd²/4 is cross-sectional area). Tensile Strain: ε = ΔL / L (dimensionless ratio). Young Modulus: E = σ / ε = (FL) / (AΔL) (in Pa, equal to the gradient of the linear proportional region of a stress–strain graph). Area under a Force–Extension graph = Elastic strain energy stored (E = ½FΔL); Area under a Stress–Strain graph = Elastic strain energy stored per unit volume (½σε). |
| Define Resistivity (ρ), state the equation for the terminal potential difference of a cell with internal resistance r, and explain how to find ε and r from a V–I graph. | Resistivity: ρ = RA / L (in Ω m). For a battery with Electromotive Force (EMF, ε) and internal resistance r connected to external load R: ε = I(R + r) = V + Ir, which rearranges to V = −rI + ε. Plotting terminal p.d. (V) on the y-axis against current (I) on the x-axis gives a straight line with y-intercept = ε (EMF) and gradient = −r. |
| State the defining equation for Simple Harmonic Motion (SHM) and the equations for displacement, velocity, maximum acceleration, and period of a mass-spring system and simple pendulum. | Definition of SHM: acceleration is directly proportional to displacement from the equilibrium position and always directed towards the equilibrium position: a = −ω²x (where ω = 2πf = 2π/T). If x = A cos(ωt), then v = ±ω√(A² − x²), v_max = ωA (at equilibrium x=0), and a_max = ω²A (at amplitude x=±A). Mass-spring period: T = 2π√(m / k). Simple pendulum period: T = 2π√(L / g). |
| State the equations for Centripetal Acceleration and Centripetal Force, and explain the forces on a vehicle over a curved bridge or banked track. | Angular speed: ω = v / r = 2πf (rad s⁻¹). Centripetal acceleration: a = v² / r = ω²r (directed towards the centre of the circle). Centripetal force: F = mv² / r = mω²r. Over the top of a convex humpback bridge of radius r: mg − R = mv² / r, so the vehicle loses contact with the road (R = 0) when v ≥ √(gr). |
| Compare Newton's Law of Gravitation, Gravitational Field Strength (g), Gravitational Potential (V), and Kepler's Third Law for circular orbits. | Gravitational Force: F = Gm₁m₂ / r². Radial Gravitational Field Strength: g = F / m = GM / r² (N kg⁻¹). Gravitational Potential (work done per unit mass to move from infinity to a point, always negative): V = −GM / r (J kg⁻¹; Escape velocity v_esc = √(2GM / R)); also g = −ΔV / Δr. Equating gravitational force to centripetal force (GMm / r² = mv² / r where v = 2πr / T) gives Kepler's 3rd Law: T² = (4π² / GM) r³. |
| Compare Coulomb's Law, Electric Field Strength (E) in radial vs uniform fields, and Electric Potential (V). | Coulomb's Law in free space: F = Q₁Q₂ / (4πε₀r²). Radial Electric Field Strength: E = F / q = Q / (4πε₀r²) (N C⁻¹ or V m⁻¹). Uniform Electric Field between parallel plates: E = V / d (a charged particle enters horizontally and follows a parabolic trajectory, just like a projectile in uniform gravity). Radial Electric Potential: V = Q / (4πε₀r) (positive for +Q, negative for −Q). |
| Define Capacitance (C), state the energy stored in a capacitor, and give the RC exponential discharge equations and time constant τ = RC. | Capacitance: C = Q / V, and for parallel plates C = ε₀εᵣA / d (in Farads, F). Energy stored: E = ½QV = ½CV² = ½Q²/C. When a capacitor discharges through a resistor R: X(t) = X₀ e^(−t/RC), where X can be Charge Q, Voltage V, or Current I. The Time Constant τ = RC is the time taken for Q, V, or I to fall to 1/e (≈ 37%) of its initial value, and the half-life is t₁/₂ = RC ln 2 ≈ 0.693 RC. |
| # | Front (Active Recall Prompt) | Back (Verified Exam Explanation) |
|---|---|---|
| 1 | Define Specific Charge of a nucleus or ion, and compare the quark compositions of a proton, neutron, and pion. | Specific charge = Total Charge ÷ Total Mass (Q / m, in C kg⁻¹). Hadrons experience the strong nuclear force and are subdivided into Baryons (3 quarks, baryon number B = +1) and Mesons (1 quark + 1 antiquark, B = 0). Proton = uud (+⅔ + ⅔ − ⅓ = +1); Neutron = udd (+⅔ − ⅓ − ⅓ = 0). Pions (π⁺ = u d̄, π⁻ = d ū, π⁰ = u ū or d d̄) are the exchange particles of the strong nuclear force; W⁺/W⁻ bosons mediate the weak nuclear force (β decay and electron capture). |
