Molecular structure and bondsCore
Scope: Core · §3.1.1 Monomers and polymers
Open revision guidance
What to know
The biochemical basis of past and present life is similar: cells use a small set of carbon-based molecule groups in related ways, providing indirect evidence of common ancestry. Carbohydrates commonly fuel respiration and form structural components of membranes/cell walls; lipids form membrane bilayers, some hormones and respiratory substrates; proteins form cell structures, enzymes, messengers and blood components; nucleic acids carry genetic information for protein production. The genetic code is shared by viruses and living organisms; the common code is evidence of evolution. Water is the most common cell component, so searches for life elsewhere include searching for liquid water. A monomer is a small unit; a polymer contains many joined monomers. Monosaccharides, amino acids and nucleotides are examples. Condensation joins molecules by forming a bond and eliminating water; hydrolysis breaks a bond using water.
Amino acids share an amine group (NH₂), carboxyl group (COOH) and variable R side chain; the 20 common amino acids differ in their R groups. Condensation forms a peptide bond; two amino acids form a dipeptide, many form a polypeptide, and a functional protein may contain one or more polypeptide chains.
What to do in an assessment
Relate a change in sequence to altered folding and therefore function
distinguish a bond-forming reaction from hydrolysis.
In plain language
Relate a change in sequence to altered folding and therefore function; distinguish a bond-forming reaction from hydrolysis.
Self-check
What reaction joins two amino acids?
Show the answer
A condensation reaction removes water and forms a peptide bond between the amino acids.

Carbohydrates and lipid testsCore
Scope: Core · §3.1.2 Carbohydrates
Open revision guidance
What to know
Monosaccharides glucose, galactose and fructose are carbohydrate monomers. Glucose has α and β isomers that differ in the orientation of the hydroxyl group on carbon 1. Condensation forms glycosidic bonds; maltose is glucose+glucose, sucrose glucose+fructose and lactose glucose+galactose.
α-glucose forms starch: unbranched amylose coils and branched amylopectin make a compact plant energy store; highly branched glycogen is a compact animal glucose store that can be hydrolysed rapidly. β-glucose forms straight cellulose chains linked by hydrogen bonds into strong microfibrils that support plant walls. Structure–function follows from branching, compactness, insolubility and the straight cellulose fibres.
Triglycerides contain glycerol and three fatty acids; phospholipids have hydrophilic heads and hydrophobic tails that form bilayers.
Reducing sugars give a coloured precipitate with heated Benedict's reagent; test a non-reducing sugar by boiling with dilute acid, cooling, neutralising with sodium hydrogencarbonate, then repeating Benedict's test. Iodine/potassium iodide changes orange-brown to blue-black with starch. For chromatography, separate monosaccharides beside known standards and compare spots/Rf values under the same conditions; a glucose dilution series and colorimeter can produce a calibration curve for an unknown.
What to do in an assessment
For reducing sugar, add Benedict's reagent and heat in a hot-water bath; blue may progress through green, yellow and orange to a brick-red precipitate as concentration rises. For non-reducing sugar, first hydrolyse with dilute acid, cool, neutralise, then add Benedict's and heat; compare with an untreated sample if establishing that sugar was non-reducing.
Add iodine/potassium iodide for starch (orange-brown to blue-black); use a blank and known standards for colorimetry or chromatography, and distinguish a qualitative positive from a measured concentration.
In plain language
Use a named test and its positive result when interpreting a food sample; iodine changes from orange-brown to blue-black with starch.
Self-check
A sugar sample stays blue with Benedict's until it is boiled with dilute acid, neutralised, then tested again and forms a brick-red precipitate. What does this show?
Show the answer
The sample contained a non-reducing sugar: acid hydrolysis produced reducing monosaccharides that then reduced Benedict's reagent. The acid must be neutralised before the second test.

Protein function and enzymesCore
Scope: Core · §3.1.4.2 Many proteins are enzymes
Open revision guidance
What to know
An enzyme's tertiary structure forms a substrate-specific active site; complementary binding forms an enzyme–substrate complex. The older lock-and-key model treated the site as rigid; the induced-fit model accounts for evidence that substrate binding changes active-site shape slightly. Enzymes lower activation energy without changing reaction equilibrium and catalyse diverse intracellular and extracellular reactions that build and operate cells, tissues and whole organisms.
Increasing enzyme concentration raises rate if substrate is available; increasing substrate raises rate until active sites are saturated. Higher temperature initially increases collisions, then denaturation lowers rate; pH alters bonds/active-site shape. Competitive inhibitors occupy active sites; non-competitive inhibitors bind elsewhere and change shape.
What to do in an assessment
When comparing enzyme data, calculate initial rate from the steepest early interval and control pH, temperature, enzyme and substrate concentrations.
In plain language
When comparing enzyme data, calculate initial rate from the steepest early interval and control pH, temperature, enzyme and substrate concentrations.
Self-check
How does a competitive inhibitor reduce enzyme activity?
Show the answer
It occupies the active site, reducing substrate binding; adding more substrate can lessen its effect by competition.

DNA replication and expressionCore
Scope: Core · §3.1.5.1 Structure of DNA and RNA
Open revision guidance
What to know
DNA and RNA are nucleotide polymers; each nucleotide has a pentose, phosphate and base. DNA has deoxyribose and A, C, G, T; RNA has ribose and A, C, G, U. Nucleotides join by phosphodiester bonds; DNA is a double helix with complementary A–T and C–G hydrogen-bonded strands, while RNA is a relatively short polynucleotide. Ribosomes are made of RNA and protein.
In cells DNA holds genetic information and RNA transfers it from DNA to ribosomes, where it directs protein production. DNA replication is semi-conservative because each copy contains one template strand and one new complementary strand. DNA's chemical simplicity led some scientists to doubt that it could carry the genetic code.
What to do in an assessment
Track strand direction and complementary bases explicitly; in double-stranded DNA, A=T and C=G. Example: if A is 28% of bases, T is 28% and G=C=(100−56)/2=22%. Explain how a base substitution may be silent, alter one amino acid, or create a stop codon.
In plain language
Track strand direction and complementary bases explicitly; explain how a base substitution may be silent, alter one amino acid, or create a stop codon.
Self-check
Why is each new DNA molecule described as semi-conservative?
Show the answer
Each contains one parental strand retained from the original molecule and one newly synthesised complementary strand.

Cell ultrastructure and microscopyCore
Scope: Core · §3.2.1.1 Structure of eukaryotic cells
Open revision guidance
What to know
Cell theory unifies biology: cells share basic features, with extra features accounting for differences and providing indirect evidence for evolution. The cell-surface membrane controls exchange; the nucleus contains one or more nucleoli and chromosomes of linear DNA bound to proteins. Mitochondria carry out aerobic respiration; chloroplasts in plants and algae photosynthesise; ribosomes synthesise proteins.
The Golgi apparatus modifies and packages molecules into Golgi vesicles; lysosomes are membrane-bound organelles that release hydrolytic enzymes. Rough ER bears ribosomes and processes/transports proteins, while smooth ER synthesises lipids. A plant vacuole contains cell sap; cell walls in plants, algae and fungi support cells. Specialised eukaryotic cells form tissues, organs and organ systems; explain adaptations by linking each named structure to its function.
What to do in an assessment
Calculate actual size as image size ÷ magnification and check the units
compare organelle abundance to the cell’s metabolic role.
In plain language
Calculate actual size as image size ÷ magnification and check the units; compare organelle abundance to the cell’s metabolic role.
Self-check
A 12 mm image at ×30 000 represents what actual length?
Show the answer
12 mm ÷ 30 000 = 0.0004 mm = 0.4 µm.

Membrane transport and cell signallingCore
Scope: Core · §3.2.3 Transport across cell membranes
Open revision guidance
What to know
The fluid-mosaic membrane has mobile phospholipids, proteins, glycoproteins and glycolipids; cholesterol restricts movement. Membrane proteins can act as receptors in cell signalling; surface molecules help the immune system recognise pathogens, cells from other individuals, abnormal body cells and toxins. Simple diffusion passes small non-polar molecules through the bilayer; facilitated diffusion uses channels/carriers down a concentration gradient; osmosis is net water movement down water-potential gradient through a partially permeable membrane.
Active transport uses ATP hydrolysis and carrier proteins to move substances against a concentration gradient. In ileum co-transport, sodium moving down its gradient couples to glucose/amino-acid uptake; increased membrane area or transporter number raises capacity.
What to do in an assessment
Explain a transport result using concentration or water potential, membrane permeability, and energy use
do not call every membrane crossing diffusion.
In plain language
Explain a transport result using concentration or water potential, membrane permeability, and energy use; do not call every membrane crossing diffusion.
Self-check
How can glucose enter an intestinal epithelial cell against its gradient?
Show the answer
Sodium–glucose co-transport uses the sodium gradient maintained by ATP-driven pumps to bring glucose into the cell.

Cell cycle, mitosis and meiosisCore
Scope: Core · §3.2.2 All cells arise from other cells
Open revision guidance
What to know
Within multicellular organisms, not all cells retain the ability to divide; eukaryotic cells that do can pass through a cell cycle. DNA replicates in interphase. In mitosis chromosomes condense (prophase), line at the equator (metaphase), sister chromatids separate (anaphase), nuclei reform (telophase), and cytokinesis usually yields two genetically identical cells. Spindle fibres attach at centromeres. Explain each stage from chromosome appearance as well as its name.
Binary fission copies a prokaryote's circular DNA and plasmids before cytoplasm divides; each daughter has one circular DNA copy and a variable number of plasmids. Viruses do not divide; host cells make particles after viral nucleic acid enters. Mitosis is normally controlled; uncontrolled cell division can form tumours/cancers, and many treatments aim to control the rate of division.
What to do in an assessment
Distinguish homologous chromosomes from sister chromatids in diagrams and state chromosome number before and after each division. In a root-tip squash, locate the meristem, recognise phase-specific chromosome arrangement and calculate mitotic index; use actual size=image size/magnification with consistent units.
In plain language
Distinguish homologous chromosomes from sister chromatids in diagrams and state the chromosome number before and after each division.
Self-check
Why does meiosis halve chromosome number?
Show the answer
Homologous chromosomes separate in meiosis I and sister chromatids in meiosis II, producing cells with one chromosome of each pair.

Exchange surfaces and mass transportCore
Scope: Core · §3.3.1 Surface area to volume ratio
Open revision guidance
What to know
As an organism grows, volume increases faster than surface area, so surface-area-to-volume ratio falls while total metabolic demand increases; simple diffusion across the body surface cannot supply large organisms quickly enough. Flat or branched body shape and specialised exchange surfaces with mass-transport systems compensate. Use surface area ÷ volume with consistent units to calculate SA:V.
Thin, moist, large exchange surfaces with maintained concentration gradients speed diffusion; ventilation and perfusion renew gases in lungs, while counter-current flow maintains oxygen gradients across fish gills.
What to do in an assessment
Use Fick’s law qualitatively: larger area and steeper gradient increase rate, while a thicker barrier decreases it. In an agar-block diffusion investigation, vary block size or external concentration, keep other factors constant and compare the fraction/volume penetrated after a fixed time; diffusion distance relative to SA:V explains the result.
In plain language
Use Fick’s law qualitatively: larger area and steeper gradient increase rate, while a thicker barrier decreases it.
Self-check
Why does blood flow help maintain oxygen diffusion into an alveolus?
Show the answer
It carries absorbed oxygen away and brings deoxygenated blood, preserving a steep concentration gradient.
Exchange surfaces, tissue fluid and mass transportCore
Scope: Core · §3.3 introductory prose and §3.3.4 opening statement (printed pp. 26, 29)
Open revision guidance
What to know
A cell or organism has an internal environment distinct from the external environment. Exchange with the surroundings occurs at exchange surfaces, but most substances entering or leaving an organism must cross cell-surface (plasma) membranes.
For cells in large multicellular organisms, tissue fluid is the immediate environment. Many cells are too distant from exchange surfaces and one another for diffusion alone to keep tissue-fluid composition within a suitable metabolic range. Mass-transport systems move substances between exchange surfaces and body regions, maintaining the final diffusion gradients that bring materials to or from individual cell membranes and helping keep tissue fluid relatively stable.
Over large distances, mass transport provides efficient movement to and from exchange surfaces. Bulk movement in a transport system and diffusion across the final cell membrane are linked but distinct steps.
What to do in an assessment
Trace a substance from the external environment across an exchange surface, through mass transport and tissue fluid, then down its final diffusion gradient across a cell plasma membrane; reverse the route for a waste product. Explain why diffusion alone is inadequate over large organism-scale distances and how delivery/removal by mass transport maintains the gradient.
In plain language
Blood or another transport fluid carries substances over long distances; tissue fluid then brings them close to cells, where diffusion across the plasma membrane completes the exchange.
Self-check
Why can a large organism not rely on diffusion alone to deliver oxygen to every cell?
Show the answer
Many cells are far from exchange surfaces and from one another, so diffusion would be too slow to maintain tissue-fluid composition. Mass transport brings oxygen close to cells and maintains the gradient for final diffusion across each cell membrane.

