Cambridge IGCSE / A-Level
IGCSE Biology 0610 Paper 4: Structure, Question Styles and Technique
Paper 4 is the Extended theory paper for IGCSE Biology 0610: 1 hour 15 minutes, 80 marks, 50 percent of your final grade. Structured questions cover the whole Extended syllabus, mixing recall, data interpretation and longer explanations. Transport, enzymes, coordination, inheritance and ecology are the areas I see most often in past sessions.
Biology Paper 4 has the most content behind it of any of my IGCSE sciences: twenty one topics, and any of them can appear. What took me longest to learn is that Paper 4 is not really a memory test. Most marks come from using the right technical terms in the right order, reading data honestly, and answering the question that was actually asked. These are my notes on how the paper is put together, which topics dominate it, and three original exam-style questions with worked answers, plus the wording habits that took my explain answers from half marks to full marks.
How Paper 4 is structured
Paper 4 is 1 hour 15 minutes, 80 marks, and 50 percent of the qualification, examining the full Extended syllabus of Core plus Supplement content. In the Extended route it sits alongside Paper 2, the 40 mark multiple choice paper worth 30 percent, and Paper 5 or 6 for practical skills at 20 percent. Expect around six long structured questions, each roughly 10 to 15 marks, and each usually built around a figure: a diagram to label or interpret, a graph, a table of experimental results, or a genetic cross. Parts escalate from 1 mark recall to explain parts worth 3 to 5 marks, and there is often an extended writing part late in the paper, frequently on something synoptic like homeostasis or a nutrient cycle. Nothing is optional and there is no question choice, so revision has to cover the whole syllabus rather than betting on topics. The mark in brackets is the contract: a 3 mark explain needs three distinct biological points, not one point written three ways.
Topic weighting: what shows up most
With twenty one topics and only 80 marks, not everything appears each session, but patterns hold. Transport in plants and in humans is close to guaranteed: xylem and phloem, translocation, the heart, and blood vessels. Enzymes appear either directly or inside digestion, respiration or biotechnology questions, always with the active site model and the effect of temperature and pH. Coordination and response is a Paper 4 favourite because so much of it is Supplement: reflex arcs, synapses, hormones and homeostasis with negative feedback. Inheritance brings the reliable genetic cross worth 4 to 6 marks, plus codominance and sex linkage on the Extended side. Ecology and human influences have grown in weight, with food webs, nutrient cycles and eutrophication turning up regularly. Photosynthesis and respiration underpin the data questions about limiting factors and gas exchange. My honest advice from practice: master the figures. The heart, the eye, the kidney and nephron, the villus and the reflex arc are drawn, labelled or interpreted constantly.
Question styles and command words
Describe means say what the data or process shows; explain means give the biological reason, and muddling the two is the most common way to lose marks on this paper. In data questions, quote figures with units and name the trend before explaining it: comparative words like increases, faster and higher score, while vague words like changes do not. Suggest signals an unfamiliar context, often a strange organism or experiment, and the examiner wants known biology applied to it, so a sensible application scores even if you have never met the organism. Genetic crosses are almost a ritual: parental genotypes, gametes clearly shown, a completed Punnett square, then phenotypes and their ratio, and every one of those stages carries a mark. Calculations are few but predictable: magnification, percentage change, and rates read from graphs. For magnification, image size divided by actual size, with both in the same units, is worth checking twice, because unit conversion is where those marks die. Extended response parts are credited point by point, so write short separate sentences.
Wording habits that earn the explain marks
Biology mark schemes are lists of accepted phrases, and the fastest improvement I made was learning to write in those phrases. Water moves by osmosis from a dilute solution to a more concentrated solution through a partially permeable membrane: that sentence has three scoring ideas in it, and each is a syllabus phrase. Active transport needs movement against a concentration gradient using energy from respiration, and missing either half loses the mark. In enzyme answers, denatured means the active site changes shape so the substrate no longer fits; enzymes are never killed. In homeostasis, name the receptor, the change detected, the effector and the corrective response, in that order. And in any question about exchange surfaces, the trio of large surface area, short diffusion distance, and a steep concentration gradient maintained by blood supply or ventilation covers most sites in the syllabus. None of this is extra content. It is the same biology, written the way the examiner is instructed to reward.