| 2 | Explain Einstein's Photoelectric Effect equation (hf = ϕ + Ek(max)) and why classical wave theory could not explain the threshold frequency. | In classical wave theory, light of any frequency should eventually transfer enough energy to eject photoelectrons if bright enough. In reality, no electrons are emitted below a threshold frequency f₀ regardless of intensity, and emission is instantaneous because each electron absorbs a single discrete photon of energy E = hf in a one-to-one interaction. Work function ϕ = hf₀ is the minimum energy to liberate a surface electron; remaining energy becomes maximum kinetic energy: hf = ϕ + ½mv²_max (stopping potential: eVs = Ek(max)). |
| 3 | State the de Broglie wavelength equation (λ = h / p) and describe the experimental evidence for wave–particle duality of electrons. | λ = h / p = h / (mv), where h = Planck's constant (6.63 × 10⁻³⁴ J s) and mv = momentum. Evidence for wave nature of electrons: when a beam of accelerated electrons (where eV = ½mv²) passes through a thin polycrystalline graphite film, it diffracts through the atomic lattice spacings and forms concentric bright and dark interference rings on a fluorescent screen. Increasing accelerating voltage increases electron speed, decreasing λ and making the diffraction rings smaller. |
| 4 | How do stationary (standing) waves form on a stretched string, and what is the equation for the first harmonic (fundamental) frequency? | A stationary wave is formed by the superposition of two progressive waves of the same frequency, wavelength, and amplitude travelling in opposite directions. Nodes are points of zero amplitude (destructive interference, phase difference π rad); antinodes are points of maximum amplitude (constructive interference). For the first harmonic on a string of length L (with nodes at both ends, L = λ/2): f₁ = (1 / 2L) √(T / μ), where T is tension (N) and μ is mass per unit length (kg m⁻¹). |
| 5 | State the equations for Young's Double-Slit Interference (w = λD / s) and a Diffraction Grating (d sin θ = nλ). | Double-slit fringe spacing: w = λD / s (where w = distance between adjacent bright fringes, λ = wavelength, D = distance from slits to screen, s = slit separation, valid when s ≪ D). Diffraction grating: d sin θ = nλ (where d = 1 / N is the slit spacing in metres for a grating with N lines per metre, θ = angle of diffraction, and n = order of maximum; maximum order is found by setting sin θ = 1 and rounding n down to the nearest integer). |
| 6 | State Snell's Law of Refraction (n₁ sin θ₁ = n₂ sin θ₂) and the two conditions required for Total Internal Reflection (TIR) in an optical fibre. | Refractive index of a medium: n = c / v. Snell's Law: n₁ sin θ₁ = n₂ sin θ₂. At the critical angle θc (where θ₂ = 90°): sin θc = n₂ / n₁ (for n₁ > n₂). Total Internal Reflection occurs when: 1) Light travels from a more optically dense core (higher n₁) towards a less optically dense cladding (lower n₂); AND 2) The angle of incidence θ₁ exceeds the critical angle θc. Step-index optical fibres use a narrow core and monochromatic light to minimise modal and material pulse dispersion. |
Module 6 (ADVANCED): A-Level Psychology & Economics: AQA Core Studies, Approaches, Biopsychology, Research Methods, Market Failure & Macroeconomic Policy
Module Scope & Exam Blueprint: High-yield A-Level Psychology (AQA/OCR) and A-Level Economics (Edexcel/AQA/OCR): Memory models, Attachment, Social Influence, Psychopathology, Biopsychology, Inferential Stats, Price/Income/Cross Elasticities, Market Failure externalities, Market Structures, and AD/AS Macro Policy.