Digestion and absorptionCore
Scope: Core · §3.3.3 Digestion and absorption
Open revision guidance
What to know
Digestion hydrolyses large biological molecules to smaller molecules that can cross cell membranes. Amylase hydrolyses starch; membrane-bound disaccharidases complete carbohydrate digestion. Lipase digests lipids, aided by bile salts that emulsify fat. Endopeptidases cut internal peptide bonds; exopeptidases remove terminal amino acids; membrane-bound dipeptidases finish at the brush border.
Ileal cells co-transport monosaccharides and amino acids with sodium. Lipid products enter micelles with bile salts, diffuse into epithelial cells, are reassembled and transported; villi/microvilli provide area, capillaries carry soluble products and lacteals receive lipid.
What to do in an assessment
To investigate a digestive enzyme, vary a named factor such as pH or bile-salt concentration, measure product formation/substrate disappearance over a fixed early interval and control enzyme/substrate amounts, temperature and time. A Visking-tubing model can test absorption: compare which digestion products cross a selectively permeable membrane over time using suitable tests, with an undigested or time-zero control. Explain a named adaptation in terms of the movement of a specific nutrient and its destination.
In plain language
Explain a named adaptation in terms of the movement of a specific nutrient and its destination.
Self-check
Where do absorbed fatty-acid products initially enter the villus?
Show the answer
They enter the lacteal and travel through the lymphatic system before reaching the blood.

Circulation and respiratory exchangeCore
Scope: Core · §3.3.4.1 Mass transport in animals
Open revision guidance
What to know
Red blood cells transport oxygen using haemoglobin, a quaternary protein that binds cooperatively: binding of the first oxygen changes its shape and makes further binding easier, producing a sigmoid dissociation curve. High oxygen partial pressure loads haemoglobin in lungs; low pressure unloads it in tissues. Raised CO₂/lower pH shifts unloading (Bohr effect); different species' haemoglobins suit different environments.
Mammalian circulation is double: vena cava → right heart → pulmonary artery → lungs → pulmonary vein → left heart → aorta → body. Renal arteries carry blood to kidneys and renal veins carry it away; coronary arteries supply the heart muscle. The heart has right/left atria and ventricles separated by a septum; atrioventricular and semilunar valves maintain one-way flow. Arteries have thick elastic/muscular walls, arterioles regulate flow, veins have wide lumens/valves, and one-cell-thick capillary walls form exchange beds.
During diastole, low chamber pressure lets blood enter and ventricles fill; atrial systole raises atrial pressure and pushes remaining blood into ventricles. Ventricular systole raises ventricular pressure, closing atrioventricular valves and opening semilunar valves to eject blood; ventricular relaxation lowers pressure, closing semilunar valves and allowing filling to resume. Capillary hydrostatic pressure drives fluid out to form tissue fluid; plasma proteins lower blood water potential, drawing much fluid back, with excess entering lymphatics.
Assess cardiovascular risk-factor evidence by checking sample size and confounders, separating association from cause, and comparing conflicting results. Cardiac output = stroke volume × heart rate, with units such as cm³ min⁻¹ or dm³ min⁻¹.
What to do in an assessment
Read a dissociation curve by locating partial pressure and saturation
compare curves only after identifying which condition shifts them.
In plain language
Read a dissociation curve by locating partial pressure and saturation; compare curves only after identifying which condition shifts them.
Self-check
Why does haemoglobin unload more oxygen in an active tissue?
Show the answer
High CO₂ lowers pH and shifts haemoglobin towards lower oxygen affinity, so oxygen is released where respiration is high.

Plant transport and water relationsCore
Scope: Core · §3.3.4.2 Mass transport in plants
Open revision guidance
What to know
Xylem transports water in stems and leaves; evaporation from leaves creates tension transmitted through cohesive water, pulling a continuous column upward (cohesion–tension). Phloem transports organic solutes. In the mass-flow hypothesis, sucrose is loaded at a source, lowering sieve-tube water potential; water enters from xylem by osmosis and raises hydrostatic pressure. At a sink, unloading raises sieve-tube water potential, water leaves and pressure falls. The resulting pressure gradient drives mass flow from source to sink.
Radioactive tracers show labelled assimilates moving through phloem; ringing removes phloem but leaves xylem, and sugar accumulates above the ring. These support phloem transport and source-to-sink movement, but do not alone prove all features of the pressure-flow hypothesis; interpret what each result supports and consider alternative explanations.
Water evaporates from leaf mesophyll and diffuses through stomata; guard-cell turgor regulates the pore. A potometer measures uptake as a proxy for transpiration, not water loss itself, because some water is used or retained.
What to do in an assessment
Cut a leafy shoot under water and fit it into an airtight potometer; introduce an air bubble into the calibrated capillary, allow the shoot to equilibrate, then record bubble movement over a known time. Use the reservoir/tap to reset the bubble and repeat; convert displacement to volume only if capillary dimensions are known.
Explain a potometer as an estimate of water uptake, not a direct measurement of transpiration; control leaf area, light, wind and temperature.
In plain language
Explain a potometer as an estimate of water uptake, not a direct measurement of transpiration; control leaf area, light, wind and temperature.
Self-check
Why is water uptake not identical to transpiration rate?
Show the answer
Some absorbed water is used in photosynthesis or growth, so uptake is only a proxy for water lost from leaves.

Photosynthesis and respiration pathwaysCore
Scope: Core · §3.5.1 Photosynthesis
Open revision guidance
What to know
Photosynthesis converts light energy to chemical energy in photoautotrophs; respiration occurs in all organisms. Their shared biochemistry is indirect evidence of common ancestry. In thylakoid membranes, light photoionises chlorophyll; electrons pass along carriers and release energy used to pump H⁺ into the thylakoid. Some electron energy is conserved in ATP and reduced NADP. Photolysis replaces lost electrons and splits water into H⁺, electrons and O₂. H⁺ returns through ATP synthase, making ATP; electrons and H⁺ reduce NADP to reduced NADP.
In the Calvin cycle in the stroma, CO₂ combines with the five-carbon RuBP; rubisco catalyses formation of an unstable six-carbon intermediate that splits into two glycerate-3-phosphate (GP) molecules. ATP and reduced NADP reduce GP to triose phosphate (TP); some TP regenerates RuBP and the rest is used to make carbohydrates and other organic molecules.
Respiration begins with glycolysis in cytoplasm. Aerobic stages in mitochondria include link reaction, Krebs cycle and oxidative phosphorylation; proton flow through ATP synthase makes ATP. Chemiosmosis links proton-gradient energy to phosphorylation in both organelles. Light intensity, carbon dioxide availability and temperature can limit photosynthetic rate; growers may use artificial light, CO₂ enrichment or temperature control, but assess yield data against energy cost and other trade-offs.
What to do in an assessment
Describe where each stage occurs and follow carbon or hydrogen carriers; state that ATP is made by substrate-level and oxidative phosphorylation.
For the light-dependent reactions distinguish photolysis, electron transport, proton-gradient formation, ATP synthesis and NADP reduction; for the Calvin cycle track carbon from CO₂ through GP and TP. To test a limiting factor, vary one named condition for an aquatic plant, algae or immobilised algal beads, measure oxygen evolution or another suitable rate proxy, control other conditions and repeat.
In plain language
Describe where each stage occurs and follow carbon or hydrogen carriers; state that ATP is made by substrate-level and oxidative phosphorylation.
Self-check
What is the role of reduced NADP in the Calvin cycle?
Show the answer
It donates hydrogen/reducing power used to reduce glycerate-3-phosphate to triose phosphate.

Energy transfer and nutrient cyclingCore
Scope: Core · §3.5.3 Energy and ecosystems
Open revision guidance
What to know
Plants synthesise organic compounds from atmospheric or aquatic CO₂. Most plant sugars are respiratory substrates; the remainder makes other biological molecules and plant biomass. Biomass is measured as dry mass or carbon mass per area; calorimetry estimates chemical energy stored in dry biomass.
Gross primary production (GPP) is the chemical-energy store in plant biomass per area or volume. Net primary production is NPP=GPP−R, where R is respiratory loss; NPP remains available for plant growth/reproduction and consumers such as herbivores and decomposers. In communities, photosynthesis products are consumed by animals, bacteria and fungi; some are used as respiratory substrates. Consumer net production is N=I−(F+R): I is energy ingested, F is energy lost in faeces and urine, and R is respiratory loss.
Primary or secondary productivity is production per area or volume per time, for example kJ ha⁻¹ year⁻¹. Energy-transfer efficiency is energy incorporated into biomass at the next trophic level ÷ energy available from the preceding level ×100. Photosynthesis, respiration and transfer of biomass/chemical energy between organisms are not 100% efficient; losses include respiration, heat, uneaten material and waste. Farming can improve transfer by simplifying food webs or reducing respiratory losses.
What to do in an assessment
Keep energy store, production and productivity distinct; calculate GPP or NPP, consumer N=I−(F+R), productivity and percentage transfer efficiency using consistent area, time and energy units. Example: if GPP=20 000 and R=8 000 kJ ha⁻¹ year⁻¹, NPP=12 000 kJ ha⁻¹ year⁻¹; if 1 200 kJ enters consumer biomass from 12 000 available, efficiency=10%.
In a biomass calorimetry estimate, dry and weigh the sample, burn it beneath a known mass of water and record water-temperature rise; energy gained by water=mass of water×specific heat capacity×temperature change, then divide by dry biomass mass for energy per gram. For 20 g water, c=4.2 J g⁻¹ °C⁻¹ and a 12 °C rise, energy=1008 J; if 0.40 g dry biomass was burned, estimated energy=2520 J g⁻¹. Insulate and repeat; heat loss and incomplete combustion make this an underestimate.
Evaluate farming changes by identifying which energy loss is reduced and any associated trade-off.
In plain language
Energy passes through food chains but much is lost at every transfer. Calculate the energy retained in plant or consumer biomass, and divide production by area and time only when finding productivity.
Self-check
Plants have GPP of 18 000 kJ m⁻² year⁻¹ and respiratory losses of 6 000 kJ m⁻² year⁻¹. Calculate NPP and state one use of it.
Show the answer
NPP=GPP−R=18 000−6 000=12 000 kJ m⁻² year⁻¹. It is available for plant growth/reproduction and can be transferred to herbivores or decomposers.

Nervous and endocrine coordinationCore
Scope: Core · §3.6.2.1 Nerve impulses (A-level only)
Open revision guidance
What to know
A myelinated motor neurone has dendrites, a cell body and a long axon surrounded by myelin, with gaps called nodes of Ranvier. At rest, unequal Na⁺ and K⁺ distributions and greater membrane permeability to K⁺ than Na⁺ create a negative interior; the sodium–potassium pump moves 3 Na⁺ out for every 2 K⁺ in, maintaining electrochemical gradients. A threshold stimulus opens voltage-gated Na⁺ channels, Na⁺ enters and depolarises the membrane; Na⁺ channels then close, K⁺ channels open and K⁺ leaves to repolarise. Action potentials are all-or-nothing.
Local currents depolarise the adjacent axon membrane to threshold, propagating the action potential. The refractory period, while channels reset, separates impulses and limits their maximum frequency. Non-myelinated axons conduct continuously; in myelinated axons, excitation jumps between nodes (saltatory conduction). Greater myelination, larger axon diameter and higher temperature within physiological limits increase conduction speed.
A receptor detects a stimulus, a coordinator processes it and an effector produces a response. Nerve impulses travel along neurones and chemical transmitters act directly on target cells, so responses are usually rapid, brief and localised. Mammalian hormones travel in blood and act only on cells with a complementary receptor; responses are usually slower, longer-lasting and more widespread.
What to do in an assessment
Interpret an action-potential graph by identifying threshold, depolarisation, repolarisation and refractory period; calculate maximum frequency as 1 ÷ refractory period after converting units (for example, 2 ms gives 500 Hz); relate myelin to saltatory conduction and explain how each stated factor affects speed.
In plain language
A nerve impulse is a brief electrical change across the neurone membrane. Myelin insulates most of an axon so the signal is regenerated at gaps, making conduction faster; a refractory period prevents the membrane firing again immediately.
Self-check
Why is there a brief refractory period after an action potential?
Show the answer
Voltage-gated channels need to reset, preventing immediate refiring and helping impulses travel in one direction.

Homeostasis and kidney functionCore
Scope: Core · §3.6.4.1 Principles of homeostasis and negative feedback (A-level only)
Open revision guidance
What to know
Homeostasis uses physiological control systems to keep the internal environment within restricted limits. Stable core temperature and blood pH support enzyme activity; stable blood glucose provides respiratory substrate and helps maintain blood water potential. Negative feedback reverses a deviation; separate mechanisms correct departures in opposite directions. Positive feedback amplifies a change rather than restoring the starting state, as in a self-amplifying clotting response.
Insulin and glucagon oppose one another in blood-glucose regulation; kidney/nephron mechanisms and ADH regulate blood water potential. For each pathway name stimulus, receptor/control centre, signal, effector and corrective response; when interpreting a loop, identify whether its output reduces or amplifies the original change.
What to do in an assessment
Compare diabetes treatments through the actual defect and risk: insulin deficiency differs from insulin resistance
evaluate evidence and consequences.
In plain language
Compare diabetes treatments through the actual defect and risk: insulin deficiency differs from insulin resistance; evaluate evidence and consequences.
Self-check
Why does an increase in ADH make urine more concentrated?
Show the answer
More aquaporins increase collecting-duct water permeability, allowing more water to return to blood.

