The International Biology Olympiad (IBO) works across seven areas: cell and molecular biology, plant and animal anatomy and physiology, ethology, genetics and evolution, ecology, and biosystematics. Cell biology, human physiology and genetics overlap heavily with IGCSE 0610 and AS & A Level 9700, but ethology, plant diversity and tree-building are almost absent from school courses. Most olympiad questions are also built around data, so reading experiments and doing quick statistics matter as much as facts.
Organelles and their jobs, prokaryote versus eukaryote, and magnification and resolution calculations.
Learn it Animal and plant cells (A Level) Prokaryotic cells and viruses (A Level) Microscopes, magnification and resolution (A Level) Animal, plant and bacterial cells (IGCSE) Cell structure lab
Structures of carbohydrates, lipids, proteins and water, the bonds that hold them, and the food tests.
Learn it Carbohydrates (A Level) Lipids (A Level) Proteins (A Level) Water (A Level) Large molecules from small ones (IGCSE) Biological molecules lab
The fluid mosaic model, diffusion, osmosis in terms of water potential, active and bulk transport.
Learn it The fluid mosaic model (A Level) Osmosis and water potential (A Level) Diffusion, active and bulk transport (A Level) Cell membranes and transport lab
School covers Vmax and Km in words; the olympiad wants the Michaelis-Menten equation, Lineweaver-Burk plots and telling inhibitor types apart from data.
Learn it How enzymes work (A Level) Factors that affect enzyme action (A Level) Enzymes lab
Practise An enzyme and its look-alike
Kinetics questions test whether you can turn a rate table or graph into a statement about how an enzyme or a drug works.
For many enzymes the initial rate $v$ depends on substrate concentration $[S]$ as
$$v = \frac{V_{max}[S]}{K_m + [S]}$$Three things to know:
A handy shortcut: $v/V_{max} = [S]/(K_m+[S])$, so $[S] = K_m$ gives 50%, $[S] = 3K_m$ gives 75% and $[S] = 9K_m$ gives 90% of $V_{max}$. Olympiad questions often add allosteric enzymes, which give an S-shaped (sigmoid) curve instead of a hyperbola.
An enzyme has $V_{max} = 120\ \mu\text{mol min}^{-1}$ and $K_m = 4$ mM. Find the rate at $[S] = 2$ mM and at $[S] = 12$ mM.
At 2 mM: $v = \frac{120 \times 2}{4 + 2} = 40\ \mu\text{mol min}^{-1}$.
At 12 mM: $v = \frac{120 \times 12}{4 + 12} = 90\ \mu\text{mol min}^{-1}$, which is 75% of $V_{max}$, as expected for $[S] = 3K_m$.
A Lineweaver-Burk line has $y$-intercept $0.02$ min $\mu$mol$^{-1}$ and $x$-intercept $-0.5$ mM$^{-1}$. With an inhibitor added, the $y$-intercept stays the same and the $x$-intercept moves to $-0.25$ mM$^{-1}$. Find $V_{max}$ and $K_m$, and name the type of inhibition.
$V_{max} = 1/0.02 = 50\ \mu\text{mol min}^{-1}$ and $K_m = 1/0.5 = 2$ mM. With inhibitor, $V_{max}$ is unchanged and apparent $K_m = 1/0.25 = 4$ mM, so the inhibition is competitive.
Going deeper: The enzyme kinetics chapter of a first-year university biochemistry textbook.
School covers glycolysis, the Krebs cycle and chemiosmosis; the olympiad adds ATP yield bookkeeping, redox carriers, and the effects of poisons and uncouplers on oxygen use and ATP output.
Learn it Glycolysis and the Krebs cycle (A Level) Oxidative phosphorylation and chemiosmosis (A Level) Energy, ATP and respiratory quotients (A Level) Anaerobic respiration (A Level) Energy and respiration lab
School covers replication, transcription, translation and the lac operon; the olympiad adds eukaryotic splicing and regulation, reading mutant operon data, and the genetic code in detail.
Learn it The structure of nucleic acids (A Level) Semi-conservative replication (A Level) Transcription and translation (A Level) Gene control (A Level) Nucleic acids and protein synthesis lab
School covers PCR, gel electrophoresis and recombinant DNA; the olympiad asks you to map restriction sites from fragment sizes and to read sequencing gels and blots.
Learn it Recombinant DNA (A Level) PCR and gel electrophoresis (A Level) Microarrays and bioinformatics (A Level) Genetic technology lab
Practise Mapping a plasmid by cutting it
Restriction maps and gels are puzzle-like data tasks that appear in almost every international paper.
A restriction enzyme cuts DNA at a specific short sequence. Gel electrophoresis then separates the pieces by size: DNA is negative, so it runs towards the positive electrode, and small fragments travel furthest. The distance moved is roughly a straight line against the logarithm of fragment size, so you can read sizes from a ladder of known fragments.
Rules for mapping:
In Sanger sequencing, each fragment ends with a labelled terminating base. Reading the gel from the smallest fragment (furthest from the well) to the largest gives the new strand from its 5' end to its 3' end. PCR doubles the target each cycle, so $n$ cycles give up to $2^n$ copies of each starting molecule.
A linear 10 kb fragment is cut by EcoRI into 3 kb and 7 kb pieces, and by BamHI into 4 kb and 6 kb pieces. Cutting with both gives 1 kb, 3 kb and 6 kb. Taking the EcoRI site to be 3 kb from the left end, where is the BamHI site?
BamHI is 4 kb from one end, so it is either 4 kb or 6 kb from the left end.
So the BamHI site is 4 kb from the left end, 1 kb to the right of the EcoRI site.
An 8 kb circular plasmid gives a single 8 kb band when cut with enzyme X alone, and a single 8 kb band with enzyme Y alone. Cutting with both gives bands at 3 kb and 5 kb. How many sites does each enzyme have, and how far apart are they?
One band from a circle means one cut, so X and Y each have one site. Two cuts in a circle give two pieces, so the sites are 3 kb apart going one way round the plasmid and 5 kb apart going the other way.
Going deeper: The DNA technology chapters of a molecular biology textbook, and past national olympiad papers that include gel data.
Bacterial growth curves and generation time, Gram staining, gene transfer between bacteria, and viral life cycles such as lytic, lysogenic and retroviral.