- Direct Deck Link: A-Level Psychology & Economics: AQA Core Studies, Approaches, Biopsychology, Research Methods, Market Failure & Macroeconomic Policy (32 Cards)
- Browse Level Sets: ADVANCED Flashcard Sets on Flashcards.world
| High-Yield Exam Concept / Question | Verified Answer & Core Rule |
|---|---|
| In Biopsychology, explain Localisation of Function in the Brain, Sperry's (1968) Split-Brain research, Plasticity/Functional Recovery, and Biological Rhythms. | Localisation: Motor cortex (frontal lobe), Somatosensory cortex (parietal lobe), Visual cortex (occipital lobe), Auditory cortex (temporal lobe), Broca's area (left frontal—speech production; damage = Broca's aphasia), Wernicke's area (left temporal—speech comprehension). Sperry (1968): corpus callosum severed in 11 epilepsy patients; images in left visual field (processed by right hemisphere) could not be named verbally (left hemisphere is language-dominant) but could be drawn with the left hand. Plasticity (Maguire 2000: London taxi drivers had larger posterior hippocampi) and functional recovery via axonal sprouting and neuronal unmasking. Rhythms: Circadian (~24h sleep-wake driven by suprachiasmatic nucleus SCN endogenous pacemaker + light exogenous zeitgeber; Siffre cave study), Infradian (>24h menstrual/SAD), Ultradian (<24h 90-minute 5-stage REM/NREM sleep cycle; Dement & Kleitman). |
| How do you choose the correct Inferential Statistical Test in AQA A-Level Psychology using the 3-criterion decision matrix? | Identify: 1) Test of Difference vs Correlation; 2) Experimental Design (Independent Groups vs Repeated Measures/Matched Pairs); 3) Level of Measurement (Nominal = categories, Ordinal = ranks/scores, Interval = equal standard units). For Nominal Difference (Independent) → Chi-Squared (χ²); Nominal Difference (Repeated) → Sign Test; Ordinal Difference (Independent) → Mann–Whitney U; Ordinal Difference (Repeated) → Wilcoxon T; Interval Difference (Independent) → Unrelated t-test; Interval Difference (Repeated) → Related t-test; Ordinal Correlation → Spearman's rho; Interval Correlation → Pearson's r. |
| Define Type I vs Type II Statistical Errors, Reliability vs Validity, and the five major Issues and Debates in A-Level Psychology. | Type I Error (false positive): rejecting a true null hypothesis because p-level is too lenient (e.g., p = 0.10). Type II Error (false negative): accepting a false null hypothesis because p-level is too stringent (e.g., p = 0.01). Reliability = consistency (test-retest, inter-observer ≥ +0.80); Validity = accuracy (internal vs external ecological/population/temporal validity). Five Issues & Debates: 1) Gender & Culture bias (alpha/beta bias, ethnocentrism, etic vs emic); 2) Free Will vs Determinism (hard/soft, biological/environmental/psychic); 3) Nature vs Nurture (interactionist diathesis-stress model); 4) Holism vs Reductionism (levels of explanation); 5) Idiographic (qualitative case studies) vs Nomothetic (general laws). |
| In A-Level Economics, define Scarcity, Opportunity Cost, the Production Possibility Frontier (PPF), and Positive vs Normative Statements. | The basic economic problem is finite scarce resources (CELL: Capital, Enterprise, Land, Labour) relative to infinite human wants. Opportunity cost is the value of the next best alternative foregone when a choice is made. A PPF shows the maximum productive potential of an economy between capital and consumer goods: points ON the curve are productively efficient (concave due to diminishing marginal returns / imperfect factor substitutability); points INSIDE represent unemployment/inefficiency; outward shifts represent long-run economic growth (increase in quantity/quality of factors of production). |