Inheritance and linkageCore
Scope: Core · §3.7.1 Inheritance
Open revision guidance
What to know
A population contains individuals with the same genes but often different allele combinations; an individual inherits alleles from one or both parents. Genotype is genetic constitution; phenotype results from genotype interacting with environment. A gene may have many alleles in a population, but each diploid individual has two at a locus: homozygous alleles are the same and heterozygous alleles differ. Dominant alleles are expressed in heterozygotes; recessive alleles are masked by a dominant allele; codominant alleles are both expressed.
A fully labelled genetic cross states parental phenotypes/genotypes, possible gametes, offspring genotypes and phenotypes, and expected ratios/probabilities. For Aa × Aa with complete dominance, the genotypes are 1 AA:2 Aa:1 aa and phenotypes 3 dominant:1 recessive. A dihybrid cross tracks two loci; independent assortment can give a 9:3:3:1 ratio for AaBb × AaBb, but linkage alters gamete proportions. In X-linked notation, show the allele on each X chromosome (e.g. XᴬXᵃ is a heterozygous female; an XY male has only one X-linked allele). For ABO, Iᴬ and Iᴮ are codominant and i is recessive. Epistasis occurs when a genotype at one locus changes or masks phenotypic expression at another; group offspring using the stated interaction, not the standard independent two-gene phenotype classes.
χ²=Σ(O−E)²/E tests whether observed phenotypic counts fit expected ratios. For a goodness-of-fit test, df=number of categories−1; compare with the supplied critical value and state whether the difference is significant, not whether the hypothesis is proven.
What to do in an assessment
Use allele symbols consistently and show parental genotypes, gametes, offspring genotypes, phenotype classes, ratios and probabilities. In sex-linked crosses place alleles on X/Y chromosomes; in codominance and epistasis group offspring by their actual expressed phenotype.
For linkage, distinguish parental from recombinant offspring, calculate recombinant percentage and compare with 50%; use observed data rather than assuming an unlinked ratio.
In plain language
Show parental genotype, gametes and offspring classes; for linkage, compare recombinant frequency with 50% and state the inference.
Self-check
Observed ratio differs from expectation but χ² is below the critical value. What conclusion follows?
Show the answer
The deviation is not statistically significant at that threshold; retain the null hypothesis that chance explains the difference.

Evolution, populations and selectionCore
Scope: Core · §3.7.3 Evolution may lead to speciation
Open revision guidance
What to know
Evolution underpins modern biology: new species arise from existing species, so species share common ancestors, represented in phylogenetic classification. Over long periods, cumulative evolutionary change has produced the diversity of living species. Common ancestry helps explain shared chemistry (including the same common amino acids), physiological pathways, cell structure, DNA as genetic material and a near-universal genetic code.
Phenotype variation reflects genetic and environmental factors; mutation is the primary source of genetic variation, while meiosis and random fertilisation add combinations. Predation, disease and competition cause differential survival and reproduction; directional, stabilising and disruptive selection alter phenotype distributions and allele frequencies over generations. Genetic drift changes allele frequencies by chance and has the greatest effect in small populations.
Evolution is a change in a population's allele frequencies. Isolation prevents gene flow; selection and drift can then make separated gene pools diverge until organisms can no longer interbreed to produce fertile offspring. Allopatric speciation follows geographic separation; sympatric speciation occurs without geographic separation when reproductive isolation develops within one area.
What to do in an assessment
Explain the population-level chain from variation and selection pressure through differential reproductive success to changed allele frequency; distinguish a change in individual phenotype from evolution across generations.
For speciation, identify the source of isolation, absence of gene flow, divergence of gene pools and eventual reproductive isolation/failure to produce fertile offspring. Model natural selection and drift with a computer simulation or repeated random samples from large and small populations; compare replicate frequency changes to show why chance has a larger effect in small populations.
In plain language
Use population evidence and the correct time scale; distinguish a change in individuals from a change in allele frequencies across generations.
Self-check
Why can a small founder population lose genetic diversity by chance?
Show the answer
A small sample may omit alleles present in the source population, so drift changes allele frequencies even without selection.

Gene regulation and biotechnologyCore
Scope: Core · §3.8.1 Alteration of the sequence of bases in DNA
Open revision guidance
What to know
Gene mutations can arise spontaneously, including during DNA replication; mutagenic agents increase mutation rate. They include substitution, addition, deletion, inversion, duplication and translocation of bases. A substitution alters one triplet: genetic-code degeneracy means it may be silent, or it may change one amino acid or create a stop codon. Adding or deleting bases not in a multiple of three changes the reading frame and all downstream triplets, potentially changing many amino acids and protein function.
Gene expression is controlled at transcription/translation; stem-cell potency, epigenetics, cancer, genome projects, recombinant DNA, probes and fingerprints are separate applications with their own evidence and ethical limits.
What to do in an assessment
For a genetic technology question, identify the target, vector, enzyme, selection step and intended product
then evaluate benefits, risks and controls.
In plain language
For a genetic technology question, identify the target, vector, enzyme, selection step and intended product; then evaluate benefits, risks and controls.
Self-check
Why does an inserted gene need a promoter that works in the host?
Show the answer
The host transcription machinery must recognise the promoter to transcribe the inserted coding sequence.

Immunity and experimental evidenceCore
Scope: Core · §3.2.4 Cell recognition and the immune system
Open revision guidance
What to know
An antigen is a molecule that elicits a specific immune response. Cell-surface molecules identify pathogens, cells from other individuals of the same species, abnormal body cells and toxins; antigen variation can hinder recognition and disease prevention. Phagocytes engulf pathogens and lysozymes digest them. Antigen-presenting cells display foreign antigen and activate helper T cells, which stimulate cytotoxic T cells, B cells and phagocytes; cytotoxic T cells destroy infected body cells.
A B cell with a complementary receptor is selected by its antigen and clones into plasma cells that secrete antibodies and memory cells. An antibody is a specific protein that binds an antigen; it has two heavy and two light polypeptide chains, disulfide bonds and variable binding regions complementary to an antigen. Antigen–antibody complexes can agglutinate bacterial cells so phagocytes engulf them. Memory cells produce a faster, stronger secondary response than the primary response.
Vaccination exposes the immune system to antigen and induces active immunity; passive immunity transfers ready-made antibodies and is temporary. Herd immunity reduces transmission when enough individuals are immune. HIV is an enveloped RNA virus with attachment proteins and a capsid; it infects helper T cells, uses reverse transcriptase to make DNA, then uses host-cell machinery to make viral components and new particles. Loss of helper T cells weakens immunity and can lead to AIDS and opportunistic infections. Antibiotics target bacterial processes, not viruses.
Monoclonal antibodies are identical antibodies with one specificity; they can target a therapeutic drug to selected cells and can be used in diagnosis. In ELISA, a target is captured by a specific antibody, an enzyme-linked antibody binds the target, and adding substrate produces a detectable colour signal. Include known positive/negative controls. Assess vaccine/monoclonal-antibody methods and evidence, and weigh ethical issues.
What to do in an assessment
Explain each immune response in causal order; for ELISA or vaccine data, identify the target, controls, measured outcome and limits, distinguish correlation from causation, and evaluate a stated ethical issue using evidence.
In plain language
The immune system recognises foreign antigens. Some lymphocytes destroy infected cells; others make specific antibodies and long-lived memory cells. Vaccines train this system before infection, while tests such as ELISA use antibody binding to detect a target.
Self-check
Why is a rapid secondary antibody response evidence of immunological memory?
Show the answer
Memory lymphocytes formed after the first exposure rapidly proliferate and differentiate when the same antigen returns.

Practical methods and mathematical reasoningCore
Scope: Core · §8.1 Apparatus and techniques (AT a–l)
Open revision guidance
What to know
AT a–c: measure mass, time, volume, temperature, length and pH with suitable equipment (for example, balance, stopwatch, graduated/volumetric glassware, ruler or micrometer, thermometer and calibrated pH meter) and record units/resolution. Calibrate quantitative instruments such as colorimeters, potometers, manometers or simple respirometers; three-way taps can route gas in a respirometer. Use pipettes, burettes, measuring cylinders and volumetric glassware appropriately. For a serial dilution, transfer a measured stock volume into a known final volume, mix, then transfer a measured aliquot into the next diluent; label each concentration and keep total volumes explicit.
AT d–g: use a light microscope at low power to locate a specimen, then high power and fine focus to examine it; prepared mounts can show gas-exchange surfaces, meiosis or striated skeletal muscle, and temporary leaf sections can show mesophyll/stomata. Calibrate an eyepiece graticule against a stage micrometer separately for each objective. Make clear, proportionate, annotated scientific drawings from observation. Use qualitative reagents and state a positive result; separate compounds by paper/thin-layer chromatography (compare Rf only under matched conditions) or electrophoresis. In a restriction-digest investigation, compare DNA fragment patterns on a gel to examine enzyme specificity; DNA migrates toward the positive electrode and smaller fragments usually travel farther.
AT h–j: use organisms safely and ethically to measure responses or physiological functions, with approved procedures and the least necessary disturbance. Human sensory investigations can compare safe temperature thresholds, responses to repeated standard touch stimuli (habituation), or the minimum spacing at which two blunt points are perceived separately; standardise stimulus, site and interval, obtain informed consent and allow withdrawal. For human-volunteer investigations such as exercise/heart-rate or repeated muscle-contraction/fatigue studies, standardise workload/rest and stop if discomfort or risk arises. For microbiology use sterile agar/broth and aseptic transfer, minimise plate exposure and follow approved incubation/disposal rules; turbidity readings can track broth-culture growth over time when conditions and instrument settings are controlled. Dissect plant or animal organs with appropriate instruments, secure the specimen, cut safely and identify structures without assuming a single required specimen.
AT k–l: select field sampling suited to the organism and question—random quadrats for sessile organisms, transects for a spatial gradient, or mark–release–recapture for motile organisms—and state sampling assumptions. Use computer models, data loggers or software to collect/process data; check sensor calibration, units, sampling interval, axes and whether transformations are justified. Manipulate an independent variable, measure a dependent variable, control confounders and use repeats; percentage change=(final−initial)/initial×100.
What to do in an assessment
In a practical answer specify a reproducible sequence, the precise instrument and measurement, test/control conditions, replication and sampling, relevant hazard/ethical safeguard, planned graph/calculation and how the result addresses the hypothesis. Explain what the chosen instrument or proxy measures and its limitation.
In plain language
In practical answers specify a method another student could reproduce, state the safety step and explain how the data test the hypothesis.
Self-check
A count rises from 40 to 50. What is the percentage increase?
Show the answer
(50−40) ÷ 40 ×100 = 25%.

Statistical evidence in biologyCore
Scope: Core · §3.4.7 Investigating diversity
Open revision guidance
What to know
Genetic diversity within or between species may be compared using frequencies of measurable/observable characteristics, DNA base sequences, mRNA base sequences, or amino-acid sequences of encoded proteins. Interpret sequence similarities and differences to suggest relationships within or between species. Direct DNA sequence comparison has replaced inferring DNA differences from observable traits; gene-technology details are not assessed.
For quantitative variation within a species, use random samples, calculate the mean and standard deviation of the mean, and interpret the mean and standard deviation together. In this diversity context, written exams do not require calculation of standard deviation; interpret given values without treating a mean difference or overlapping spreads alone as a significance test.
What to do in an assessment
Explain which kind of diversity evidence is being compared; interpret a sequence difference as evidence for a possible relationship, not proof of its cause. For quantitative samples, identify random sampling, compare the mean and standard deviation of the mean, and state what the data support without claiming an untested significance.
In plain language
Compare biological variation using observable traits or molecular sequences. For measurements within a species, use random samples and interpret the mean together with its standard deviation; a pattern may suggest a relationship but does not by itself prove a cause.
Self-check
Two species have similar DNA and amino-acid sequences. What conclusion is justified, and what cannot be concluded from similarity alone?
Show the answer
The similarities support a possible close evolutionary relationship. Similarity alone does not prove a particular cause, direction of ancestry or the statistical significance of a measured difference.

Genetic crosses and linkage evidenceCore
Scope: Core · §3.7.1 Inheritance
Open revision guidance
What to know
A test cross reveals an unknown dominant genotype by crossing with a homozygous recessive; offspring ratios distinguish likely genotypes.
Linked loci are inherited together more often than unlinked loci; recombinant frequency estimates the distance between loci.
What to do in an assessment
Show allele symbols and gametes, calculate the recombinant fraction and state how sample size affects confidence.
In plain language
Show allele symbols and gametes, calculate the recombinant fraction and state how sample size affects confidence.
Self-check
What does a recombinant frequency close to 50% suggest?
Show the answer
The loci assort independently or are so far apart on the same chromosome that crossing-over occurs frequently.

