Eleven lessons covering the eight topic areas of A-level Biology, from biological molecules and cells to gene expression, with practical skills and interactive explorers.
A study guide to A-level Biology, not a full course or textbook. The paper structure follows the AQA 7402 pages; other boards (Edexcel, OCR, WJEC) organize topics differently. Biology content is standard A-level knowledge for study, and you should check your own specification. Lab numbers are practice values.
['GCSE Biology or Combined Science at higher tier']
Course outline
Biological molecules
Describe carbohydrates, lipids, proteins and how enzymes work.
Nucleic acids, DNA replication and protein synthesis
Explain DNA and RNA structure, replication and the production of proteins.
Cell structure, membranes and transport
Link cell parts to function and explain how substances cross membranes.
Cell division and the immune system
Describe mitosis, the cell cycle and immune responses.
Exchange surfaces and mass transport
Explain gas exchange, digestion and transport in animals and plants.
Energy transfers: photosynthesis and respiration
Compare the stages of photosynthesis and respiration and ATP production.
Responding to the environment: nerves, hormones and homeostasis
Describe nervous and hormonal control and blood glucose regulation.
Inheritance, variation and the Hardy-Weinberg principle
Predict inheritance and use allele frequencies in populations.
Evolution, populations and ecosystems
Explain natural selection, speciation and ecosystem energy flow and diversity.
Control of gene expression and gene technology
Explain mutations, regulation of transcription and tools such as PCR.
Practical skills, mathematics and exam technique
Plan investigations, handle data and structure exam answers.
Sources and curriculum note
Reviewed October 7, 2026. Confirm the specification for your board and exam year.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Biological molecules
Learning goal: Describe carbohydrates, lipids, proteins and how enzymes work.
Carbohydrates are made of sugar units. Monosaccharides such as glucose join by glycosidic bonds in condensation reactions to form disaccharides and polysaccharides. Starch and glycogen store energy, and cellulose gives cell walls strength. Hydrolysis uses water to break bonds.
Triglycerides contain glycerol and three fatty acids joined by ester bonds. Saturated fatty acids have no carbon-carbon double bonds, and unsaturated ones have at least one. Phospholipids have a hydrophilic head and hydrophobic tails, so they form bilayers in membranes.
Proteins are chains of amino acids joined by peptide bonds. The primary structure is the sequence, the secondary structure is folding such as alpha helices, and the tertiary structure is the three-dimensional shape held by hydrogen, ionic and disulfide bonds. Some proteins have a quaternary structure of several chains.
Enzymes are biological catalysts that lower the activation energy of reactions. The substrate binds to the active site, whose shape depends on the protein's tertiary structure. Temperature, pH, substrate concentration and enzyme concentration all change the rate. Competitive inhibitors bind at the active site, and non-competitive inhibitors bind elsewhere and change its shape.
Worked example
Explain a rate change.
Name the factor
Link to structure
Say what happens to collisions or shape
Conclude
Practice problem and solution
A protein has 120 amino acids in one chain. How many peptide bonds does it contain? Enter a number.
n - 1 = 120 - 1 = 119.
Mental model: Condensation builds polymers, hydrolysis breaks them, and enzyme shape decides how it works.
Common trap: Saying that enzymes are killed by heat instead of denatured.
2. Nucleic acids, DNA replication and protein synthesis
Learning goal: Explain DNA and RNA structure, replication and the production of proteins.
DNA is a double helix of two antiparallel polynucleotide chains. Each nucleotide has deoxyribose, a phosphate group and one of the bases adenine, thymine, cytosine or guanine. Adenine pairs with thymine and cytosine with guanine by hydrogen bonds. RNA is a single strand with ribose and uracil instead of thymine.
DNA replication is semi-conservative: each new molecule has one original strand and one new strand. DNA helicase breaks hydrogen bonds, and DNA polymerase joins nucleotides to the exposed template strands. The Meselson and Stahl experiment gave the evidence using nitrogen isotopes.
Transcription makes messenger RNA from a DNA template using RNA polymerase. In eukaryotes the pre-mRNA is spliced to remove introns before the mRNA leaves the nucleus. Translation takes place at a ribosome, where transfer RNA anticodons pair with mRNA codons and amino acids join by peptide bonds.
The genetic code is triplet, non-overlapping and degenerate, meaning several codons can code for the same amino acid. A gene mutation changes the base sequence and may change the protein.
Worked example
Predict a sequence.
Write template
Pair bases
Swap T for U if RNA
Read triplets
Practice problem and solution
A gene has 300 bases coding for amino acids in triplets (ignore stop codons). How many amino acids does it code for? Enter a number.
300 / 3 = 100.