| State the formulas and interpretations for Price Elasticity of Demand (PED), Income Elasticity of Demand (YED), Cross Elasticity of Demand (XED), and Price Elasticity of Supply (PES). | Elasticity = (%Δ Quantity) ÷ (%Δ Price or Income). 1) PED: |PED| > 1 is price elastic (lowering price INCREASES total revenue TR = P×Q); |PED| < 1 is price inelastic (raising price INCREASES TR); |PED| = 1 maximises TR; determined by SPLAT (Substitutes, Percentage of income, Luxury/necessity, Addictiveness, Time). 2) YED: YED > 0 = Normal good (YED > 1 = Luxury good); YED < 0 = Inferior good. 3) XED: XED > 0 = Substitutes; XED < 0 = Complements. 4) PES: always positive; determined by PSSST (Production lag, Stocks, Spare capacity, Substitutability of factors, Time: momentary/short-run/long-run). |
| Explain the three functions of the Price Mechanism (Rationing, Incentive, Signalling) and distinguish Consumer Surplus from Producer Surplus. | Adam Smith's 'invisible hand' allocates scarce resources through prices: 1) Signalling: rising prices signal excess demand/scarcity to consumers and producers; 2) Incentive: higher prices create higher profit margins, incentivising firms to expand output; 3) Rationing: higher prices ration scarce goods to consumers with the highest willingness and ability to pay. Consumer surplus is the triangle below the demand curve and above equilibrium price P*; Producer surplus is the triangle above the supply curve and below P*. |
| Contrast Negative Production Externalities (e.g., carbon emissions) and Positive Consumption Externalities (merit goods, e.g., healthcare/education) using MSB/MSC diagrams. | Socially optimum equilibrium occurs where Marginal Social Cost = Marginal Social Benefit (MSC = MSB), whereas the free market produces where Marginal Private Cost = Marginal Private Benefit (MPC = MPB). 1) Negative production externality (MSC > MPC due to external pollution costs): free market overproduces (Q_m > Q_opt) and underprices, creating a welfare loss triangle pointing towards Q_opt. 2) Positive consumption externality (MSB > MPB due to external third-party benefits + information failure): free market underconsumes (Q_m < Q_opt), creating a potential welfare gain triangle pointing towards Q_opt. |
| Compare indirect taxes, subsidies, maximum/minimum prices, tradable pollution permits, and state provision as remedies for Market Failure, and explain Government Failure. | 1) Indirect tax (specific vs ad valorem Pigouvian tax): internalises negative externalities by shifting Supply left (MPC → MSC); incidence falls mostly on consumers if demand is price inelastic. 2) Subsidy: shifts Supply right to boost merit goods. 3) Maximum price (ceiling set BELOW P*, e.g., rent caps) causes excess demand/shortages; Minimum price (floor set ABOVE P*, e.g., minimum wage/alcohol unit pricing) causes excess supply. 4) Public goods (non-excludable and non-rivalrous, e.g., street lighting/flood defences) suffer from the free-rider problem and require direct state provision. Government Failure occurs when state intervention causes a net welfare loss due to unintended consequences, regulatory capture, information gaps, or high administrative costs. |
| # | Front (Active Recall Prompt) | Back (Verified Exam Explanation) |
|---|---|---|
| 1 | Compare the Multi-Store Model of Memory (Atkinson & Shiffrin, 1968) with the Working Memory Model (Baddeley & Hitch, 1974). | Multi-Store Model (MSM): unitary stores—Sensory Register (iconic/echoic, <0.5s, huge capacity) → Attention → Short-Term Memory (STM: acoustic coding, 7±2 items, 18–30s duration) → Maintenance/Prolonged Rehearsal → Long-Term Memory (LTM: semantic coding, unlimited capacity/duration). Working Memory Model (WMM) replaces unitary STM with an active multi-component system: Central Executive (attentional supervisor) controlling the Phonological Loop (phonological store + articulatory control process), Visuo-Spatial Sketchpad (visual cache + inner scribe), and Episodic Buffer (added 2000). |