Population genetics and allele frequencyCore
Scope: Core · §3.7.2 Populations and Hardy–Weinberg
Open revision guidance
What to know
A population is organisms of the same species occupying a particular place at a particular time that can potentially interbreed. Its gene pool contains the alleles present; allele frequency is the proportion of copies of an allele among all copies of that gene. Frequency can change through selection, migration, mutation or genetic drift.
For two alleles with frequencies p and q, p+q=1 and expected genotype frequencies are p²+2pq+q²=1. Under Hardy–Weinberg assumptions, allele frequencies and expected genotype frequencies remain unchanged from generation to generation. The model assumes a large randomly mating population without selection, mutation, migration or genetic drift; it is an expectation, not a description guaranteed for every real population.
What to do in an assessment
If q²=0.09 for a recessive phenotype, q=√0.09=0.3 and p=0.7; expected heterozygote frequency is 2pq=0.42. Check that allele/genotype totals sum to one and discuss which assumption could explain a departure.
In plain language
Calculate frequency from allele counts and identify which assumption is violated before explaining a departure from expectation.
Self-check
Why is a small population more affected by genetic drift?
Show the answer
Random sampling of a few breeders can shift allele frequencies substantially when each individual represents a large fraction of the gene pool.

Plant signalling and tropismsCore
Scope: Core · §3.6.1.1 Survival and response
Open revision guidance
What to know
A stimulus is an internal or external change; a specific receptor detects it, a coordinator processes it and an effector produces a response. Nerve impulses release a chemical messenger directly onto a target for a usually rapid, short-lived, local response; mammalian hormones travel in blood and usually cause slower, longer-lasting, widespread responses. Plants coordinate growth with hormone-like factors including indoleacetic acid (IAA).
Different IAA concentrations affect root and shoot cell elongation differently: roots are more sensitive, so a concentration that stimulates shoot elongation may inhibit root elongation. Unequal IAA distribution produces differential growth. In a shoot, IAA on the shaded side stimulates elongation toward light; in a root, gravity-driven IAA redistribution inhibits elongation on the lower side, bending the root down, whereas it stimulates lower-side shoot elongation and bends the shoot up.
A tropism is directional growth; taxis is directional movement and kinesis is a change in movement rate or turning with stimulus intensity. A protective three-neurone reflex runs receptor → sensory neurone → relay neurone in CNS → motor neurone → effector and can act before conscious processing.
What to do in an assessment
For IAA-response practicals, compare concentrations and measure root or shoot elongation against a suitable control; for tropism experiments control seedling age and stimulus direction, then measure direction/curvature. Explain a response by linking IAA redistribution and tissue-specific growth.
In plain language
For a seedling experiment, control light direction and age, measure curvature and use a covered or unilateral-light comparison.
Self-check
Why does a shoot bend towards unilateral light?
Show the answer
Auxin accumulates on the shaded side, stimulating greater elongation there so the shoot curves towards the light.

Evolutionary trees and classificationCore
Scope: Core · §3.4.5 Species and taxonomy
Open revision guidance
What to know
A species is a group whose members can interbreed to produce fertile offspring. Courtship behaviour is a necessary precursor to successful mating and helps organisms recognise members of their own species. Taxonomic groups nest species within genus, family, order, class, phylum, kingdom and domain; each group is a taxon (plural taxa), and a binomial scientific name gives genus then species.
Phylogenetic classification arranges organisms by evolutionary origins and relationships in a nested hierarchy without overlapping groups. A node on a phylogenetic tree represents a common ancestor; taxa sharing a more recent node are more closely related. Immunological evidence and genome sequencing have helped clarify relationships. Memorising the details of taxonomic systems such as three-domain or five-kingdom classifications is not required.
What to do in an assessment
Read the branching order rather than the page position; compare sequence differences using a stated method and avoid treating a living species as an ancestor.
In plain language
Read the branching order rather than the page position; compare sequence differences using a stated method and avoid treating a living species as an ancestor.
Self-check
On a tree, what does a shared recent branching point indicate?
Show the answer
The taxa share a more recent common ancestor than taxa whose lineages split earlier.

Practical validity and uncertaintyCore
Scope: Core · §8.3 Practical skills assessed in written papers (PS 1.1–4.1)
Open revision guidance
What to know
Validity asks whether a method tests the intended relationship; reliability concerns repeatability, and precision concerns the spread or resolution of measurements.
Random error creates scatter; systematic error shifts measurements consistently and is not removed by repeating alone.
What to do in an assessment
Identify the actual source of uncertainty, predict its effect and propose an improvement that addresses it rather than merely adding more repeats.
In plain language
Identify the actual source of uncertainty, predict its effect and propose an improvement that addresses it rather than merely adding more repeats.
Self-check
Why will repeating not correct a zero error in a balance?
Show the answer
A zero error is systematic and shifts every reading; the balance must be zeroed or readings corrected.

Water potential and plant cellsCore
Scope: Core · §3.2.3 Transport across cell membranes
Open revision guidance
What to know
Water moves down a water-potential gradient through partially permeable membranes; cell walls resist expansion and generate turgor pressure.
In a hypertonic solution, plant cells lose water and the plasma membrane pulls away from the wall; in a dilute solution, cells become turgid.
What to do in an assessment
State which compartment has the higher water potential and connect net movement to cell appearance.
In plain language
State which compartment has the higher water potential and connect net movement to cell appearance.
Self-check
What happens to a plant cell in a solution with lower water potential than its cytoplasm?
Show the answer
Water leaves by osmosis; the vacuole shrinks and the protoplast may pull away from the cell wall.

Data presentation and biological calculationsCore
Scope: Core · §6.1 Arithmetic and numerical computation (MS 0.1–0.5)
Open revision guidance
What to know
Rates are changes in a measured quantity per unit time; use the initial linear region where later substrate depletion or product accumulation changes the rate.
Logarithmic scales and percentage change can clarify wide ranges, but axes, units and transformations must be stated.
What to do in an assessment
Show substitution, unit conversion and rounding; explain whether plotted points are raw observations, means or derived rates.
In plain language
Show substitution, unit conversion and rounding; explain whether plotted points are raw observations, means or derived rates.
Self-check
Why use the initial gradient to estimate an enzyme reaction rate?
Show the answer
At the start, substrate is abundant and product has not accumulated, so the gradient best estimates initial reaction rate.

Lipid structure and functionsCore
Scope: Core · §3.1.3 Lipids
Open revision guidance
What to know
A triglyceride forms when one glycerol and three fatty acids join by condensation, making three ester bonds and releasing three water molecules; hydrolysis reverses this using water. A saturated fatty acid has no carbon–carbon double bonds; an unsaturated fatty acid has at least one C=C bond, fewer hydrogens and a bend that disrupts tight packing.
In a phospholipid, a phosphate-containing group replaces one fatty acid of a triglyceride. Its hydrophilic head and two hydrophobic tails make it amphipathic and form bilayers in water. Triglycerides are compact energy stores and provide insulation; unsaturated tails generally lower melting point and increase membrane fluidity relative to saturated tails.
What to do in an assessment
Recognise saturated and unsaturated fatty-acid diagrams by looking for C=C bonds and kinks; relate tail packing to physical properties. For the emulsion test, shake the sample with ethanol, add water and identify a cloudy white emulsion as lipid.
In plain language
A triglyceride is glycerol plus three fatty acids; a phospholipid replaces one fatty acid with a phosphate-containing head. Because heads mix with water but tails do not, phospholipids naturally make a two-layer cell membrane. Unsaturated fatty acids contain C=C bonds that bend their tails and change packing.
Self-check
A sample forms a cloudy white layer after shaking with ethanol and adding water. What does this support, and how would a diagram show an unsaturated fatty acid?
Show the answer
The emulsion test supports the presence of lipid. An unsaturated fatty acid has one or more C=C bonds and a kinked hydrocarbon chain; a saturated chain has no C=C bonds.
Protein structure, bonding and the biuret testCore
Scope: Core · §3.1.4.1 General properties of proteins
Open revision guidance
What to know
Amino-acid sequence is primary structure; hydrogen bonds stabilise alpha helices and beta-pleated sheets (secondary structure). R-group interactions, including hydrogen and ionic bonds and disulfide bridges, fold polypeptides into tertiary structure; multiple chains can form quaternary structure.
A functional protein's 3-D shape determines binding and activity. The biuret reagent gives a lilac/purple positive result when peptide bonds are present; a negative sample remains blue.
What to do in an assessment
When explaining a protein's function, follow the chain from amino-acid sequence through bonding and folding to the active or binding shape; for a test, state reagent and observed colour. To identify amino acids in a mixture, separate it by chromatography beside known standards and compare spot positions/Rf values under the same solvent and stationary-phase conditions.
In plain language
A protein is a chain folded into a particular shape. Its sequence determines which parts attract or repel; the resulting shape lets it bind the right molecule. The biuret test detects peptide bonds, not a particular kind of protein.
Self-check
A mutation changes one amino acid in a membrane protein. Explain one possible route from this change to loss of transport function.
Show the answer
The altered side chain can change ionic, hydrogen-bond or disulfide interactions, changing tertiary folding. If the channel or carrier's shape changes, its substrate may no longer bind or pass through efficiently.
ATP, water and inorganic ionsCore
Scope: Core · §3.1.6 ATP; §3.1.7 Water; §3.1.8 Inorganic ions
Open revision guidance
What to know
ATP is adenine, ribose and three phosphates. ATP hydrolase catalyses ATP + water → ADP + Pi; hydrolysis can be coupled to cell work, while released phosphate can phosphorylate a substrate. ATP synthase catalyses condensation of ADP and Pi to resynthesise ATP during respiration or photosynthesis.
Water is a major cell component and a metabolite in condensation and hydrolysis. Its polarity makes it an important solvent for metabolic reactions; hydrogen bonding gives it relatively high heat capacity, which buffers temperature change, and high latent heat of vaporisation, which enables cooling with little water loss. Cohesion supports continuous water columns in plant transport vessels and creates surface tension at an air–water boundary. Ions occur dissolved or bound in molecules; examples include phosphate in ATP/DNA, iron in haemoglobin, sodium in co-transport and hydrogen ions in pH regulation.
What to do in an assessment
For a biological role, link the chemical property to its consequence (for example water's high heat capacity slows temperature change); distinguish ATP hydrolysis from its resynthesis and identify the ion's actual molecular or transport role.
In plain language
ATP is a rapidly reusable energy-coupling molecule, not a long-term energy store. Water's polarity makes it a useful solvent and its hydrogen bonds resist temperature change. Inorganic ions matter because cells use their charge and chemical position for specific jobs.
Self-check
Why can ATP hydrolysis power active transport, and why does a hydrogen-ion gradient contain potential energy?
Show the answer
ATP hydrolysis releases usable free energy and Pi can phosphorylate a transport protein, changing its shape. A proton gradient stores electrochemical potential because concentration and charge differ across a membrane; protons moving down it can drive ATP synthase.

Semi-conservative DNA replicationCore
Scope: Core · §3.1.5.2 DNA replication
Open revision guidance
What to know
Helicase unwinds DNA and breaks hydrogen bonds between complementary bases. Each exposed strand acts as a template; free DNA nucleotides pair by A–T and C–G.
DNA polymerase catalyses condensation between adjacent nucleotides, forming the sugar-phosphate backbone. Each product has one original strand and one new strand, so replication is semi-conservative.
What to do in an assessment
Explain replication in order: unwind, expose templates, complementary pairing, polymerase joins nucleotides; use the Meselson–Stahl density evidence to distinguish conservative, semi-conservative and dispersive models.
In plain language
The two old DNA strands separate. Each guides construction of a matching new partner, so the two resulting DNA molecules each keep one old strand and gain one new one.
Self-check
How did Meselson and Stahl's two generations of growth in light nitrogen distinguish semi-conservative replication from conservative and dispersive models?
Show the answer
After one generation, all DNA formed an intermediate-density hybrid band, excluding conservative replication, which predicts separate heavy and light bands. After two generations, one hybrid and one light band appeared in roughly equal amounts, matching semi-conservative replication and excluding dispersive replication, which predicts a single increasingly light/intermediate band.