Mental model: DNA replicates semi-conservatively; transcription makes mRNA; translation makes the polypeptide.
Common trap: Mixing up transcription and translation.
3. Cell structure, membranes and transport
Learning goal: Link cell parts to function and explain how substances cross membranes.
Eukaryotic cells have a nucleus, mitochondria, endoplasmic reticulum, Golgi apparatus and ribosomes. Plant cells also have chloroplasts, a cell wall and a vacuole. Prokaryotes have no nucleus, and their DNA lies free in the cytoplasm. Electron microscopes give higher resolution than light microscopes.
The fluid mosaic model describes membranes as phospholipid bilayers with proteins embedded. Channel and carrier proteins move substances across. Cholesterol affects fluidity. Simple diffusion moves small, non-polar molecules down their concentration gradient. Facilitated diffusion uses proteins. Active transport moves substances against a gradient using ATP and carrier proteins.
Osmosis is the net movement of water from a region of higher water potential to a region of lower water potential across a partially permeable membrane. Pure water has a water potential of 0, and solutions have negative values. A cell in a solution with a lower water potential loses water.
The surface area to volume ratio falls as cells get bigger, which limits exchange. That is why large organisms need exchange surfaces and transport systems.
Worked example
Explain a swelling cell.
Compare water potentials
Direction of water
Pressure effect
Conclude
Practice problem and solution
A cell is 2 micrometers across in a drawing that shows it 40 mm across. What is the magnification? Enter a number.
40 mm = 40000 micrometers, and 40000 / 2 = 20000.
Mental model: Membranes control movement; osmosis follows water potential; large size needs exchange surfaces.
Common trap: Saying water moves from high to low concentration of solute.
4. Cell division and the immune system
Learning goal: Describe mitosis, the cell cycle and immune responses.
The cell cycle has interphase, when DNA is replicated, followed by mitosis and cytokinesis. Mitosis has prophase, metaphase, anaphase and telophase, and produces two genetically identical nuclei. Control checkpoints stop cells dividing if DNA is damaged. Cancer arises when control fails and cells divide uncontrollably.
Meiosis produces four genetically different haploid cells. Independent assortment of chromosomes and crossing over between homologous chromosomes create variation. Fertilization restores the diploid number.
The immune system recognizes antigens on pathogens. Phagocytes engulf pathogens. In the specific response, T helper cells activate B cells, which divide and make plasma cells that release antibodies, and memory cells give a faster, larger secondary response. Antibodies bind to specific antigens, which can lead to agglutination and destruction.
Vaccines expose the body to antigens so memory cells form without disease. Herd immunity protects those who cannot be vaccinated when a large enough share of the population is immune. HIV infects T helper cells, which weakens immune responses.
Worked example
Explain vaccination.
Antigen exposure
Memory cells form
Fast response later
Reduced disease
Practice problem and solution
A cell with 46 chromosomes divides by mitosis. How many chromosomes does each daughter cell have? Enter a number.
Learning goal: Explain gas exchange, digestion and transport in animals and plants.
Efficient exchange surfaces have a large surface area, a short diffusion distance and a steep concentration gradient. Fick's law says the rate of diffusion is proportional to surface area times concentration difference divided by diffusion distance. Alveoli, fish gills and leaf mesophyll show these features.
In humans, air passes through the trachea, bronchi and bronchioles to the alveoli. Ventilation maintains the gradient. Fish use countercurrent flow in the gills, which keeps a gradient along the whole length of the lamella. Insects use tracheae and spiracles.
The mammalian heart pumps blood in a double circulation. Cardiac output is heart rate times stroke volume. Arteries carry blood at high pressure, veins have valves, and capillaries are one cell thick for exchange. Haemoglobin carries oxygen, and the oxygen dissociation curve shifts right when carbon dioxide is high, the Bohr effect.
In plants, xylem moves water and minerals upward by transpiration pull and cohesion-tension. Phloem moves sugars from sources to sinks by mass flow. Stomata open and close to balance gas exchange and water loss.
Worked example
Explain adaptation.
Name the feature
Link to Fick's law
Describe effect
Conclude
Practice problem and solution
A heart beats 70 times per minute with a stroke volume of 80 cm3. What is the cardiac output in cm3 per minute? Enter a number.
70 x 80 = 5600.
Mental model: Large area, short distance and steep gradient speed exchange; transport systems carry substances far.
Common trap: Forgetting to link each feature to the rate.
6. Energy transfers: photosynthesis and respiration
Learning goal: Compare the stages of photosynthesis and respiration and ATP production.