Learn it Prokaryotic cells and viruses (A Level) Why bacteria are used (IGCSE) Antibiotics: penicillin and resistance (A Level)
Microbes are quick to grow and easy to count, so they are a favourite setting for questions on growth, genetics and experimental design.
Growth. In a closed culture bacteria show a lag phase (adjusting, little division), a log or exponential phase (steady doubling), a stationary phase (nutrients run out, births equal deaths) and a death phase. In the log phase, with generation (doubling) time $g$:
$$N = N_0 \times 2^{t/g}$$Plot $\log N$ against time and the log phase is a straight line, which is how you read $g$ from data.
Gram stain. Gram-positive cells have a thick peptidoglycan wall that keeps the purple stain. Gram-negative cells have a thin wall inside an outer membrane and stain pink. The outer membrane is one reason some antibiotics do not reach Gram-negative cells.
Gene transfer. Bacteria swap genes without sex in three ways: transformation (taking up free DNA), transduction (a virus carries DNA from one cell to another) and conjugation (DNA, often a plasmid, passes through a pilus). This is how antibiotic resistance spreads.
Viruses. In the lytic cycle a phage replicates at once and bursts the cell. In the lysogenic cycle its DNA joins the host chromosome and is copied quietly with it until a trigger switches it to lytic. Retroviruses such as HIV carry RNA and use reverse transcriptase to make DNA, which is inserted into the host genome.
A culture starts with 500 bacteria, has no lag phase and a generation time of 20 minutes. How many cells are there after 2 hours of log-phase growth?
2 hours is 120 minutes, which is $120/20 = 6$ generations. $N = 500 \times 2^6 = 500 \times 64 = 32\,000$ cells.
During log-phase growth a culture rises from 1000 to 128 000 cells in 3.5 hours. What is the generation time?
$128\,000/1000 = 128 = 2^7$, so there were 7 generations in 210 minutes. Generation time $= 210/7 = 30$ minutes.
Going deeper: The bacterial growth and virology chapters of an introductory microbiology textbook.
School covers xylem, phloem and leaf structure; the olympiad adds meristems, secondary growth, and telling monocot and dicot roots, stems and leaves apart in sections.
Learn it Xylem and phloem: structure and distribution (A Level) Leaf structure (IGCSE) Xylem and phloem (IGCSE)
Water uptake and the Casparian strip, cohesion-tension, transpiration and xerophytes, and mass flow in the phloem.
Learn it From soil to xylem, and up the stem (A Level) Transpiration and xerophytic leaves (A Level) Translocation (A Level) Transpiration (IGCSE) Transport in plants lab
School covers the light-dependent stage, the Calvin cycle and limiting factors; the olympiad adds photorespiration, C4 and CAM pathways, and reading light-response curves.
Learn it The light-dependent stage (A Level) The Calvin cycle (A Level) Limiting factors and investigations (A Level) Photosynthesis lab
Practise Sun leaves and shade leaves
Comparing C3, C4 and CAM plants is the standard way olympiads test whether you understand why Rubisco limits photosynthesis.
Rubisco fixes CO2, but it can also bind O2. When it does, the plant makes a compound it must recycle, losing CO2 and ATP. This is photorespiration. It gets worse in hot, bright, dry conditions, when stomata close, CO2 inside the leaf falls and O2 builds up.
Light-response curves. Net CO2 uptake is negative in the dark (respiration only), crosses zero at the light compensation point, rises, then levels off at saturation. Gross photosynthesis = net uptake + respiration. Shade leaves have a lower compensation point and saturate at lower light than sun leaves.
A plant fixes CO2 at 20 units in air (21% O2) and at 30 units in 2% O2, all else equal. A second plant fixes 25 units in both. What fraction of the first plant's potential fixation is lost in air, and what type is each plant likely to be?
The first plant loses $(30 - 20)/30 = 1/3$, about 33%, to photorespiration in air, so it behaves as a C3 plant. The second plant is unaffected by O2, because its Rubisco sits in a high-CO2 space, so it is likely a C4 plant.
In the dark a leaf releases CO2 at 2 $\mu$mol m$^{-2}$ s$^{-1}$. In bright light its net uptake is 10 $\mu$mol m$^{-2}$ s$^{-1}$. Assuming respiration is the same in light, what is the gross rate of photosynthesis?
Gross = net + respiration = $10 + 2 = 12\ \mu$mol m$^{-2}$ s$^{-1}$.
Going deeper: The photosynthesis chapter of a university plant physiology textbook.
School covers auxin, gibberellin and ABA in a few responses; the olympiad expects all five main hormone groups, phytochrome, and predicting flowering from day and night lengths.
Learn it Control and coordination in plants (A Level) Tropic responses (IGCSE) Stomata: homeostasis in plants (A Level)
Light-break experiments on flowering are classic olympiad puzzles: one rule about night length solves them all.
The five hormone groups to know:
| Hormone | Main effects |
|---|---|
| Auxin | Cell elongation, phototropism, apical dominance, root initiation |
| Gibberellin | Stem elongation, breaking seed dormancy (switches on amylase in cereal seeds) |
| Cytokinin | Cell division, delays leaf ageing; with auxin it steers tissue culture (more auxin: roots, more cytokinin: shoots) |
| Abscisic acid | Stomatal closure in drought, seed dormancy |
| Ethylene | Fruit ripening, leaf fall |
Phytochrome is a light-sensing pigment with two forms. Red light (about 660 nm) converts Pr to Pfr; far-red light (about 730 nm) converts Pfr back to Pr. In the dark, Pfr slowly turns back into Pr. The last flash in a sequence decides the outcome.
Photoperiodism. What plants really measure is the length of the uninterrupted night. A short-day plant flowers when the night is longer than a critical length; a long-day plant flowers when the night is shorter than its critical length. A brief red flash in the middle of the night splits it into two short nights; a far-red flash straight after cancels the red one.
A short-day plant has a critical night length of 11 h. Does it flower under (a) 12 h light, 12 h dark; (b) 10 h light, 14 h dark with a red flash after 7 h of dark; (c) as (b) but the red flash is followed at once by far-red?
(a) Night 12 h is longer than 11 h: it flowers.
(b) The red flash splits the night into 7 h and 7 h. Neither is longer than 11 h: no flowering.
(c) Far-red came last, so the flash has no effect and the night counts as 14 h: it flowers.
A long-day plant has a critical night length of 9 h. It is grown with 8 h light and 16 h dark, with a red flash halfway through the dark period. Will it flower?