Prokaryotic cells and virusesCore
Scope: Core · §3.2.1.2 Structure of prokaryotic cells and of viruses
Open revision guidance
What to know
Prokaryotes are smaller than eukaryotes, lack membrane-bound organelles and a nucleus, and have small ribosomes, a free short circular DNA molecule and a murein cell wall. Many also have plasmids, a capsule or flagella; detailed structural differences beyond these are not required.
Viruses are acellular and non-living particles with genetic material, a protein capsid and attachment proteins. They do not divide; after nucleic acid enters a host cell, host machinery makes new virus particles.
What to do in an assessment
Compare a diagram only on supported features; explain why antibiotics targeting bacterial processes do not treat viruses, and distinguish bacterial binary fission from viral replication in a host.
In plain language
A bacterium is a small living cell with cytoplasm, membrane and circular DNA but no nucleus. A virus is not a cell: it carries genetic material in a protein coat and can make copies only by using a host cell.
Self-check
A drawing shows a circular DNA loop, 70S ribosomes and a murein wall, but no nucleus. Identify the cell type and explain why it is not a virus.
Show the answer
It is prokaryotic: those are bacterial cell features. A virus is acellular and consists of genetic material, capsid and attachment protein; it has neither cytoplasm nor ribosomes.
Microscopy, magnification and cell fractionationCore
Scope: Core · §3.2.1.3 Methods of studying cells
Open revision guidance
What to know
Magnification enlarges an image; resolution is the ability to distinguish two close points. Optical microscopes can view living cells but resolve less detail than electron microscopes because electrons have a much shorter wavelength than visible light. TEM reveals internal ultrastructure in thin sections; SEM shows surface detail, but electron microscopy uses non-living prepared specimens and can introduce artefacts. Scientists needed time and improved methods to distinguish genuine organelles from artefacts introduced during preparation; an image must be interpreted as evidence, not assumed to be a direct copy of a living cell.
Cell fractionation homogenises tissue in cold, isotonic, buffered solution, then filters out large debris. Differential centrifugation/ultracentrifugation spins the filtrate at increasing speeds: larger/denser organelles sediment first into a pellet, while smaller organelles remain in the supernatant until a faster spin. Cold limits enzyme damage, isotonic medium prevents osmotic lysis, and buffer stabilises pH.
What to do in an assessment
Use M = image size ÷ actual size and rearrange before substituting. Example: a 30 mm image of a 10 µm organelle gives 30,000 µm ÷ 10 µm = ×3000. State units consistently and explain what each microscope can resolve.
In plain language
An image can be large but blurry: magnification and resolution are not the same. To isolate organelles, break cells gently, remove debris, then spin faster so smaller structures sediment after larger ones.
Self-check
An organelle is 8 µm long and its image is 24 mm. Calculate magnification and name one reason for keeping the homogenisation medium cold and isotonic.
Show the answer
24 mm = 24,000 µm; magnification = 24,000 ÷ 8 = ×3000. Cold slows hydrolytic enzymes; isotonic medium prevents organelles swelling or shrinking from water movement.

Gas-exchange surfaces and ventilationCore
Scope: Core · §3.3.2 Gas exchange
Open revision guidance
What to know
Single cells exchange across their surface; insects use spiracles, tracheae and tracheoles; fish use filaments and lamellae with counter-current water/blood flow; dicot leaves exchange through stomata and moist mesophyll. Large area, thin barriers and maintained gradients speed diffusion. In terrestrial insects, closing spiracles reduces evaporative water loss but also restricts oxygen entry and carbon-dioxide exit; opening or ventilating them improves gas exchange while exposing moist tracheal surfaces to water loss. Xerophytes reduce uncontrolled evaporation with a thick waxy cuticle. Sunken stomata, leaf hairs or rolled leaves trap humid, still air near stomata, reducing the water-vapour gradient but also slowing carbon-dioxide diffusion through the boundary layer. Drought-induced stomatal closure further conserves water while restricting carbon-dioxide entry for photosynthesis. These examples show the structural and functional compromise between gas exchange and water conservation.
In humans, trachea, bronchi, bronchioles and alveoli conduct air. Alveoli provide a moist, thin epithelial surface, large area and dense capillary supply. During inspiration the diaphragm contracts/flattens and external intercostals lift the ribs, increasing thoracic volume and lowering pressure; quiet expiration is largely elastic recoil, while internal intercostals contract in forced expiration to reduce thoracic volume. Ventilation renews alveolar air and blood flow maintains the exchange gradient.
What to do in an assessment
Explain a named adaptation by identifying how it changes area, diffusion distance, gradient or water conservation. Calculate pulmonary ventilation rate (PVR)=tidal volume×breathing rate and rearrange, for example tidal volume=PVR÷breathing rate; use disease/pollution/smoking data to distinguish correlation from causation and evaluate how experimental evidence can support restrictions on risk exposures.
In plain language
Fast exchange needs lots of thin, moist surface and a fresh supply on both sides. Breathing refreshes alveolar air; blood flow carries oxygen away. Structures that save water may deliberately reduce the exposed wet surface.
Self-check
A xerophytic plant has rolled, hairy leaves and closes its stomata during drought. Explain how these features conserve water and the gas-exchange cost.
Show the answer
Rolled leaves and hairs trap humid still air, reducing the water-vapour gradient and water loss; a thick waxy cuticle limits evaporation through the epidermis. Closing stomata further conserves water but also reduces carbon-dioxide entry, potentially slowing photosynthesis; the boundary layer can also slow CO₂ diffusion.

Genes, chromosomes and the genetic codeCore
Scope: Core · §3.4.1 DNA, genes and chromosomes
Open revision guidance
What to know
Prokaryotic DNA is short, circular and not associated with proteins; eukaryotic nuclear DNA is long, linear and associated with histones, while mitochondrial and chloroplast DNA is short, circular and not protein-associated. A gene is a base sequence on a particular DNA molecule that occupies a fixed locus and codes for a polypeptide or functional RNA.
A three-base DNA triplet specifies an amino acid. The code is universal, non-overlapping and degenerate (several triplets can specify one amino acid). Eukaryotic genes may have coding exons separated by introns; much nuclear DNA is non-coding, including repeated sequences between genes.
What to do in an assessment
Keep gene, allele, locus, chromosome, genome and proteome distinct. When given a coding table, read successive triplets in the stated direction and do not assume the supplied strand is the coding rather than template strand.
In plain language
A chromosome is a long DNA molecule packaged with protein. A gene is one stretch of that DNA with instructions for a protein or functional RNA; the full DNA set is the genome, whereas the proteins a cell can make form its proteome.
Self-check
Why can a base substitution in one coding triplet leave a protein's amino-acid sequence unchanged?
Show the answer
The genetic code is degenerate, so the altered triplet may still specify the same amino acid. This is a silent substitution; it can nevertheless affect expression or splicing in some contexts.
Transcription and translationCore
Scope: Core · §3.4.2 DNA and protein synthesis
Open revision guidance
What to know
RNA polymerase joins complementary RNA nucleotides using a DNA template. Eukaryotic pre-mRNA is spliced to remove introns; prokaryotic transcription produces mRNA directly. mRNA is a single-stranded codon-bearing copy that binds a ribosome; tRNA folds into a characteristic cloverleaf shape with an anticodon at one end and an amino-acid attachment site at the other.
At the ribosome, codon–anticodon pairing positions amino acids for peptide-bond formation; ribosomes move along mRNA and ATP supports the process. The genome is the complete set of genes in a cell, while the proteome is the full range of proteins it can produce and varies with cell type.
What to do in an assessment
For sequence questions, transcribe the complementary RNA from the stated template, split it into triplets in the correct direction, then use the provided genetic-code table to infer amino acids. Do not learn specific codon/amino-acid pairings for recall. When interpreting experiments on nucleic acids as genetic material, track which labelled component enters a cell or is inherited by progeny and relate that evidence to the proposed role.
In plain language
Transcription makes a portable RNA copy of a gene. During translation, the ribosome reads that copy three bases at a time and tRNAs bring the amino acids that are joined into a protein.
Self-check
A template DNA sequence is 3′-TAC GGA-5′. What mRNA is made, and what must you do before naming its amino acids?
Show the answer
The complementary mRNA is 5′-AUG CCU-3′ (U pairs with A). Read its codons using the genetic-code table supplied in the question; no codon identities are required from memory.
Mutation, meiosis and genetic variationCore
Scope: Core · §3.4.3 Genetic diversity can arise as a result of mutation or during meiosis
Open revision guidance
What to know
Gene mutations include base substitutions and deletions, which can arise spontaneously during DNA replication; mutagenic agents increase mutation rate. Degeneracy makes some substitutions silent. Non-disjunction in meiosis can change chromosome number. Meiosis has two nuclear divisions and normally produces four haploid, genetically different cells.
Homologous chromosomes independently segregate in meiosis I; crossing-over between homologues creates additional allele combinations. Random fertilisation combines different gametes. Without crossing-over, n chromosome pairs can form 2ⁿ combinations per parent; random union of two gametes gives 2ⁿ × 2ⁿ zygotic combinations.
What to do in an assessment
Given a diploid parent diagram, draw the replicated homologous pairs, show their independent orientation/separation in meiosis I, then sister-chromatid separation in meiosis II; label chromosome and chromatid counts at each stage. Compare mitosis (one division, two genetically identical diploid daughters) with meiosis (two divisions, usually four genetically different haploid daughters), and identify meiosis in an unfamiliar life cycle by its production of haploid gametes/spores. Example: n=3 gives 2³=8 gamete combinations per parent and 8×8=64 pairings after random fertilisation, before crossing-over.
In plain language
Meiosis halves chromosome number and shuffles maternal and paternal versions of each chromosome. Crossing-over swaps matching sections; random fertilisation then combines two independently varied gametes.
Self-check
A diploid cell has three homologous pairs. Ignoring crossing-over, how many chromosome combinations can one parent pass on, and why are meiosis products still more varied in reality?
Show the answer
2³ = 8 combinations. Crossing-over creates additional recombinant chromatids, and random fertilisation combines one of many possible gametes from each parent.

Natural selection and adaptationCore
Scope: Core · §3.4.4 Genetic diversity and adaptation
Open revision guidance
What to know
Genetic diversity is the number of different alleles of genes in a population and enables natural selection. Random mutation creates new alleles; many are harmful, but in a particular environment an allele may improve reproductive success. Carriers pass it to offspring, so its frequency can rise over generations; individuals do not evolve in response to need.
Directional selection shifts a trait distribution toward one extreme (for example antimicrobial resistance); stabilising selection favours an intermediate (for example human birth mass). These processes alter population frequencies. Adaptations may be anatomical, physiological or behavioural and make populations better suited to their environment.
What to do in an assessment
Explain the full population-level chain: existing variation → selection pressure → differential reproductive success → inheritance → changed allele frequency over generations. In a graph, compare distributions and sample size, not only individual examples.
In plain language
Antibiotics do not make a bacterium choose to become resistant. Resistant variants may already exist; treatment kills more susceptible bacteria, leaving resistant survivors to reproduce and pass on their alleles.
Self-check
After antibiotic exposure, the proportion of resistant bacteria rises from 2% to 70%. Give a selection-based explanation and one reason this observation alone may not establish the mutation's timing.
Show the answer
The resistant bacteria survived treatment more often and reproduced, increasing the resistance allele's frequency. The rise is compatible with selection of pre-existing variants or new mutations; the before/after proportions alone do not show when the mutation arose.

Community biodiversity and Simpson's indexCore
Scope: Core · §3.4.6 Biodiversity within a community
Open revision guidance
What to know
Biodiversity can be considered at scales from a small local habitat to the whole Earth. Species richness is the number of species; Simpson's index of diversity also accounts for how individuals are distributed among species. A community with evenly represented species generally has greater diversity than one dominated by a single species.
AQA's formula is d = N(N−1) / Σ n(n−1), where N is all individuals and n is the count for each species. Farming may reduce biodiversity; conservation decisions weigh food production against habitat and species protection.
What to do in an assessment
Calculate each n(n−1), add the terms, calculate N(N−1), then divide; state which community has the higher index and support the comparison with species abundance data.
In plain language
Counting species alone can hide dominance: two habitats may each have four species but one may be almost entirely one species. Simpson's index combines the total with each species count to reflect evenness as well as richness.
Self-check
Counts are 4, 3 and 3. Calculate Simpson's index using d=N(N−1)/Σn(n−1).
Show the answer
N=10, so N(N−1)=90. The denominator is 4×3 + 3×2 + 3×2 = 24. Therefore d=90/24=3.75; the index is dimensionless.

Respiration and chemiosmosisCore
Scope: Core · §3.5.2 Respiration (A-level only)
Open revision guidance
What to know
Glycolysis occurs anaerobically in the cytoplasm: ATP phosphorylates glucose to glucose phosphate, which is converted to triose phosphate. Oxidation of triose phosphate forms pyruvate and reduced NAD, with a net ATP gain. Without oxygen, pyruvate is reduced to lactate (animals) or ethanol (yeast/plants); reduced NAD is oxidised back to NAD so glycolysis can continue.
With aerobic respiration, pyruvate enters the mitochondrial matrix by active transport. In the link reaction it is oxidised to acetate, reducing NAD; acetate combines with coenzyme A to form acetyl-CoA. Acetyl-CoA combines with a four-carbon molecule, releases coenzyme A and forms a six-carbon molecule entering the Krebs cycle. Oxidation–reduction and decarboxylation in the cycle produce reduced coenzymes, CO₂ and ATP by substrate-level phosphorylation.
Reduced coenzymes transfer electrons down the electron-transfer chain in the inner mitochondrial membrane; their energy pumps protons across the membrane. Protons flowing down the electrochemical gradient through membrane ATP synthase drive ADP + Pi → ATP (oxidative phosphorylation). Breakdown products of lipids and amino acids can also enter the Krebs cycle.
What to do in an assessment
Trace carbon and hydrogen/electrons separately: carbon from pyruvate is released as CO₂ in aerobic stages; hydrogen/electrons travel on reduced coenzymes to the electron-transfer chain. Distinguish ATP made directly by substrate-level phosphorylation from ATP made by oxidative phosphorylation, which depends on electron transfer, proton pumping and ATP synthase.
In plain language
Respiration transfers energy from substrates into ATP. In aerobic respiration, electrons from reduced coenzymes power proton pumping; protons then flow back through ATP synthase, which makes ATP.
Self-check
Why must reduced NAD be re-oxidised during anaerobic glycolysis, and how does the electron-transfer chain contribute to ATP production aerobically?
Show the answer
Re-oxidation regenerates NAD needed to accept hydrogen during glycolysis, allowing ATP production to continue without oxygen. Aerobically, electron transfer powers proton pumping; proton flow down the electrochemical gradient through ATP synthase drives phosphorylation of ADP.