Photosynthesis traps light energy in chloroplasts. In the light-dependent reactions on the thylakoid membranes, light excites electrons in chlorophyll, water is split and oxygen is released, and ATP and reduced NADP are made. In the light-independent reactions (the Calvin cycle) in the stroma, carbon dioxide combines with RuBP, catalyzed by rubisco, and is reduced to triose phosphate using ATP and reduced NADP.
Respiration releases energy from glucose to make ATP. Glycolysis in the cytoplasm splits glucose to pyruvate with a net gain of 2 ATP. In aerobic respiration pyruvate enters the mitochondrion, forms acetyl coenzyme A in the link reaction, and passes through the Krebs cycle in the matrix, which makes reduced coenzymes and carbon dioxide. Oxidative phosphorylation on the inner membrane uses the electron transport chain and chemiosmosis, with oxygen as the final acceptor, to make most of the ATP.
Anaerobic respiration regenerates NAD so glycolysis continues: lactate in animals and ethanol and carbon dioxide in yeast. It makes far less ATP than aerobic respiration. The respiratory quotient is carbon dioxide produced divided by oxygen used, about 1.0 for carbohydrate and lower for fats.
Limiting factors for photosynthesis include light intensity, carbon dioxide concentration and temperature. At any time the factor in shortest supply limits the rate.
Worked example
Identify a limit.
Change one factor
Rate stops rising
That factor is not limiting
Another is
Practice problem and solution
A respiring tissue uses 50 cm3 of oxygen and produces 40 cm3 of carbon dioxide. What is the respiratory quotient? Enter a number.
40 / 50 = 0.8.
Mental model: Light makes ATP and reduced NADP; the Calvin cycle uses them; respiration makes ATP in four stages.
Common trap: Saying plants do photosynthesis and animals respire only.
7. Responding to the environment: nerves, hormones and homeostasis
Learning goal: Describe nervous and hormonal control and blood glucose regulation.
A neurone's resting membrane potential is about -70 millivolts, set by the sodium-potassium pump and membrane permeability. A stimulus above threshold opens sodium channels and depolarizes the membrane, then potassium channels open and the membrane repolarizes. The action potential is all-or-nothing and travels faster in myelinated axons, where it jumps between nodes of Ranvier.
At a synapse, an action potential triggers calcium entry, and neurotransmitter is released, diffuses across the cleft and binds receptors on the next cell. Summation of inputs decides whether a new action potential starts. Reflex arcs give fast, automatic responses.
Hormones are chemical messengers carried in the blood. Blood glucose is controlled by negative feedback. High glucose makes beta cells in the pancreas release insulin, which increases uptake and storage. Low glucose makes alpha cells release glucagon, which stimulates glycogen breakdown. Type 1 diabetes is an autoimmune loss of insulin production, and type 2 involves reduced response to insulin.
Plants respond with growth regulators such as auxins in phototropism. Homeostasis keeps internal conditions in a narrow range using receptors, a control center and effectors.
Worked example
Explain a response.
Name the stimulus
Identify the receptor
Describe the effector
Conclude
Practice problem and solution
A nerve impulse travels 120 m/s along a 0.6 m neurone. How long does it take in milliseconds? Enter a number.
0.6 / 120 = 0.005 s = 5 ms.
Mental model: Action potentials depend on ions; hormones and negative feedback keep conditions stable.
Common trap: Describing negative feedback as making a change bigger.
8. Inheritance, variation and the Hardy-Weinberg principle
Learning goal: Predict inheritance and use allele frequencies in populations.
A gene is a length of DNA coding for a protein, and alleles are alternative forms. A dominant allele is expressed in heterozygotes, and a recessive one only in homozygotes. Codominance gives both phenotypes together. Sex-linked genes lie on the X chromosome. Genetic diagrams predict the ratio of offspring, such as 3 : 1 for a monohybrid cross between heterozygotes.
A chi-squared test compares observed with expected counts: chi-squared = sum of (O - E)^2 / E. Compare with the critical value at p = 0.05 for the degrees of freedom, which equals the number of classes minus 1. If the value is below the critical value, differences could be due to chance.
The Hardy-Weinberg equations are p + q = 1 and p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele and q the recessive allele. They assume a large population, random mating, no mutation, no migration and no selection. If q^2 is the frequency of the recessive phenotype then q is the square root, and the carrier frequency is 2pq.
Real populations seldom meet all the assumptions, so Hardy-Weinberg gives a baseline against which to measure change.
Worked example
Use Hardy-Weinberg.
Find q squared
Square root for q
p = 1 - q
2pq carriers
Practice problem and solution
The recessive phenotype frequency is 0.04. What is the carrier frequency 2pq? Enter a number.