Yes. The red flash splits the night into two 8 h periods. The longest uninterrupted dark is 8 h, shorter than the 9 h critical night, so the long-day plant flowers.
Going deeper: The growth regulators and photomorphogenesis chapters of a plant physiology textbook.
Alternation of haploid and diploid generations in mosses, ferns and seed plants, the male and female gametophytes of flowering plants, and double fertilisation.
Learn it Flowers and pollination (IGCSE) Fertilisation and germination (IGCSE) Reproduction in plants lab
Counting chromosomes in different plant tissues is a quick way for an olympiad to test whether you really know a life cycle.
Every land plant alternates between two multicellular generations:
The groups differ in which generation is big and long-lived:
| Group | Dominant generation | Note |
|---|---|---|
| Mosses | Gametophyte | Sporophyte grows on it and depends on it |
| Ferns | Sporophyte | Small free-living gametophyte (prothallus); sperm swim in water |
| Seed plants | Sporophyte | Gametophytes tiny: pollen grain (male), embryo sac (female) |
Double fertilisation happens only in flowering plants. The pollen tube delivers two sperm nuclei. One fuses with the egg to give the $2n$ zygote (the embryo). The other fuses with the two polar nuclei of the central cell to give the $3n$ endosperm, the food store of the seed. The seed coat and fruit wall are maternal tissue, so they are $2n$ with the mother's genotype only.
A flowering plant has $2n = 24$. Give the chromosome number in (a) a pollen tube nucleus, (b) an endosperm cell, (c) an embryo cell, (d) a seed coat cell.
(a) 12, haploid male gametophyte. (b) 36, triploid from one sperm plus two polar nuclei ($12 \times 3$). (c) 24, diploid zygote. (d) 24, diploid maternal tissue.
A fern has $2n = 32$. How many chromosomes are in (a) a spore, (b) a prothallus cell, (c) a cell from the leafy frond?
(a) 16, spores come from meiosis. (b) 16, the prothallus is the haploid gametophyte. (c) 32, the frond is the diploid sporophyte.
Going deeper: The plant diversity and reproduction chapters of a general university biology textbook.
The parts of the gut, digestive enzymes and where they act, absorption in the small intestine, and a balanced diet.
Learn it The digestive system (IGCSE) Chemical digestion (IGCSE) Absorption (IGCSE) Human nutrition lab
School covers alveoli and the dissociation curve with the Bohr shift; the olympiad adds oxygen content calculations, fetal haemoglobin and myoglobin, and gas exchange in fish, insects and birds.
Learn it Alveoli and gas exchange (A Level) Transport of oxygen and carbon dioxide (A Level) Breathing and exercise (IGCSE) Gas exchange lab
Practise Oxygen bookkeeping in a cyclist
Physiology questions at olympiad level are often bookkeeping: how much oxygen is carried, delivered and used.
Dissociation curves. Haemoglobin's curve is S-shaped because binding one O2 makes the next bind more easily. $P_{50}$ is the partial pressure at 50% saturation. A curve shifted to the right (higher $P_{50}$) unloads more O2 in tissues: this is the Bohr shift, caused by more CO2, more acid or a higher temperature. Curves to the left grab O2 more tightly: fetal haemoglobin (takes O2 from the mother's blood), myoglobin (an O2 store in muscle, hyperbolic curve) and animals at high altitude.
Oxygen content. Each gram of fully saturated haemoglobin carries about 1.34 mL of O2. So
$$\text{O}_2\text{ content} = [\text{Hb}] \times 1.34 \times \text{saturation}$$and the O2 used by the body is cardiac output times the arterial minus venous content (the Fick principle). Dissolved O2 in plasma is small and usually ignored.
Other animals. Fish gills use countercurrent flow: water and blood run in opposite directions, so blood always meets water with more O2 and can pick up most of it. Insects pipe air straight to tissues in tracheae. Birds have air sacs that give one-way flow through rigid lungs.
Blood has 150 g L$^{-1}$ of haemoglobin. Arterial blood is 98% saturated and venous blood 70%. Cardiac output is 5 L min$^{-1}$. How much O2 do the tissues take up each minute?
Full capacity: $150 \times 1.34 = 201$ mL L$^{-1}$.
Arterial: $201 \times 0.98 \approx 197.0$ mL L$^{-1}$. Venous: $201 \times 0.70 = 140.7$ mL L$^{-1}$.
Difference: about 56.3 mL per litre. Uptake: $56.3 \times 5 \approx 281$ mL min$^{-1}$.
In hard exercise an athlete's cardiac output is 20 L min$^{-1}$ and the tissues remove 150 mL of O2 from each litre of blood. What is the O2 consumption?
$20 \times 150 = 3000$ mL min$^{-1}$, that is 3.0 L of O2 per minute.
Going deeper: The respiration and circulation chapters of an animal (comparative) physiology textbook.
Double circulation, blood vessels, the cardiac cycle and its pressure changes, and tissue fluid formation.
Learn it The heart and the cardiac cycle (A Level) A closed double circulation (A Level) Blood cells, water and tissue fluid (A Level) The heart (IGCSE) Transport in mammals lab
School covers ultrafiltration, reabsorption and ADH; the olympiad adds clearance and filtered-load calculations, the countercurrent multiplier, and how different animals excrete nitrogen.
Learn it The kidney: filtering and reclaiming (A Level) Osmoregulation and ADH (A Level) The kidney (IGCSE) Homeostasis lab
Practise What the kidney clears
Clearance turns urine and plasma numbers into a verdict on what the kidney does to a substance, which is ideal for data questions.
The clearance of a substance is the volume of plasma the kidneys clear of it per minute:
$$C = \frac{U \times V}{P}$$where $U$ is the urine concentration, $V$ the urine flow rate and $P$ the plasma concentration.
The loop of Henle is a countercurrent multiplier: the ascending limb pumps out salt but holds water, building a salt gradient in the medulla. Longer loops make a steeper gradient, so desert mammals make very concentrated urine.
Nitrogen waste. Ammonia is toxic but cheap; it needs lots of water (most fish). Urea is less toxic (mammals, adult amphibians). Uric acid is nearly insoluble and saves water (birds, reptiles, insects).
Plasma inulin is 0.5 mg mL$^{-1}$, urine inulin 60 mg mL$^{-1}$ and urine flow 1.0 mL min$^{-1}$. Substance X, which is freely filtered, has plasma 2 mg mL$^{-1}$ and urine 12 mg mL$^{-1}$. Find the GFR and say what happens to X.