Nutrient cycling and fertiliser impactsCore
Scope: Core · §3.5.4 Nutrient cycles (A-level only)
Open revision guidance
What to know
Saprobionts secrete digestive enzymes onto dead material and absorb soluble products; their saprobiotic nutrition decomposes organic matter and returns mineral nutrients. Mycorrhizal fungi associate with roots and increase uptake of water and inorganic ions. Nitrogen-fixing bacteria convert atmospheric N₂ to ammonium compounds; ammonification converts organic nitrogen in dead material and waste to ammonium ions; nitrifying bacteria oxidise ammonium to nitrite and then nitrate; plants take up nitrate and build organic nitrogen; denitrifying bacteria convert nitrate back to N₂, especially in oxygen-poor soil. Individual bacterial species need not be named.
The phosphorus cycle has no major atmospheric phase: weathering releases phosphate from rocks to soil/water, plants absorb it, food chains transfer it, decomposers return it, and sedimentation/geological uplift replenish rock stores. Harvesting removes nitrogen and phosphate; natural or artificial fertilisers replace them. Nitrate leaching enriches water, stimulating algal growth; when algae die, microbial decomposition increases oxygen demand and may cause hypoxia and aquatic-animal deaths.
What to do in an assessment
In a nitrogen-cycle diagram label the substance at each step (N₂, organic nitrogen, NH₄⁺, NO₂⁻ or NO₃⁻), the process and bacterial/decomposer role; distinguish nitrate leaching from denitrification. Trace fertiliser use through eutrophication to declining dissolved oxygen.
To test a mineral's effect on plant growth, grow comparable seedlings with a complete nutrient solution and a solution lacking only the named mineral; control light, water, pH, temperature and growth time, use replicates and measure growth or dry mass.
In plain language
Matter is recycled. Decomposers return nutrients from dead organisms to soil; microbes convert nitrogen between forms plants can or cannot use. Fertiliser can replace crop losses, but runoff can overfeed water ecosystems.
Self-check
Explain how nitrate fertiliser can lead to low dissolved oxygen and fish deaths in a lake.
Show the answer
Nitrate runoff enriches the water, encouraging algal growth. Algae die and are decomposed by aerobic microorganisms; their respiration uses dissolved oxygen, so oxygen concentration may fall below the level needed by fish.

Pacinian corpuscles and retinal receptorsCore
Scope: Core · §3.6.1.2 Receptors (A-level only)
Open revision guidance
What to know
A Pacinian corpuscle is a pressure receptor with a sensory neurone ending surrounded by concentric layers of connective tissue. Pressure deforms the layers and the ending, opening stretch-mediated sodium channels; sodium influx creates a generator potential. A sufficiently large generator potential triggers nerve impulses, and receptors respond only to specific stimuli.
Rods contain rhodopsin and are sensitive in dim light but do not distinguish colour. Cones contain different optical pigments, work best in brighter light and enable colour vision. Rod/cone connections to optic-nerve neurones explain acuity: convergence of several rods increases sensitivity but reduces precise localisation, while less convergence from cones improves visual acuity.
What to do in an assessment
Trace stimulus → receptor transduction → generator potential → impulse; when comparing vision, separate light sensitivity, colour detection and spatial resolution rather than treating them as one property.
In plain language
A receptor changes a stimulus into an electrical signal. Pressure distorts a Pacinian corpuscle and opens sodium channels. In the eye, rods detect low light while cone type and wiring determine colour and sharpness.
Self-check
Why can a retinal region with many converging rods be light-sensitive but have poorer spatial resolution?
Show the answer
Several rods may feed into one optic-nerve pathway, summing weak signals so dim light is detected. Because their signals merge, the brain cannot identify exactly which rod was stimulated as precisely as with less convergence.
Myogenic heart and heart-rate controlCore
Scope: Core · §3.6.1.3 Control of heart rate (A-level only)
Open revision guidance
What to know
The heart is myogenic: the sinoatrial node (SAN) spontaneously initiates a wave of electrical excitation across the atria. The atrioventricular node (AVN) delays it so atria empty before ventricles contract; the bundle of His and Purkyne tissue conduct excitation through ventricular walls for coordinated ventricular contraction.
Chemoreceptors in the medulla, carotid bodies and aortic bodies detect blood CO₂/pH changes; pressure receptors in the carotid sinus and aortic arch detect pressure changes. Impulses reach the cardiac centre in the medulla, where autonomic output adjusts SAN activity: sympathetic stimulation raises heart rate and parasympathetic (vagus) stimulation lowers it. Cardiac output is heart rate × stroke volume.
What to do in an assessment
Use the order SAN → atria → AVN delay → bundle of His/Purkyne tissue → ventricles when interpreting a cardiac trace; calculate and label output units. Example: 72 beats min⁻¹ × 70 cm³ beat⁻¹ = 5040 cm³ min⁻¹ = 5.04 dm³ min⁻¹. To test a named factor such as exercise, obtain informed consent, record a resting baseline, standardise workload/duration, use the same pulse-measurement method and recovery intervals, repeat safely and stop if a participant is unwell.
In plain language
The heart starts its own rhythm. A short delay at the AV node lets the atria finish pushing blood into the ventricles before a coordinated ventricular contraction.
Self-check
Why is the excitation delayed at the AVN, and what is cardiac output for a rate of 80 min⁻¹ and stroke volume of 65 cm³?
Show the answer
The delay allows atrial systole to complete before ventricular systole. Cardiac output = 80 × 65 = 5200 cm³ min⁻¹, or 5.2 dm³ min⁻¹.
Cholinergic synapses and neuromuscular junctionsCore
Scope: Core · §3.6.2.2 Synaptic transmission (A-level only)
Open revision guidance
What to know
A cholinergic synapse has a presynaptic knob with mitochondria and acetylcholine vesicles, a narrow synaptic cleft and a postsynaptic membrane with specific receptors. An arriving impulse opens presynaptic calcium channels; calcium entry causes vesicles to fuse with the presynaptic membrane and release acetylcholine. Acetylcholine diffuses across, binds postsynaptic receptors and opens sodium channels; Na⁺ influx depolarises the postsynaptic membrane. Acetylcholinesterase hydrolyses acetylcholine; choline is taken back up and reused.
Transmission is one-way because vesicles release transmitter only from the presynaptic side and receptors are on the postsynaptic side; release and diffusion create a synaptic delay. Temporal summation combines repeated inputs at one synapse; spatial summation combines inputs from several synapses. An inhibitory synapse releases transmitter that hyperpolarises the postsynaptic membrane (for example by opening channels that move it farther from threshold), reducing the chance of an action potential.
At a neuromuscular junction, the postsynaptic cell is a skeletal-muscle fibre rather than another neurone: acetylcholine binds receptors on the folded motor end-plate membrane, depolarises it and triggers a muscle action potential and contraction. Both junction types use calcium-triggered acetylcholine release and acetylcholinesterase, but the target cell and resulting response differ.
What to do in an assessment
Infer an unfamiliar drug effect from its target: blocking calcium entry/release, receptor binding, acetylcholinesterase breakdown or choline recycling changes signal strength or duration. Compare a cholinergic neurone-to-neurone synapse with a neuromuscular junction by identifying the target cell, postsynaptic structure and response.
In plain language
A synapse converts an electrical signal to a chemical one and back again. A neurotransmitter crosses the tiny gap, binds to the next cell, then is removed so the message does not continue indefinitely.
Self-check
A substance inhibits acetylcholinesterase. Predict what happens to acetylcholine at the cleft and to postsynaptic stimulation.
Show the answer
Acetylcholine persists longer because breakdown slows. Receptors remain stimulated for longer, so postsynaptic excitation may be prolonged; the exact whole-organism effect depends on the target and other control.
Skeletal-muscle contractionCore
Scope: Core · §3.6.3 Skeletal muscles are stimulated to contract by nerves and act as effectors
Open revision guidance
What to know
Skeletal muscles pull but do not push, so antagonistic pairs move an incompressible skeleton. A muscle contains bundles of long fibres; each fibre contains myofibrils made of repeating sarcomeres between Z-lines, with thin actin and thick myosin filaments. During contraction actin slides past myosin, shortening sarcomeres; filaments do not shorten, and the A-band stays constant while the I-band and H-zone narrow. A motor-neurone impulse at the neuromuscular junction starts a muscle action potential that causes calcium release from the sarcoplasmic reticulum; calcium shifts tropomyosin away from actin-binding sites.
Myosin heads form actin–myosin bridges and pull actin inward; ATP binding detaches a head and ATP hydrolysis re-cocks it. Repeated cycles shorten the myofibril. Calcium is pumped away when stimulation ends. Phosphocreatine rapidly regenerates ATP. Slow fibres have abundant mitochondria, myoglobin and capillaries for aerobic ATP production and resist fatigue; fast fibres produce rapid, powerful contractions but fatigue sooner.
What to do in an assessment
Explain contraction as repeated cross-bridge cycling: calcium exposes actin sites, myosin binds and pulls, ATP detaches and resets the head, and the filaments slide rather than shorten. Compare slow/fast fibre structure and properties against the task and fatigue evidence.
In plain language
A muscle shortens because myosin heads repeatedly grip and pull actin filaments. Calcium exposes the grip sites and ATP lets the heads release and reset for another pull.
Self-check
Why does a lack of ATP cause actin and myosin to remain attached, and how does calcium normally permit contraction?
Show the answer
ATP binding is needed to detach a myosin head from actin; without ATP cross-bridges remain attached. Calcium moves tropomyosin away from actin's myosin-binding sites so cycling can begin.

Kidneys, ADH and blood water potentialCore
Scope: Core · §3.6.4.3 Control of blood water potential (A-level only)
Open revision guidance
What to know
Osmoregulation controls blood water potential. Hypothalamic osmoreceptors detect its change and the posterior pituitary releases ADH. ADH raises the water permeability of the distal convoluted tubule and collecting ducts, so more water is reabsorbed to blood and a smaller, more concentrated urine volume is produced; when ADH falls, less water is reabsorbed and urine is more dilute.
At the renal corpuscle, high glomerular hydrostatic pressure forces water and small solutes through the filtration barrier into Bowman's capsule; blood cells and large plasma proteins remain in the capillaries. The proximal convoluted tubule selectively reabsorbs all glucose in a healthy kidney and reabsorbs water. In the loop of Henle, water leaves the descending limb while the ascending limb moves sodium ions into the medulla and is much less permeable to water; this establishes the medullary sodium/water-potential gradient. Water then moves by osmosis from the distal tubule and collecting duct down that gradient.
What to do in an assessment
Follow water from filtrate to blood down the medullary water-potential gradient; distinguish glomerular filtration from selective reabsorption. In an ADH graph, connect hormone concentration to distal/collecting-duct permeability, water reabsorbed, urine volume and urine concentration.
In plain language
The kidney first filters small molecules from blood, then returns useful substances. ADH adjusts how much water the final nephron reabsorbs, helping keep blood water potential within limits.
Self-check
After dehydration, explain the pathway from blood change to concentrated urine.
Show the answer
Blood water potential falls; hypothalamic osmoreceptors signal release of more ADH from the posterior pituitary. Nephron collecting ducts become more permeable, more water is reabsorbed into the medulla and blood, and a smaller volume of concentrated urine leaves.
Blood-glucose control and feedbackCore
Scope: Core · §3.6.4.2 Control of blood glucose concentration (A-level only)
Open revision guidance
What to know
Blood glucose is influenced by carbohydrate absorption, use in respiration and storage/release by the liver, and changes with feeding and physical activity. Pancreatic β cells release insulin when glucose is high: insulin binds target-cell surface receptors, promotes inclusion of glucose-channel proteins in the cell surface and activates enzymes converting glucose to glycogen (glycogenesis). Low glucose stimulates α cells to release glucagon; glucagon binds target-cell receptors and activates liver enzymes for glycogenolysis and gluconeogenesis from glycerol and amino acids. Adrenaline also binds receptors and activates glycogen-to-glucose enzymes.
Glucagon and adrenaline act through a second-messenger pathway: hormone–receptor binding activates adenylate cyclase, which forms cyclic AMP (cAMP); cAMP activates protein kinase, which activates downstream enzymes. Insulin and glucagon oppose one another in negative feedback. Type 1 diabetes results from loss/destruction of pancreatic β-cell insulin production and is controlled with insulin; type 2 involves insulin resistance, with genetic and lifestyle risk factors, and may be managed by diet and other treatment, including medication or insulin when needed.
What to do in an assessment
For a graph or clinical scenario, identify the deviation, pancreatic cell/hormone, target response and direction of correction. Distinguish blood-glucose concentration from glucose detected in urine; evaluate diet/treatment evidence and claims about rising type 2 diabetes incidence by checking study design, confounders and interests of health advisers or food-industry sources rather than inferring cause from association alone.
In plain language
After a meal, insulin helps remove excess glucose from blood and store it. Between meals, glucagon helps release glucose from liver stores. Their opposing actions keep blood glucose within a range rather than at one unchanging value.
Self-check
After a carbohydrate-rich meal, blood glucose rises. Give the corrective response and explain how glucagon raises blood glucose between meals.
Show the answer
β cells release insulin; target cells include more glucose-channel proteins in their surface membranes, cells take up glucose and liver/muscle form glycogen, lowering blood concentration. Between meals, glucagon binds liver-cell receptors and its cAMP/protein-kinase pathway activates glycogenolysis and gluconeogenesis, releasing glucose.