| 2 | How do Loftus & Palmer (1974) and Gabbert et al. (2003) explain the inaccuracy of Eyewitness Testimony (EWT), and how does the Cognitive Interview improve recall? | Loftus & Palmer (1974): Leading questions alter speed estimates of a car crash ('smashed' = 40.5 mph vs 'contacted' = 31.8 mph) and cause false memory of broken glass via response-bias and memory substitution. Gabbert et al. (2003): 71% of witnesses recalled items they hadn't seen after post-event discussion (memory conformity/source monitoring error). Geiselman's Cognitive Interview uses 4 techniques: 1) Report everything; 2) Reinstate the context; 3) Reverse the order; 4) Change perspective. |
| 3 | Contrast Asch's (1951) Conformity baseline and variations with Milgram's (1963) Obedience study and situational variables. | Asch (1951, line judgement with confederates): 36.8% mean conformity on critical trials (75% conformed at least once); conformity dropped with a dissenter (to 5.5%) and rose with task difficulty (informational social influence, ISI) vs fear of rejection (normative social influence, NSI). Milgram (1963, electric shock generator): 65% obeyed to 450 V (100% to 300 V). Situational variables reducing obedience: Proximity (same room = 40%, touch proximity = 30%), Location (run-down office = 47.5%), and Uniform (ordinary clothes = 20%), explained by the shift from Autonomous to Agentic State and Legitimacy of Authority. |
| 4 | Summarise Bowlby's Monotropic Theory of Attachment, Lorenz & Harlow's animal studies, and Ainsworth's Strange Situation classifications. | Lorenz (1935): goslings imprinted on the first moving object seen during a critical period (12–17 hours). Harlow (1958): infant rhesus monkeys preferred the cloth-covered 'contact comfort' mother over the wire milk-dispensing mother. Bowlby's Monotropic Theory: attachment is innate/adaptive, relies on social releasers, forms one primary bond (monotropy) within a critical period (~2.5 years), creating an Internal Working Model for future relationships (Maternal Deprivation Hypothesis → affectionless psychopathy in 44 Thieves study). Ainsworth's Strange Situation: Type B Secure (66%, moderate distress, easily comforted), Type A Insecure-Avoidant (22%, avoids caregiver), Type C Insecure-Resistant (12%, ambivalent/angry). |
| 5 | Compare the Behavioural (Mowrer's Two-Process Model), Cognitive (Beck's Negative Triad & Ellis's ABC Model), and Biological explanations/treatments of Phobias, Depression, and OCD. | Phobias: Mowrer's Two-Process Model—acquired by classical conditioning (Little Albert) and maintained by negative reinforcement/avoidance (operant conditioning); treated via Systematic Desensitisation (anxiety hierarchy + reciprocal inhibition + counterconditioning) or Flooding. Depression: Beck's cognitive vulnerability (faulty processing, negative self-schemas, Negative Triad of self/world/future) & Ellis's ABC Model (Activating event → Irrational Belief → Emotional Consequence); treated with CBT. OCD: Polygenic inheritance (COMT and SERT genes) + low serotonin + hyperactive orbitofrontal cortex; treated with SSRIs (e.g., fluoxetine, which blocks presynaptic serotonin reuptake). |
| 6 | Compare the six core Approaches in A-Level Psychology: Learning (Behaviourist & SLT), Cognitive, Biological, Psychodynamic, and Humanistic. | 1) Behaviourist (Pavlov, Skinner): blank slate, observable behaviour shaped by classical & operant conditioning. 2) Social Learning Theory (Bandura Bobo doll): vicarious reinforcement, mediational processes (Attention, Retention, Motor reproduction, Motivation) & identification. 3) Cognitive: internal mental processes, computer analogy, schemas, cognitive neuroscience. 4) Biological: genotype/phenotype, neurochemistry, evolution. 5) Psychodynamic (Freud): unconscious mind, tripartite personality (Id, Ego, Superego), defence mechanisms, psychosexual stages. 6) Humanistic (Maslow, Rogers): free will, hierarchy of needs, self-actualisation, congruence between self-concept and ideal self, unconditional positive regard. |
3. 4-Phase Active Recall & Spaced Repetition Study Plan
| Study Phase | Actionable Protocol |
|---|---|
| Phase 1: Foundation & Core Definitions (Days 1–7) | Complete Modules 1 & 2 (basic). Focus on foundational terminology, statutory definitions, core formulas, and classification rules using daily 20-minute active-recall sessions. |