GFR $= (60 \times 1.0)/0.5 = 120$ mL min$^{-1}$.
Clearance of X $= (12 \times 1.0)/2 = 6$ mL min$^{-1}$, far below 120, so X is reabsorbed.
Filtered load $= 120 \times 2 = 240$ mg min$^{-1}$; excreted $= 12$ mg min$^{-1}$; reabsorbed $= 228$ mg min$^{-1}$.
GFR is 125 mL min$^{-1}$ and the glucose transport maximum is 375 mg min$^{-1}$. Above what plasma glucose concentration will glucose start to appear in urine?
Glucose appears when the filtered load passes 375 mg min$^{-1}$: $P = 375/125 = 3.0$ mg mL$^{-1}$.
Going deeper: The kidney chapters of a human or comparative physiology textbook.
School covers resting and action potentials, synapses and the sliding filament model; the olympiad adds equilibrium potentials from ion concentrations, conduction speed and refractory periods.
Learn it Resting and action potentials (A Level) The cholinergic synapse (A Level) Striated muscle and contraction (A Level) Control and coordination lab
Practise Potentials across a nerve membrane
The Nernst equation links ion gradients to membrane potentials, and international papers expect you to use it.
If a membrane lets only one ion through, the potential settles where the electrical pull balances the concentration gradient. This is the equilibrium potential, given by the Nernst equation. At 37 °C it simplifies to
$$E_{ion} \approx \frac{61.5\text{ mV}}{z}\log_{10}\frac{[\text{ion}]_{out}}{[\text{ion}]_{in}}$$where $z$ is the charge on the ion ($+1$ for K+ and Na+, $-1$ for Cl-). The answer is the inside relative to the outside.
A neurone has K+ at 5 mM outside and 140 mM inside. Calculate $E_K$ at 37 °C.
$E_K = 61.5 \times \log_{10}(5/140) = 61.5 \times \log_{10}(0.0357)$.
$\log_{10}(0.0357) = -1.447$, so $E_K = 61.5 \times (-1.447) \approx -89$ mV.
Cl- is 110 mM outside and 10 mM inside a cell. Calculate $E_{Cl}$ at 37 °C.
$E_{Cl} = \frac{61.5}{-1}\log_{10}(110/10) = -61.5 \times 1 = -61.5$ mV.
Going deeper: The membrane potential chapter of a neuroscience or cell physiology textbook.
Negative feedback, blood glucose control, the menstrual cycle hormones, and the endocrine system compared with the nervous system.
Learn it Principles of homeostasis (A Level) Cell signalling and blood glucose (A Level) Nervous and endocrine systems (A Level) Sex hormones and the menstrual cycle (IGCSE)
Phagocytes, B and T lymphocytes, antibody structure, vaccination and the types of immunity.
Learn it Phagocytes and antigens (A Level) Lymphocytes and the immune response (A Level) Antibodies and monoclonal antibodies (A Level) Vaccination and types of immunity (A Level) Immunity lab
Cleavage, gastrulation and the three germ layers, induction by an organiser, and genes that set up the body plan.
Development links cell biology, genetics and evolution, and grafting experiments make good data questions.
After fertilisation the zygote divides quickly without growing (cleavage), making a ball of smaller cells that becomes a hollow blastula. In gastrulation cells move inwards through an opening, the blastopore, and form three germ layers:
| Germ layer | Gives rise to |
|---|---|
| Ectoderm (outer) | Epidermis, nervous system, lens of the eye |
| Mesoderm (middle) | Muscle, bone, blood, heart, kidneys, gonads |
| Endoderm (inner) | Lining of the gut and lungs, liver, pancreas |
Induction. One group of cells can signal to its neighbours and change their fate. In amphibians the dorsal lip of the blastopore acts as an organiser: grafted onto another embryo it can induce a second body axis. The mesoderm under the ectoderm induces it to form the neural tube.
Body-plan genes. Gradients of signalling molecules tell cells where they are. Hox genes are switched on in order along the head-to-tail axis and give each region its identity. A mutation can turn one body part into another, for example legs growing where antennae should be. Programmed cell death (apoptosis) sculpts structures such as the gaps between fingers.
The dorsal lip of the blastopore is cut from one newt embryo and grafted onto the opposite (ventral) side of a second embryo. The host grows a second, partly complete body axis, made mostly of host cells. What does this show?
The grafted tissue did not build the new axis itself, since most of it is host tissue. It sent signals that changed what the host cells became. So the dorsal lip is an organiser that induces neighbouring cells to form axial structures.
Name the germ layer that gives rise to (a) the lens of the eye, (b) the liver, (c) the kidney.
(a) Ectoderm. (b) Endoderm. (c) Mesoderm.
Going deeper: The development chapter of a general university biology textbook, then an introductory developmental biology text.
Tinbergen's four questions, fixed action patterns and sign stimuli, taxes and kineses, and the main kinds of learning.
Behaviour questions reward clear definitions and careful reading of simple experiments, which you can learn quickly.
Four questions. Any behaviour can be explained in four ways: its mechanism (what triggers it), its development (how it appears in the life of the animal), its function (how it helps survival or reproduction) and its evolution (how it arose). Many questions ask which kind of explanation a statement gives.
Innate behaviour needs no learning:
Learning:
Woodlice in a choice chamber move at 12 mm s$^{-1}$ with 8 turns per minute on the dry side, and at 4 mm s$^{-1}$ with 2 turns per minute on the humid side. The chamber gives no gradient of direction. After 20 minutes most are on the humid side. Is this a taxis or a kinesis?
A kinesis. The woodlice do not head towards the humid side. They simply move fast in dry air and slow down in humid air, so they spend longer, and pile up, in the humid half. The response depends on intensity, not direction.
A sea anemone stops pulling in its tentacles after being touched gently 20 times. A sharp touch then makes it pull them in at once. Name the type of behaviour, and explain why the sharp touch shows it is not muscle fatigue.
Habituation. The anemone can still contract fully when the stimulus changes, so its muscles are not tired: it has learned to ignore the harmless repeated touch.
Going deeper: The opening chapters of an introductory animal behaviour textbook.
Optimal foraging, prey choice from energy and handling time, and the costs and benefits of defending a territory.
Olympiads like cost-benefit questions where a short calculation predicts what an animal should do.