Ecosystems, population estimates and successionCore
Scope: Core · §3.7.4 Populations in ecosystems (A-level only)
Open revision guidance
What to know
A community plus its non-living environment is an ecosystem, a dynamic system that may be very small or large. A species' niche reflects adaptations to abiotic and biotic conditions. Carrying capacity is the population size an ecosystem supports and varies with physicochemical factors and intraspecific/interspecific competition or predation. Random quadrats suit sessile or slow organisms; a belt transect tests distribution along a gradient. Frequency or percentage cover can estimate abundance of sessile species.
Mark–release–recapture estimates a motile population as n₁×n₂/m, assuming marks persist, marked individuals mix with the population, marking does not change survival/capture, the population is closed/stable and individuals have an equal chance of capture. Primary succession proceeds from pioneer colonisation toward a changing climax community; later species modify conditions, which can make them more suitable for new species and less suitable for earlier ones. Conservation may manage succession.
What to do in an assessment
For 40 marked animals released, then 50 caught with 10 marked, estimate N=(40×50)/10=200; state assumptions and uncertainty. Evaluate a conservation choice by balancing human resource use, sustainability and evidence for biodiversity effects; compare conflicting claims, sampling methods, uncertainty and the time scale of the data before judging a conclusion.
In plain language
Ecologists sample because counting every organism is rarely possible. Random placement avoids selecting convenient spots; marking and recapturing gives an estimate only if the marked animals have mixed and the assumptions are reasonable.
Self-check
A first catch marks 30 animals. Later, 45 are caught and 9 are marked. Calculate the estimate and identify one assumption that could make it biased.
Show the answer
N=(30×45)/9=150. If marks fade or marked animals avoid traps, fewer marked individuals are recaught, so the estimate is too high; other assumptions include mixing and a closed population.

Cell specialisation and stem cellsCore
Scope: Core · §3.8.2.1 Most of a cell’s DNA is not translated (A-level only)
Open revision guidance
What to know
Cells with the same genome specialise because they translate only some genes. Totipotent cells can divide to produce any type of body cell and occur only for a limited time in early mammalian embryos. Embryonic pluripotent cells can divide in unlimited numbers and form many cell types; multipotent and unipotent stem cells in mature mammals have progressively narrower potential. Cardiomyocytes are an example of cells formed from a unipotent stem-cell lineage.
Pluripotent cells may be used to treat human disorders. Induced pluripotent stem (iPS) cells can be produced from adult somatic cells by using appropriate protein transcription factors to reprogramme gene expression. Potential therapies include replacing damaged cells, but risks include immune rejection, uncontrolled division/tumours, difficulty directing differentiation and ethical concerns about embryo use. Aseptic culture of cauliflower explants is an example of tissue-culture technique, distinct from the clinical evidence needed to establish a stem-cell treatment.
What to do in an assessment
Evaluate a proposed stem-cell treatment by balancing evidence of benefit, clinical safety, immune compatibility, source, consent/ethics and alternatives; do not treat laboratory potential as proven clinical effectiveness. For tissue-culture practical contexts, keep explants, instruments and culture medium sterile and compare growth under controlled conditions.
In plain language
Most body cells share DNA, but each cell switches on only some genes. Stem cells retain the ability to make other cell types; the broader their potential, the greater both their therapeutic promise and the challenge of controlling what they become.
Self-check
A trial reports that stem cells formed heart-like cells in a dish. Give one reason this is not enough to conclude that the treatment is safe and effective.
Show the answer
Cells in a dish do not show whether transplanted cells integrate and improve heart function in people. Evidence is still needed on dose, survival, correct cell identity, tumour risk, immune rejection, long-term outcomes and controlled clinical benefit.
Transcription control, epigenetics and RNA interferenceCore
Scope: Core · §3.8.2.2 Regulation of transcription and translation (A-level only)
Open revision guidance
What to know
In eukaryotes, specific transcription factors move from cytoplasm into the nucleus and stimulate or inhibit target-gene transcription; oestrogen acting through its receptor can initiate transcription. Epigenetic changes alter heritable gene function without changing DNA base sequence, influence development and are relevant to diseases including cancer.
Environmental conditions can change epigenetic marks: increased DNA methylation or decreased acetylation of associated histones tends to inhibit transcription by making DNA less accessible. RNA interference (RNAi) uses small RNA molecules to bind target mRNA and inhibit its translation in eukaryotes and some prokaryotes; environmental and genetic factors can both affect phenotype.
What to do in an assessment
Interpret gene-expression data by naming the measured variable, comparing controls and treatment, and distinguishing a change in expression from a change in DNA sequence. Evaluate evidence for relative genetic/environmental effects on phenotype; correlation between exposure and expression alone does not establish causation.
In plain language
Epigenetic control changes how accessible a gene is, rather than rewriting its letters. More methylation or less histone acetylation usually makes a gene harder to transcribe; RNA interference acts later by stopping a message being translated.
Self-check
A treatment increases methylation near a gene and its mRNA level falls. Explain a plausible mechanism and one limitation of that conclusion.
Show the answer
Increased methylation may reduce transcription-factor/RNA-polymerase access, lowering transcription and mRNA. The association alone does not prove methylation caused the fall; a controlled intervention, appropriate controls and replication would strengthen causal inference.
Gene expression and cancerCore
Scope: Core · §3.8.2.3 Gene expression and cancer (A-level only)
Open revision guidance
What to know
Benign tumours remain localised; malignant tumours invade nearby tissue and can spread. Oncogenes can promote cell division when abnormally active; tumour-suppressor genes normally restrain division or support repair and can contribute to cancer when inactivated.
Abnormal methylation can silence tumour-suppressor genes or alter oncogene expression. Increased oestrogen is associated with development of some breast cancers. Cancer risk evidence may involve inherited variants, environment and interactions between them.
What to do in an assessment
When interpreting risk data, identify denominator and comparison group, distinguish relative from absolute risk, and avoid claiming causation from a correlation. Explain how understanding a gene pathway suggests—but does not guarantee—a prevention or treatment strategy.
In plain language
Cancer can arise when the controls on cell division fail. A permanently active growth signal or a switched-off brake can let cells divide; tumour behaviour depends on invasion and spread, not only on a lump's size.
Self-check
A study finds higher cancer incidence in people exposed to a chemical. What evidence would strengthen a causal claim, and how can a tumour-suppressor gene be involved?
Show the answer
A dose-response pattern, temporality, suitable control of confounders, replication and mechanistic evidence strengthen—but do not by themselves guarantee—a causal conclusion. If a tumour-suppressor gene is silenced or damaged, the cell may lose a restraint on division or DNA repair.

Genome projects and proteomesCore
Scope: Core · §3.8.3 Using genome projects (A-level only)
Open revision guidance
What to know
Genome projects sequence the DNA of organisms. For a simple organism, identifying coding sequences can help predict its proteome and potential antigens for vaccine research. Automated methods have made sequencing faster and continue to develop.
In complex organisms, non-coding DNA, introns and regulatory genes mean genome sequence alone does not directly specify which proteins a cell makes; expression depends on cell type, developmental stage and environment.
What to do in an assessment
Explain both the use and limit of a genome result: sequence can identify candidate genes/variants, but protein expression, function and phenotype require additional evidence.
In plain language
A genome is the instruction set, not a list of every protein a living cell is currently making. Gene regulation determines which instructions are used and proteins may be modified after translation.
Self-check
Why does sequencing every base in a human genome not by itself reveal the proteome of a liver cell?
Show the answer
Much DNA is non-coding or regulatory, and the liver cell expresses only a subset of genes at a given time. Its proteins also depend on RNA processing, translation and subsequent modification, so sequence alone is insufficient.
Recombinant DNA and gene transferCore
Scope: Core · §3.8.4.1 Recombinant DNA technology (A-level only)
Open revision guidance
What to know
A target DNA fragment can be made from mRNA by reverse transcriptase to form complementary DNA (cDNA), cut from DNA with restriction enzymes, or synthesised in a gene machine. In PCR, heating separates DNA strands, primers bind target sequences as the mixture cools, and a thermostable DNA polymerase extends the primers using free nucleotides; repeated cycles amplify the fragment in vitro. Culturing transformed host cells amplifies inserted DNA in vivo.
Restriction endonucleases cut the target and vector at matching recognition sites; DNA ligase joins the fragments. Promoter and terminator regions are added so the host can transcribe the inserted sequence. Vectors transform host cells, and marker genes identify likely GM cells (specific marker genes need not be recalled). The universal genetic code and shared transcription/translation machinery let transferred genes be expressed in another organism. Recombinant DNA can also underpin gene therapy: a functional gene may be delivered to target cells to restore a missing or defective gene product.
What to do in an assessment
Describe the method in causal order: obtain target → amplify if needed → cut target/vector with compatible restriction endonuclease → ligate → transform host → identify transformed cells with a marker → culture or express product. Relate recombinant DNA to gene therapy and evaluate medical, agricultural and industrial benefits against safety/containment, environmental, animal-welfare, financial, social and ownership concerns; weigh humanitarian benefits and opposition using evidence.
In plain language
To make a cell produce a new protein, place the coding sequence into a carrier DNA molecule, get that carrier into a host cell and identify successful cells. PCR copies DNA in a tube; culturing transformed cells copies it inside living hosts. Gene transfer can also deliver a functional gene to cells in gene therapy.
Self-check
Why can a gene inserted into a bacterium fail to produce the intended human protein even if the bacteria take up the vector, and how could recombinant DNA be used in gene therapy?
Show the answer
The gene may lack a promoter the bacterium recognises, be in the wrong orientation or reading frame, or require RNA processing/modifications bacteria cannot provide. Transformation alone does not prove correct expression. In gene therapy a functional DNA sequence may be delivered to affected cells so they can produce the missing or functional protein.
DNA probes, screening and personalised medicineCore
Scope: Core · §3.8.4.2 Differences in DNA between individuals of the same species can be exploited for identification and diagnosis of heritable conditions
Open revision guidance
What to know
A labelled DNA probe has a base sequence complementary to a target allele. Under suitable conditions it hybridises to that sequence, and its label reveals whether the target is present. Probes can screen for heritable conditions, likely drug responses or health risks.
Results may inform genetic counselling and personalised medicine, but a risk-associated allele does not guarantee disease. Screening raises questions about consent, privacy, discrimination, false results and access to follow-up care.
What to do in an assessment
For a screening decision, distinguish test accuracy from disease certainty and consider prevalence, false positives/negatives, informed choice, clinical actionability and consequences for relatives.
In plain language
A DNA probe is a labelled matching sequence: if the patient's DNA contains the complementary target, the probe binds and its label can be detected. A positive risk marker changes probability; it is not a certain prediction.
Self-check
A screening probe detects a risk allele in an asymptomatic patient. Why should counselling avoid saying that the patient will develop the condition?
Show the answer
The allele may increase risk without being sufficient to cause disease; penetrance, other genes and environment can affect outcome. The result must be explained probabilistically with test limitations and possible choices.
Genetic fingerprinting and electrophoresisCore
Scope: Core · §3.8.4.3 Genetic fingerprinting (A-level only)
Open revision guidance
What to know
Variable number tandem repeats (VNTRs) differ among individuals. PCR can amplify DNA fragments containing VNTRs; electrophoresis separates charged fragments by size through a gel, producing a banding pattern. A smaller DNA fragment usually migrates farther through the gel.
Similar profiles can support relatedness or identification, but interpretation depends on sample quality, matching loci and the probability of coincidental matches. Uses include forensic investigation, diagnosis, population variation and breeding.
What to do in an assessment
Compare bands at the same loci; count matching and non-matching bands and account for lane orientation/size markers. A match supports a relationship but is not by itself proof of identity or guilt.
In plain language
DNA fragments act like a barcode: gel electrophoresis sorts them by length and creates visible bands. The pattern can be compared, but a strong conclusion also needs reliable sampling and a known chance-match rate.
Self-check
Two samples share all visible bands at the tested loci. What conclusion is justified, and what cannot be concluded from the gel alone?
Show the answer
Their profiles are consistent at those loci and may support a common source or close relationship. The gel alone does not establish unique identity, exclude contamination, or prove a person's involvement; sample quality and match probability matter.