| Phase 2: Mechanisms, Applied Problem-Solving & Frameworks (Days 8–16) | Master Modules 3 & 4 (intermediate). Practice multi-step mechanisms, comparative tables, case applications, and quantitative shortcuts while reviewing Phase 1 leeches. |
| Phase 3: Advanced Vignettes, Edge Cases & High-Difficulty Synthesis (Days 17–24) | Complete Modules 5 & 6 (advanced). Drill high-discrimination exam traps, clinical/legal vignettes, and multi-topic synthesis cards under timed conditions. |
| Phase 4: Spaced-Repetition Consolidation & Pre-Exam Sprint (Final 7 Days) | Filter ausschließlich for cards marked 'Again' or 'Hard' across all 192 cards, interleave all 6 decks, and simulate timed exam blocks. |
4. Top 5 High-Yield Exam Strategies & Scoring Tips
- Master Synoptic & Practical Application in A-Level Biology (Paper 3 Essay / Practical Endorsement): Over 30% of A-Level Biology marks test AO2 (Application to unfamiliar contexts) and AO3 (Experimental analysis & statistical evaluation). Always state the null hypothesis, check whether standard deviation/95% confidence error bars overlap, and link 25-mark synoptic essays (AQA Paper 3) across at least 4 distinct specification topics.
- Draw Curly Arrows with Precision in A-Level Organic Chemistry: Every curly arrow represents the movement of a pair of electrons (unless fishhook radicals are specified). Always start your curly arrow directly on a lone pair or the centre of a covalent bond, and point it directly to the target atom or bond—examiners penalise sloppy arrow origins and missing dipole charges (
δ+ / δ−). - Check Radians Mode & Derive Units in A-Level Physics and Maths: In A-Level Pure Mathematics, all calculus (differentiation and integration) of trigonometric functions assumes angles are in radians. In A-Level Physics, always convert prefixes (
fm = 10⁻¹⁵ m,nm = 10⁻⁹ m,MeV = 1.60 × 10⁻¹³ J) before substituting into equations. - Balance AO1 and AO3 in AQA Psychology 16-Mark Essays: In a 16-mark AQA Psychology essay, AO1 (Description) is worth only 6 marks while AO3 (Evaluation) is worth 10 marks. Structure 3–4 PEEL evaluation paragraphs using named research support, methodological critiques (validity/reliability), real-world therapeutic/economic applications, and Issues & Debates.
- Anchor Every 25-Mark A-Level Economics Essay in Diagrams and Elasticities: Never leave a diagram floating without explanation—reference exact coordinates (
P₁ → P₂,Q₁ → Q₂,Y_fe,MSC = MSB) in your prose, and evaluate policies by testing how the outcome changes whenPED,PES, spare capacity, or the Marginal Propensity to Import (MPM) changes.
5. Frequently Asked Questions (FAQ)
Do these A-Level flashcards cover AQA, OCR (A & B), Edexcel, and Cambridge International (CIE) specifications?
Yes. All 6 modules are engineered around the Ofqual common core for UK A-Level Biology, Chemistry, Physics, Mathematics, Psychology, and Economics, plus the corresponding Cambridge International (CAIE 9700/9701/9702/9708/9709) and Edexcel International A-Level (IAL) syllabuses.
How should Year 12 (AS) vs Year 13 (A2) students use this 6-deck curriculum?
Modules 1 and 2 (basic) cover the Year 12 / AS-Level foundations in Biological Molecules, Cell Membranes, Immunity, Physical Chemistry, Bonding, Equilibria, and Periodicity. Modules 3, 4, 5, and 6 (intermediate and advanced) cover full Year 13 / A2 topics including Respiration, Photosynthesis, Nervous/Renal Homeostasis, Organic Synthesis, NMR, Fields, Calculus, Psychology, and Economics.
What separates an A* candidate from a Grade B candidate in UK A-Levels?
Grade B candidates know the textbook definitions (AO1) but lose marks on multi-step mathematical derivations, spectroscopy synthesis problems, statistical test interpretations, and synoptic evaluation (AO2/AO3). Active recall with mechanism and evaluation cards builds automaticity for the hardest 25% of exam marks.
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