Behavioural ecology assumes natural selection favours behaviour that gives the best balance of benefits and costs. For feeding, the measure is usually net energy gained per unit time.
Prey choice. Each prey type has an energy value $E$ and a handling time $h$ (catching, opening and eating it). Its profitability is $E/h$. A predator should always take the most profitable prey. It should also take a less profitable prey type when met only if that prey's $E/h$ is higher than the overall rate it gets by ignoring it and searching for the better prey:
$$\text{rate (best prey only)} = \frac{E_1}{S_1 + h_1}$$where $S_1$ is the average search time to find one item of the best prey. The surprising result: whether to eat the poorer prey depends on how common the better prey is, not on how common the poorer prey is.
Patches. When food in a patch runs down, an animal should leave once its intake rate falls to the average rate for the whole habitat. Longer travel between patches means staying longer in each.
Territories. Defending an area costs energy, time and injury risk. A territory is worth defending only if the extra food or mates it secures outweigh these costs. Defence is not worth it when food is very scarce (not enough to gain) or very plentiful (no need to defend).
Prey A gives 60 J and takes 10 s to handle; prey B gives 20 J and takes 5 s. Finding an A takes 20 s on average. When the predator meets a B, should it eat it?
Profitabilities: A is $60/10 = 6$ J s$^{-1}$, B is $20/5 = 4$ J s$^{-1}$.
Rate from eating only A: $60/(20 + 10) = 2$ J s$^{-1}$.
B's profitability, 4 J s$^{-1}$, beats 2 J s$^{-1}$, so the predator should eat B when it meets one.
Prey A becomes common, so the search time for A falls to 2 s. Should the predator still eat B?
No. The rate from A alone is now $60/(2 + 10) = 5$ J s$^{-1}$, higher than B's 4 J s$^{-1}$, so it should ignore B and wait for A.
Going deeper: The foraging and territory chapters of a behavioural ecology textbook.
Coefficients of relatedness, Hamilton's rule, eusocial insects and haplodiploidy, reciprocal altruism and dominance hierarchies.
Altruism looks like a puzzle for natural selection, and Hamilton's rule gives a short numerical answer that olympiads love.
An altruistic act lowers the actor's own reproduction but helps another individual. It can still spread if it helps relatives, who share copies of the same genes. Hamilton's rule says a gene for altruism is favoured when
$$rB \gt C$$where $r$ is the relatedness of actor to recipient, $B$ the benefit to the recipient and $C$ the cost to the actor, both measured in offspring.
| Relationship (diploid) | $r$ |
|---|---|
| Parent and offspring | 0.5 |
| Full siblings | 0.5 |
| Half siblings, grandparent, uncle or aunt | 0.25 |
| First cousins | 0.125 |
Haplodiploidy. In ants, bees and wasps, males come from unfertilised eggs and are haploid. Sisters get an identical set of genes from their father, so full sisters have $r = 0.75$, more than a mother and her daughter (0.5). This may help explain why sterile female workers raise sisters. A female's relatedness to her brother is only 0.25.
Other routes to cooperation: reciprocal altruism (helping those who help you back, which needs repeated meetings and memory of cheats) and dominance hierarchies, which cut the cost of constant fighting.
A ground squirrel's alarm call costs the caller 0.2 offspring on average. Each full sibling nearby gains 0.1 offspring. Is calling favoured with 3 siblings nearby? With 5?
With 3: $rB = 0.5 \times (3 \times 0.1) = 0.15$, which is less than $C = 0.2$, so calling is not favoured.
With 5: $rB = 0.5 \times 0.5 = 0.25$, which is more than 0.2, so calling is favoured.
A honeybee worker could, in principle, raise either a daughter of her own or a full sister. Using relatedness, which is better for her genes, and how related is she to a brother?
A full sister ($r = 0.75$) is better than a daughter ($r = 0.5$). She is related to a brother by only $r = 0.25$.
Going deeper: The social behaviour chapter of a behavioural ecology textbook.
Signal channels and honest signals, the honeybee dance, sexual selection by competition and choice, and how parental care shapes mating systems.
Sexual selection and signalling explain many odd traits, and the bee dance is a favourite decoding task.
Signals can be visual, acoustic, chemical (pheromones) or by touch. Each channel suits some settings: sound travels in the dark and round obstacles, pheromones last after the sender leaves. A signal stays honest when it is costly to fake; for example, a bright tail that only a healthy male can afford (the handicap idea).
The honeybee waggle dance. On the vertical comb, the angle of the waggle run from straight up equals the angle of the food from the direction of the sun, measured the same way round. The length of the waggle run increases with distance. A dance straight down means fly directly away from the sun.
Sexual selection. Eggs are large and few, sperm small and many (anisogamy), so the sex that invests more, usually females, tends to be choosy. Intrasexual selection is competition within one sex (antlers, fights); intersexual selection is choice by the other sex (displays, songs).
Mating systems follow parental care and resources: monogamy is common where two parents are needed (many birds); polygyny (one male, several females) where a male can control resources or females; polyandry (one female, several males) where males care for the young, as in some wading birds.
A forager's waggle run points 40° clockwise from vertical and lasts 1.5 s. The sun's compass bearing is 120°, and for this colony 1 s of waggling means about 1 km. Where is the food?
The food is 40° clockwise from the sun's direction: bearing $120 + 40 = 160$°. Distance about $1.5 \times 1 = 1.5$ km.
A bee dances with its waggle run pointing straight down the comb. In which direction is the food?
Directly away from the sun, 180° from the sun's bearing.
Going deeper: The communication and sexual selection chapters of an animal behaviour textbook.
Chromosome behaviour in mitosis and meiosis, how meiosis makes variation, and DNA content at each stage.
Learn it Chromosome behaviour in mitosis (A Level) Meiosis and genetic variation (A Level) Mitosis (IGCSE) The mitotic cell cycle lab
Monohybrid and dihybrid crosses, codominance, multiple alleles, sex linkage, epistasis and the chi-squared test.
Learn it Monohybrid and dihybrid crosses (A Level) The chi-squared test (A Level) Monohybrid crosses (IGCSE) Inheritance lab
School covers drawing and reading pedigrees; the olympiad asks for exact risks, such as the 2/3 carrier chance of an unaffected sibling and multi-generation X-linked risks.