Biological data handling and statisticsCore
Scope: Core · §6.2 Handling data (MS 1.1–1.11)
Open revision guidance
What to know
Organise observations in frequency tables with headings, units and consistent decimal places. Use bar charts for categories, histograms for grouped continuous frequency data, and scatter diagrams for paired continuous variables; label scales/axes and show uncertainty when appropriate. Mean=sum/n; median is the middle ordered value and mode the most common. Range=max−min. For a sample, standard deviation s=√[Σ(x−x̄)²/(n−1)] describes spread around the mean; use a specified formula or calculator convention and report calculated precision no greater than the least accurate measurement.
Select a statistical test to fit the data and question: χ² compares observed categorical counts with expected counts (χ²=Σ(O−E)²/E; goodness-of-fit df=categories−1); Student's t-test assesses whether two sample means differ; a correlation coefficient assesses association between paired variables. State the null hypothesis, use any required test calculation and the appropriate degrees of freedom/critical-value table, and conclude in biological context. Correlation does not demonstrate causation.
Uncertainty applies to measurements and calculated results. Percentage error=absolute uncertainty ÷ measured value ×100. For sums/differences, add absolute uncertainties; for products/quotients, add fractional or percentage uncertainties. Use repeat spread and instrument resolution to discuss uncertainty, accuracy and precision; do not imply a result is more precise than its measurements.
What to do in an assessment
For a test, state H₀, select the method, show substitutions, find degrees of freedom/critical value and conclude whether the difference/association is significant at the stated probability. For observations 12, 14, 16, mean=14; deviations are −2, 0, +2, so sample SD=√[(4+0+4)/(3−1)]=2. Keep mean and spread distinct; §3.4.7 does not require calculation of SD for its within-species variation context, but MS 1.10 does require SD calculation more generally.
In plain language
Statistics help decide whether a pattern is larger than random scatter, not whether it matters biologically or what caused it. Pick a method that matches the kind of data, and state exactly what the result does and does not support.
Self-check
Observed categories are 18 and 22; expected counts are 20 and 20. Calculate χ² and state what else is needed to decide significance.
Show the answer
χ²=Σ(O−E)²/E = (18−20)²/20 + (22−20)²/20 = 0.2+0.2=0.4. Compare with the critical value for the correct degrees of freedom and chosen probability level; the statistic alone does not establish significance.
Biological equations, units and algebraCore
Scope: Core · §6.3 Algebra (MS 2.1–2.5)
Open revision guidance
What to know
Use =, <, <<, >>, >, ∝, ≈ and ~ accurately to express equality, inequality, much less/greater, proportionality and approximation. Substitute values with units into supplied biological equations and rearrange while preserving dimensions; convert before calculating where units differ.
A logarithmic scale represents quantities across orders of magnitude; use powers/logarithms for microbial growth data. For Simpson's index, use the exact equation and definitions provided: AQA §3.4.6 uses d=N(N−1)/Σn(n−1), while the mathematical exemplification may present D=1−Σ(n/N)². Do not mix the two forms. For magnification, M=image/actual and actual=image/M. Check whether a rate unit is per second, minute or year.
What to do in an assessment
Example: cardiac output CO=heart rate×stroke volume; if CO=4.8 dm³ min⁻¹ and rate=80 min⁻¹, stroke volume=4.8/80=0.060 dm³=60 cm³ beat⁻¹. Sense-check the size and cancel units.
For counts 4, 3 and 3 (N=10), the §3.4.6 form gives d=90/24=3.75; the alternative supplied form gives D=1−[(4/10)²+(3/10)²+(3/10)²]=0.66. The interpretation is in relation to the specified index; do not substitute one formula's expected scale for the other.
In plain language
An equation is a relationship between measured quantities. Rearranging asks the same relationship a different way; writing units at every step reveals many errors before the final answer.
Self-check
A cell image is 18 mm wide at ×6000. Calculate its actual width in micrometres.
Show the answer
Actual size = image/magnification = 18 mm/6000 = 0.003 mm. Since 1 mm=1000 µm, this is 3 µm.
Graph choice, gradients and biological ratesCore
Scope: Core · §6.4 Graphs (MS 3.1–3.6)
Open revision guidance
What to know
Translate between a graph, table, equation and verbal relationship. Plot independent variable on x and measured dependent variable on y; choose a scatter graph for paired continuous data, bar chart for categories and histogram for grouped continuous frequency.
For a straight line, gradient is Δy/Δx and y=mx+c. A tangent estimates instantaneous rate on a curve; an intercept is the value where the graph crosses an axis. Keep plotted scales, units, uncertainty bars and trend claims explicit.
What to do in an assessment
For an enzyme curve, draw a tangent at the requested time and calculate Δproduct/Δtime. Example: a tangent rise of 6 µmol over 2 min gives 3 µmol min⁻¹. Do not use a distant secant for an instantaneous rate.
In plain language
The slope tells how quickly one quantity changes as another changes. On a curve, the slope changes from point to point, so a tangent gives the rate at the selected point.
Self-check
A tangent on a concentration–time graph rises 8 mmol dm⁻³ over 4 min. What is the rate and its unit?
Show the answer
Rate=8/4=2 mmol dm⁻³ min⁻¹. The units combine the y-axis concentration unit with the x-axis time unit.
Geometry, surface area and volumeCore
Scope: Core · §6.5 Geometry and trigonometry (MS 4.1)
Open revision guidance
What to know
Use circle circumference 2πr and area πr². A rectangular prism has volume lwh and surface area 2(lw+lh+wh); a cylinder has volume πr²h and total surface area 2πr²+2πrh; a sphere has volume 4πr³/3 and surface area 4πr². Area has squared units and volume cubed units.
For similar shapes, if linear dimensions scale by k, surface area scales by k² and volume by k³; this explains why small cells have a larger surface-area-to-volume ratio than geometrically similar large cells.
What to do in an assessment
Convert dimensions to a common unit before calculating. Example: a cube of side 2 mm has area 6×2²=24 mm² and volume 2³=8 mm³, so SA:V=24:8=3:1 mm⁻¹.
In plain language
Surface area is the boundary available for exchange; volume is the amount of material needing resources. As a similar object grows, its volume increases faster than its surface area.
Self-check
A cube has side 3 mm. Calculate its surface area-to-volume ratio.
Show the answer
Surface area=6×3²=54 mm²; volume=3³=27 mm³. Ratio=54/27=2 mm⁻¹, or 2:1 when expressed as a simple ratio.
Arithmetic, ratios, percentages and scaleCore
Scope: Core · §6.1 Arithmetic and numerical computation (MS 0.1–0.5)
Open revision guidance
What to know
Convert units before substitution (1 dm³=1000 cm³; 1 mm=1000 µm; 1 cm³=1000 mm³); label rates with quantity per time. Use decimals and standard form without losing significant figures, e.g. 0.0050=5.0×10⁻³.
Calculate ratios, fractions, percentage yield = actual/theoretical ×100 and percentage change = change/original ×100. Estimate first, use powers/logarithms for very large populations, and sense-check whether a result is biologically plausible. For a population that doubles each generation, N=N₀×2ᵍ after g generations (for example, 1 000 cells after four doublings gives 1 000×2⁴=16 000); if asked to infer generations, rearrange to g=log₂(N/N₀).
What to do in an assessment
Show a conversion before using a formula; keep units through each line and round only at the end to the precision supported by data. Example: 24 of 30 expected seeds germinate gives 24/30×100=80%.
In plain language
A biology calculation is not complete until its units and scale make sense. A result can have correct arithmetic but still be wrong if millimetres were treated as micrometres or a percentage used the wrong starting value.
Self-check
A culture increases from 2.0×10⁴ to 5.0×10⁴ cells. Calculate percentage increase and express the final count in ordinary form.
Show the answer
Increase=3.0×10⁴; percentage increase=(3.0×10⁴)/(2.0×10⁴)×100=150%. Final count=50,000 cells.
pH, logarithms and numerical scaleCore
Scope: Core · §6.1 Arithmetic and numerical computation (MS 0.5)
Open revision guidance
What to know
pH=-log₁₀[H⁺], where [H⁺] is in mol dm⁻³. A tenfold increase in hydrogen-ion concentration lowers pH by one unit; pH is logarithmic, not a linear measure of acidity.
Use a calculator carefully with powers of ten and brackets. For [H⁺]=1.0×10⁻⁶ mol dm⁻³, pH=6; if concentration becomes 1.0×10⁻⁵ mol dm⁻³, pH=5.
What to do in an assessment
Write the concentration in standard form, enter the negative base-10 logarithm, and retain the stated significant figures. When conditions change, compare the concentration ratio before describing the pH shift.
In plain language
One pH unit represents a tenfold change in hydrogen-ion concentration. A solution at pH 4 has ten times the hydrogen-ion concentration of one at pH 5.
Self-check
Calculate pH for [H⁺]=1.0×10⁻⁴ mol dm⁻³ and state how its hydrogen-ion concentration compares with pH 6.
Show the answer
pH=−log₁₀(1.0×10⁻⁴)=4. It has 100 times the hydrogen-ion concentration of pH 6 because the difference is two logarithmic units.

Practical skills assessed in written papersCore
Scope: Core · §8.3 Practical skills assessed in written papers (PS 1.1–4.1)
Open revision guidance
What to know
At least 15% of A-level marks assess practical skills in written papers. PS 1.1–1.2 require solving a practical problem and applying biological knowledge; PS 2.1–2.4 require commenting on design/methods, presenting data, evaluating results against errors/uncertainties and identifying independent, dependent and controlled variables. PS 3.1–3.3 require plotting/interpreting graphs, processing data mathematically and considering margins of error, accuracy and precision; PS 4.1 requires understanding suitable instruments and techniques.
A valid method tests the stated hypothesis and controls confounding variables; reliability is consistency on repetition, accuracy is closeness to a true value and precision is closeness among measurements. Specify an operational measurement and calibrated instrument, repeats/sample size, relevant controls and a safe/ethical procedure. In results tables include headings, units and consistent precision; plot the independent variable on x and dependent variable on y, choose a suitable graph and state the calculation/test used.
Evaluate conclusions against variation, anomalies, measurement resolution and systematic/random error. Explain whether uncertainty could change the conclusion, distinguish a practical improvement from a reason for it, and avoid claiming causation from an uncontrolled correlation.
What to do in an assessment
For an unfamiliar practical, identify the question and variables, apply the relevant biology to predict a measurable response, write a reproducible procedure, justify controls/repeats/equipment and safety, plan a table/graph and calculation, then evaluate the result and conclusion with named errors and uncertainty. Tie each proposed improvement to a stated limitation.
In plain language
Exam questions can ask you to design or critique an experiment without performing it. Every improvement should solve a stated problem; 'repeat it' alone cannot fix a biased sample, poor calibration or uncontrolled variable.
Self-check
A colorimeter investigation uses different cuvettes and one reading per concentration. Give two specific improvements and explain the problem each addresses.
Show the answer
Use the same clean cuvette orientation (or matched cuvettes) to reduce optical-path differences; take replicate readings at each concentration and calculate a mean to reveal scatter and improve precision. Calibrate/blank the colorimeter to address instrument baseline.

Written paper scope and extended biological argumentCore
Scope: Core · Assessment at a glance, specification pp. 9–10
Open revision guidance
What to know
Paper 1 assesses topics 1–4 in 2 hours for 91 marks (35%): 76 marks of short and long answers and 15 marks of extended response. Paper 2 assesses topics 5–8 in 2 hours for 91 marks (35%): 76 marks of short and long answers and 15 marks of comprehension. Paper 3 assesses topics 1–8 in 2 hours for 78 marks (30%): 38 marks of structured questions including practical techniques, 15 marks of critical analysis of experimental data, and one 25-mark essay selected from two titles. Relevant practical skills can appear across the written papers.
Biology assessment rewards accurate knowledge, application to unfamiliar contexts and analysis/evaluation of evidence. A strong essay selects relevant knowledge across topics, uses precise biological links, organises a coherent argument and avoids unsupported claims or irrelevant lists.
What to do in an assessment
For an essay, spend time selecting the title you can answer with relevant breadth; plan a sequence of linked paragraphs, define terms, explain mechanisms and make comparisons. Apply the same evidence chain in structured answers: claim → biological mechanism/data → justified conclusion, with units and limits.
In plain language
The first two papers focus on different halves of the course, but Paper 3 can connect any topic and includes an essay. The essay is marked for relevant biological knowledge and a clear, accurate argument—not for simply listing facts.
Self-check
A Paper 3 essay title invites discussion of energy transfers. Why is a linked argument across photosynthesis, respiration and ecosystems stronger than three disconnected topic lists?
Show the answer
A linked argument explains how light energy is stored in biomass by photosynthesis, transferred to ATP through respiration, and lost at trophic transfers. Connections answer the title directly; isolated lists may contain facts without showing their relevance or relationship.

Revision points, with the explanation beside them

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language

Self-check

What to know

What to do in an assessment

In plain language