Learning goal: Explain natural selection, speciation and ecosystem energy flow and diversity.
Natural selection acts on variation. Individuals with advantageous alleles survive and reproduce more, so allele frequencies change over generations. Directional selection favors one extreme, stabilizing selection favors the mean and disruptive selection favors both extremes. Antibiotic resistance in bacteria is a clear example.
Speciation happens when populations become reproductively isolated. In allopatric speciation a geographical barrier separates them, and in sympatric speciation isolation occurs without a physical barrier. Genetic drift is a random change in allele frequency that has larger effects in small populations.
In ecosystems, gross primary productivity is the total energy fixed by producers and net primary productivity is gross minus respiration. Only a small share of energy passes between trophic levels. Nutrient cycles include nitrogen, carbon and phosphorus. Succession leads from pioneer species to a climax community.
Simpson's index of diversity measures species richness and evenness. D = 1 - sum of (n/N)^2, where n is the count of each species and N is the total. A higher value shows greater diversity. Biodiversity can be threatened by habitat loss, and conservation balances human needs with protection.
Worked example
Explain resistance.
Mutation exists
Antibiotic selects
Resistant survive
Spread
Practice problem and solution
A producer fixes 10000 kJ and uses 4000 kJ in respiration. What is NPP in kJ? Enter a number.
10000 - 4000 = 6000.
Mental model: Selection changes allele frequencies; isolation can lead to speciation; productivity and diversity can be measured.
Common trap: Saying individuals evolve instead of populations.
10. Control of gene expression and gene technology
Learning goal: Explain mutations, regulation of transcription and tools such as PCR.
Gene mutations include substitutions, deletions and insertions. A substitution may change one amino acid or none because the code is degenerate. A deletion or insertion that is not a multiple of three bases causes a frameshift. Mutagenic agents include radiation and some chemicals.
Cells in one organism have the same genes but express different ones. Transcription factors bind to promoter regions and switch genes on or off. Epigenetic changes such as methylation of DNA and modification of histones affect expression without changing the base sequence. RNA interference uses small RNAs to block translation.
Stem cells can differentiate. Totipotent cells can form any cell type, and pluripotent cells form most. Gene technology includes the polymerase chain reaction, which makes many copies of DNA by cycles of heating to separate strands, cooling for primers to bind, and warming for DNA polymerase to extend. Each cycle doubles the number of copies. Gel electrophoresis separates fragments by size, and DNA probes find specific sequences.
Genetic engineering moves genes between organisms, for example to make human insulin in bacteria. Ethical and safety questions need careful discussion with evidence.
Worked example
Assess a mutation.
Type
Where
Effect on protein
Effect on organism
Practice problem and solution
Starting with 1 DNA molecule, how many copies are there after 5 PCR cycles? Enter a number.
2^5 = 32.
Mental model: Mutations change sequences, transcription factors regulate genes, PCR doubles DNA each cycle.
Common trap: Thinking all mutations are harmful.
11. Practical skills, mathematics and exam technique
Learning goal: Plan investigations, handle data and structure exam answers.
AQA A-level Biology has required practical activities, such as the effect of enzyme concentration, osmosis in plant tissue, microscopy, chromatography and the effect of temperature on membrane permeability. These skills are assessed in the written exams and are also reported separately by schools. Know variables, controls, accuracy, precision, repeatability and reproducibility.
Mathematical skills include magnification = image size / actual size, percentages, rates, logarithms and statistical tests. Use units, and convert before you calculate. Chi-squared, the Student's t-test, correlation coefficients and the Simpson's index are in the specification. Graphs need labeled axes with units and a sensible scale. Draw lines of best fit through the trend.
The three papers are each 2 hours. Paper 1 covers topics 1 to 4 and Paper 2 topics 5 to 8, each 91 marks and 35 percent of the A-level. Paper 3 can assess any topic, with 78 marks and 30 percent, including a data-analysis section and one essay from two titles. Check the AQA page for the current format.
Use command words: describe, explain, suggest, calculate. In extended responses, plan a sequence, use key terms and link cause to effect. Do not repeat the question; make each point count.
Worked example
Check a method.
Control variable
Repeat
Measure precisely
Draw conclusion
Practice problem and solution
A 3:1 cross gives 160 of 200 offspring dominant. Expected dominant is 150 and expected recessive is 50. Chi-squared = (160-150)^2/150 + (40-50)^2/50. Enter the value to one decimal place.
100/150 + 100/50 = 0.667 + 2 = 2.667, which rounds to 2.7. The critical value for 1 degree of freedom at p = 0.05 is 3.84, so the fit is accepted.
Mental model: Know the three papers, the practical skills and how to write structured, evidence-based answers.