Learn it Pedigrees, codominance and sex linkage (IGCSE) Monohybrid and dihybrid crosses (A Level)
Genetic counselling questions test probability as much as biology, and one missed fraction ruins the answer.
Two rules: multiply probabilities of independent events that must all happen; add probabilities of separate ways the same outcome can happen.
The 2/3 rule. Two unaffected carrier parents ($Aa \times Aa$) have children in the ratio 1 $AA$ : 2 $Aa$ : 1 $aa$. If you already know a child is unaffected, the $aa$ quarter is ruled out, so the chance that child is a carrier is $2/3$, not $1/2$.
Steps for a risk question:
X-linked recessive. A carrier mother passes the allele to half her children, so each son has a $1/2$ chance of being affected and each daughter a $1/2$ chance of being a carrier. An affected father passes it to every daughter and to no son.
Harder questions update risks when extra facts are known (for example, several unaffected sons lower the chance the mother is a carrier). That uses Bayes' theorem; learn it after these basics.
Ana's brother has an autosomal recessive disorder; her parents are unaffected and Ana is unaffected. She has a child with an unrelated man whose chance of being a carrier is $1/30$. What is the chance the child is affected?
Both of Ana's parents must be carriers, and Ana is unaffected, so P(Ana carrier) $= 2/3$.
P(child affected) $= \frac{2}{3} \times \frac{1}{30} \times \frac{1}{4} = \frac{2}{360} = \frac{1}{180}$.
A woman's maternal uncle has an X-linked recessive disorder. Her grandfather, her mother and she are unaffected, and nothing else is known. What is the chance her first son is affected?
Her grandmother must be a carrier (the uncle got his X from her). P(mother carrier) $= 1/2$, P(woman carrier) $= 1/2 \times 1/2 = 1/4$, P(son affected) $= 1/4 \times 1/2 = 1/8$.
Going deeper: The human genetics chapter of a genetics textbook, especially its worked problems on pedigrees.
School covers linkage and crossing over in words; the olympiad wants recombination frequencies, map distances and gene order from test-cross data.
Learn it Monohybrid and dihybrid crosses (A Level) Meiosis and genetic variation (A Level)
Practise Body colour and wing shape in fruit flies
Mapping from test crosses is one of the most common calculation types in olympiad genetics.
Genes on the same chromosome tend to be inherited together (linked). Crossing over in meiosis can separate them, making recombinant gametes. The further apart two genes are, the more often a crossover falls between them.
Cross a double heterozygote with a double recessive (a test cross). Every offspring shows the alleles of the heterozygote's gamete, so you can count gamete types directly.
Three-point crosses. With three linked genes there are eight classes. The largest pair is parental; the smallest pair are double crossovers. Compare a double crossover with the parental type: the gene that has swapped is the one in the middle. Then find each pair's distance; for genes far apart, the sum of the two short distances is more accurate than the direct count, because double crossovers hide.
A heterozygous $AaBb$ fly, from true-breeding $AABB \times aabb$ parents, is test-crossed. The offspring are: AB 410, ab 395, Ab 92, aB 103. Are the genes linked, and how far apart are they?
The parental classes (AB and ab) are far more common than the others, so the genes are linked. Recombinants: $92 + 103 = 195$ out of $1000$, so RF $= 19.5\%$ and the genes are 19.5 cM apart.
Genes P and Q are 12 cM apart. A $PQ/pq$ individual is test-crossed with $pq/pq$ and gives 500 offspring. How many do you expect to show the Pq phenotype?
12% of 500 = 60 recombinants, split equally between Pq and pQ, so about 30 Pq offspring.
Going deeper: The linkage and mapping chapters of an introductory genetics textbook.
Deletions, duplications, inversions and translocations, non-disjunction in meiosis I and II, sex chromosome aneuploidy, and polyploidy in plant evolution.
Learn it Meiosis and genetic variation (A Level) Gene mutations (A Level)
Chromosome changes link meiosis, human genetics and speciation, so they turn up across several kinds of question.
Changes in structure: a deletion loses a segment; a duplication copies one; an inversion flips a segment end to end; a translocation moves a segment to a different chromosome.
Changes in number (aneuploidy) usually come from non-disjunction, when chromosomes fail to separate:
Examples: trisomy 21 (Down syndrome), XO (Turner syndrome, female), XXY (Klinefelter syndrome, male). In mammals all X chromosomes but one are inactivated as Barr bodies, so the number of Barr bodies is the number of X chromosomes minus one.
Polyploidy means whole extra sets. A hybrid of two species is often sterile, because its chromosomes have no partners to pair with in meiosis. If its chromosome number doubles (allopolyploidy), every chromosome has a partner, the plant becomes fertile, and it is at once a new species, unable to breed with either parent. Bread wheat arose this way.
Species A has $2n = 14$ and species B has $2n = 18$. They form a hybrid. How many chromosomes does the hybrid have, why is it sterile, and what happens if its chromosome number doubles?
The hybrid gets $7 + 9 = 16$ chromosomes. These are two unmatched sets, so they cannot pair in meiosis and gametes fail. After doubling it has 32 chromosomes, every one with a partner; meiosis works and it makes gametes with 16. It is fertile but isolated from both parents.
How many Barr bodies are found in the cells of (a) an XXY man, (b) an XXX woman, (c) an XO woman?
(a) 1. (b) 2. (c) 0.
Going deeper: The chromosome mutation chapter of an introductory genetics textbook.
School covers the Hardy-Weinberg equation; the olympiad adds X-linked frequencies, selection against a recessive allele, drift, inbreeding and heterozygote advantage.
Learn it The Hardy-Weinberg principle (A Level) Natural selection, drift and resistance (A Level) Selection and evolution lab
Practise Selection on an island population
Allele-frequency calculations are where evolution becomes quantitative, and they appear at every olympiad tier.
In Hardy-Weinberg equilibrium (large population, random mating, no selection, migration or mutation) the allele frequencies $p + q = 1$ give genotype frequencies $p^2 + 2pq + q^2 = 1$. Going beyond school:
In a population, 8% of males have an X-linked recessive form of colour blindness. Assuming equilibrium, what percentage of females are colour-blind, and what percentage are carriers?
Males: $q = 0.08$, so $p = 0.92$.
Colour-blind females: $q^2 = 0.0064$, that is 0.64%.
Carrier females: $2pq = 2 \times 0.92 \times 0.08 = 0.1472$, about 14.7%.
A recessive lethal allele has frequency 0.1, and every $aa$ individual dies before breeding. What is its frequency in the next generation?
$q_1 = \frac{0.1}{1 + 0.1} = \frac{0.1}{1.1} \approx 0.091$.
Going deeper: The population genetics chapter of an evolution or genetics textbook.
Variation, natural and artificial selection, evidence for evolution, and allopatric and sympatric speciation.
Learn it Evolution and speciation (A Level) Selective breeding (A Level) Natural selection (IGCSE) Adaptive features (IGCSE)
School covers food chains, pyramids and energy loss; the olympiad adds gross and net primary productivity, transfer and assimilation efficiencies, and comparing ecosystems.
Learn it Energy flow (IGCSE) Food chains, webs and pyramids (IGCSE) Organisms and their environment lab
The carbon and nitrogen cycles, the microbes that drive them, and how human activity changes them.
School covers the sigmoid growth curve; the olympiad adds exponential and logistic equations, carrying capacity, r and K strategies, survivorship curves and life tables.
Learn it Populations (IGCSE)
Practise A life table for a mountain lizard
Population equations and life tables give ecology questions their numbers, and they are short to learn.
Exponential growth. With unlimited resources, $\frac{dN}{dt} = rN$, where $r$ is the per-capita growth rate. The population doubles in a fixed time.
Logistic growth. Resources run short as $N$ nears the carrying capacity $K$:
$$\frac{dN}{dt} = rN\,\frac{K - N}{K}$$This gives the S-shaped curve. Growth is fastest at $N = K/2$, which is why harvesting a population at about half its carrying capacity gives the largest sustainable yield.
Life strategies. r-selected species breed early and make many small offspring with little care (good in unstable habitats). K-selected species breed later with few, well-cared-for offspring (stable, crowded habitats).
Life tables. $l_x$ is the fraction of newborns surviving to age $x$; $m_x$ is the average number of daughters produced at age $x$. The net reproductive rate
$$R_0 = \sum l_x m_x$$is the average number of daughters per female over her life. $R_0 \gt 1$ means the population grows, $R_0 \lt 1$ means it shrinks. Plotting $\log l_x$ against age gives survivorship curves: type I (most die old, as in humans), type II (steady risk, as in many birds), type III (most die young, as in oysters).
A population grows logistically with $r = 0.4$ per year and $K = 1000$. Find the growth rate when $N = 500$ and when $N = 900$.
At 500: $0.4 \times 500 \times \frac{500}{1000} = 100$ per year.
At 900: $0.4 \times 900 \times \frac{100}{1000} = 36$ per year. Growth slows as $N$ nears $K$.
A short-lived lizard has $l_0 = 1.0$, $l_1 = 0.4$, $l_2 = 0.2$ and none survive past age 2. It has $m_0 = 0$, $m_1 = 2$ and $m_2 = 3$. Find $R_0$. Is the population growing?
$R_0 = 1.0 \times 0 + 0.4 \times 2 + 0.2 \times 3 = 0 + 0.8 + 0.6 = 1.4$. This is above 1, so the population is growing.
Going deeper: The population ecology chapters of an ecology textbook.
Quadrats, transects, mark-release-recapture, Simpson's diversity index and correlation in field data.
Learn it Sampling and measuring biodiversity (A Level) Transects and correlation (A Level) Classification, biodiversity and conservation lab
Practise Counting fish and weighing diversity
Fundamental and realised niches, competitive exclusion and resource partitioning, predator-prey cycles, and mutualism, commensalism and parasitism.
Practise Fungi, fertiliser and a fair test
Field experiments that remove one species are a standard olympiad data type, and they all turn on the idea of a niche.
Interactions by effect. Mark each species $+$, $-$ or $0$:
| Interaction | Signs | Example |
|---|---|---|
| Competition | $-/-$ | Two plants sharing light |
| Predation, parasitism, herbivory | $+/-$ | Fox and rabbit, tick and deer |
| Mutualism | $+/+$ | Mycorrhizal fungi and plant roots |
| Commensalism | $+/0$ | An epiphyte growing on a tree branch |
Niches. The fundamental niche is the full range of conditions and resources a species could use alone. The realised niche is the smaller part it actually uses when competitors and predators are present. Removing a competitor and seeing a species spread shows the difference.
Competitive exclusion. Two species that need exactly the same limiting resource cannot coexist for long; one is pushed out. Species coexist by resource partitioning, for example feeding at different heights, and may evolve to differ more where they overlap (character displacement).
Predator-prey cycles. Prey numbers rise, then predators rise with a time lag, prey fall, then predators fall. On a graph the predator peaks come a little after the prey peaks.
A keystone species has an effect far larger than its numbers suggest; removing it can change the whole community.
A barnacle species S lives on a rocky shore from 0.5 m to 1.5 m above low tide. Species T lives below it. When T is removed, S settles and survives from 0 m to 1.5 m. Describe S's fundamental and realised niches on this shore.
Fundamental niche: 0 m to 1.5 m, the range S can live in without T. Realised niche: 0.5 m to 1.5 m, the range it occupies when T is present. Competition from T keeps S out of the lowest zone.
Classify each pair by signs: (a) a cleaner fish removing parasites from a larger fish, (b) a tapeworm in a dog's gut, (c) an orchid growing on a tree branch without harming it.
(a) $+/+$, mutualism. (b) $+/-$, parasitism. (c) $+/0$, commensalism.
Going deeper: The community ecology chapters of an ecology textbook.
Primary and secondary succession and how communities change during it, and the main world biomes set by temperature and rainfall.
Succession data (species and biomass over time) are a common olympiad graph, and biomes give the context for adaptation questions.
Succession is the change in a community over time.
As succession goes on, soil deepens, total biomass and the amount of organic matter rise, and the community becomes more stable. Species richness usually rises, often peaks in the middle stages, then may fall as a few dominant species (often tall trees) shade out others. The final stable community is the climax, set mainly by climate.
Biomes are large regions with similar climate and vegetation:
| Biome | Climate |
|---|---|
| Tropical rainforest | Hot and wet all year |
| Savanna | Hot, with a long dry season |
| Desert | Very dry, hot or cold |
| Temperate deciduous forest | Warm summers, cold winters, rain all year |
| Boreal forest (taiga) | Long cold winters, short summers |
| Tundra | Very cold, frozen subsoil, low growth |
In an abandoned field, species richness was 8, 25, 40 and 32 species at 0, 5, 20 and 60 years, and plant biomass was 0.5, 3, 20 and 120 t ha$^{-1}$. Explain the two trends.
Biomass rises throughout, because larger, longer-lived plants (shrubs, then trees) take over and store more organic matter. Richness rises as more species arrive and soil improves, peaks around year 20 when early and late species overlap, then falls as tall trees shade out light-demanding plants.
Which shows primary succession, and which recovers faster: plants colonising a new lava flow, or a forest regrowing after a fire?
The lava flow is primary succession (no soil). The burnt forest is secondary succession and recovers faster, because soil, seeds and roots survive.
Going deeper: The ecosystems and biomes chapters of an ecology textbook.
Habitat loss, pollution, extinction risks, and in-situ and ex-situ conservation.
Learn it Extinction and conservation (A Level) Conservation (IGCSE) Pollution (IGCSE) Human influences on ecosystems lab
Domains and kingdoms, the binomial system, the features of the main groups and using dichotomous keys.
Learn it Species, domains and kingdoms (A Level) Classification and dichotomous keys (IGCSE) Features of organisms (IGCSE) Characteristics and classification lab
Reading trees, clades and paraphyletic groups, shared derived characters, choosing a tree by parsimony, and molecular clocks.
Practise Counting steps on three trees
IBO practicals and theory papers often hand you a character table and ask you to build or judge a tree.
A phylogenetic tree shows how species are related by common descent. Each branch point (node) is a common ancestor. Rotating branches around a node does not change the tree; only the branching order matters.
Parsimony. For each possible tree, count the fewest changes of character state needed to explain the data. The tree needing the fewest changes is preferred. Changes that happen more than once (convergence) or reverse add steps.
Molecular clocks. If DNA changes build up at a roughly steady rate, the number of differences between two species estimates how long ago they split, once the clock is calibrated with fossils.
Four characters (0 = ancestral, 1 = derived) are scored: outgroup O 0000, A 1100, B 1110, C 1001. Count the steps for each rooted tree ((A,B),C), ((A,C),B) and ((B,C),A), and pick the most parsimonious.
Character 1 (1 in A, B, C): 1 step on every tree. Character 3 (only B) and character 4 (only C): 1 step each on every tree. Character 2 (A and B): 1 step on ((A,B),C) but 2 steps on the other two.
Totals: ((A,B),C) = 4 steps, the other two = 5 steps. So ((A,B),C) is most parsimonious. Only character 2, shared and derived, decided it.
The traditional group 'reptiles' includes lizards, snakes, turtles and crocodiles but leaves out birds, which share a common ancestor with crocodiles. Is 'reptiles' in this sense a clade?
No. It contains the common ancestor but not all its descendants (birds are left out), so it is paraphyletic.
Going deeper: The phylogeny chapter of an evolution textbook, then practice building trees from character tables.
The main animal phyla, identified by symmetry, germ layers, body cavity, protostome or deuterostome development, segmentation and chordate features.
Learn it Features of organisms (IGCSE)
Identifying an animal from its body plan is tested in both theory and the practical exams.
Sort animals with a few body-plan questions:
| Phylum | Key features |
|---|---|
| Cnidaria | Radial, two layers, stinging cells, one gut opening |
| Platyhelminthes | Flat, bilateral, no coelom, one gut opening |
| Nematoda | Round, unsegmented, pseudocoelom, gut with two openings |
| Annelida | Segmented worms, true coelom |
| Mollusca | Muscular foot, mantle, often a shell |
| Arthropoda | Jointed legs, chitin exoskeleton, segmented |
| Echinodermata | Five-part radial adults, bilateral larvae, tube feet |
| Chordata | Notochord, dorsal hollow nerve cord, pharyngeal slits, tail behind the anus |
Arthropods are the largest phylum, so learn their classes: insects (six legs, three body parts, often wings), arachnids (eight legs, two body parts), crustaceans (two pairs of antennae, mostly aquatic) and myriapods (many legs on many segments). Vertebrates are the chordates with a backbone.
An animal is bilateral and triploblastic, has no body cavity, is flattened, and its gut has only one opening. Which phylum is it in?
Platyhelminthes (flatworms): bilateral and triploblastic but acoelomate, with a single gut opening.
Adult starfish are radially symmetrical, yet starfish are placed closer to chordates than to molluscs. Give two reasons.
They are deuterostomes (the blastopore becomes the anus), like chordates; and their larvae are bilaterally symmetrical, so the radial adult form evolved later.
Going deeper: The animal diversity chapters of a general university biology textbook, with a zoology lab manual for dissection.
The four main plant groups, telling monocots from dicots, and the features of fungi, lichens and the main protist groups.
Learn it Features of organisms (IGCSE)
Practical exams often ask you to sort real specimens, so you need quick features for each group.
Plants fall into four main groups:
| Group | Vascular tissue | Seeds | Example |
|---|---|---|---|
| Bryophytes | No | No (spores) | Mosses, liverworts |
| Ferns and allies | Yes | No (spores) | Ferns, horsetails |
| Gymnosperms | Yes | Naked, often in cones | Pines, cycads |
| Angiosperms | Yes | Enclosed in a fruit | Flowering plants |
Monocots and dicots. Monocots have one seed leaf, parallel leaf veins, flower parts in threes, scattered vascular bundles in the stem and fibrous roots (grasses, lilies). Dicots (eudicots) have two seed leaves, net veins, flower parts in fours or fives, vascular bundles in a ring and often a taproot (beans, roses).
Fungi are heterotrophs with cell walls of chitin. They grow as threads (hyphae) forming a mycelium, digest food outside the body and absorb it, and spread by spores. Yeasts are single-celled fungi. A lichen is a partnership of a fungus with an alga or cyanobacterium.
Protists are the eukaryotes that are not plants, animals or fungi: algae, amoebae, ciliates such as Paramecium, and parasites such as Plasmodium. They are not one clade.
A plant has parallel leaf veins, flowers with six petals, and vascular bundles scattered through the stem. Is it a monocot or a dicot?
A monocot: parallel veins, flower parts in multiples of three (six petals) and scattered bundles all point the same way.
An organism has walls made of chitin, no chlorophyll, grows as threads through rotting wood and releases spores. Which kingdom?
Fungi.
Going deeper: The plant and fungal diversity chapters of a general university biology textbook, and a field guide for practice.