Worked questions, step by step
Cylinders of potato of equal mass were placed in sucrose solutions of different concentrations for two hours. The percentage changes in mass were: 0.0 mol/dm3, +12 percent; 0.2 mol/dm3, +4 percent; 0.4 mol/dm3, -3 percent; 0.6 mol/dm3, -10 percent. (a) Explain the result at 0.0 mol/dm3. (b) Estimate the concentration of the potato cell sap and explain your reasoning. [5]
- (a) In distilled water the external solution is more dilute than the cell sap, so water moves into the cells by osmosis through the partially permeable cell membranes, increasing the mass.
- Mention all three elements: osmosis, the direction in terms of concentration, and the partially permeable membrane. That is usually the shape of the 3 marks.
- (b) The cell sap concentration is the point where the potato neither gains nor loses mass, because the concentrations inside and outside are equal.
- The change in mass crosses zero between 0.2 and 0.4 mol/dm3, so a sensible estimate from these numbers is about 0.3 mol/dm3.
- Justify with the data: at 0.2 the potato still gains mass and at 0.4 it loses mass, so the balance point must lie between them.
Answer: (a) Water enters the cells by osmosis, from the more dilute external solution to the more concentrated sap, through partially permeable membranes. (b) About 0.3 mol/dm3, where percentage change in mass would be zero.
Where marks slip: Estimate questions want you to interpolate and say why. Quoting the two data points either side of zero is what turns a guess into a scoring answer.
Try one yourself: Sketch the graph of these results, then predict and explain the appearance of the cells at 0.6 mol/dm3 using the terms plasmolysis and turgor.
In mice, black coat (B) is dominant to brown coat (b). Two black mice are crossed and produce eight offspring: six black and two brown. (a) State the genotypes of the parents and explain how you know. (b) Draw a Punnett square for the cross and state the expected ratio of phenotypes. (c) Suggest why the observed numbers do not exactly match the expected ratio. [6]
- (a) Brown offspring must be bb, so each parent must carry one b allele. Both parents show black coats, so both are Bb, heterozygous.
- (b) Gametes from each parent are B or b. The Punnett square gives the genotypes BB, Bb, Bb and bb.
- Phenotypes: BB and both Bb mice are black, bb is brown, so the expected ratio is 3 black to 1 brown.
- (c) Fertilisation is random, and eight offspring is a small sample, so chance produces deviations from the expected 3 to 1 ratio.
Answer: (a) Both parents are Bb; the brown (bb) offspring must have received one b allele from each parent. (b) 3 black to 1 brown. (c) Random fertilisation and a small sample size.
Where marks slip: Set out genotypes, gametes, offspring genotypes and phenotypes as separate labelled lines. Each line scores, and examiners cannot award what they cannot find.
Try one yourself: Try the follow-up: one brown offspring is crossed with a heterozygous black mouse. Predict the ratio, then rewrite the whole cross for a codominant case such as red, white and roan cattle.
After a meal, a healthy person's blood glucose concentration rises and then returns to normal within about two hours. (a) Name the organ that detects the rise and the hormone it releases. (b) Explain how this hormone returns the blood glucose concentration to normal. (c) Explain why this control system is described as negative feedback. [6]
- (a) The pancreas detects the rise in blood glucose concentration and releases insulin.
- (b) Insulin travels in the blood plasma to target organs, mainly the liver and muscles.
- It increases the uptake of glucose by cells, and in the liver glucose is converted to glycogen for storage, so the blood glucose concentration falls.
- (c) The change, rising glucose, triggers a response that reverses the change, bringing the level back towards the normal set point.
- When glucose falls back to normal, insulin secretion decreases, so the correction switches itself off. A change producing the opposite, corrective effect is negative feedback.
Answer: (a) Pancreas; insulin. (b) Insulin increases glucose uptake by cells and drives conversion of glucose to glycogen in the liver, lowering blood glucose. (c) The response opposes the original change and shuts down once normal levels return.
Where marks slip: Name the storage molecule, glycogen, and keep it clearly separate from glucagon. Confusing glycogen and glucagon in one answer is the most common own goal on this topic.
Try one yourself: Write the mirror version for a fall in blood glucose using glucagon, then the equivalent negative feedback answer for body temperature control on a hot day.
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Solve my question →Quiz me on this topicQuestions students ask
How long is Biology 0610 Paper 4 and what does it cover?
It is 1 hour 15 minutes, 80 marks, and 50 percent of the IGCSE. It examines the whole Extended syllabus, Core plus Supplement, across all twenty one topics, and is sat alongside Paper 2 multiple choice and a practical paper.
Is Paper 4 harder than the Core theory paper?
It examines extra Supplement content and asks for more depth, but it also unlocks the full grade range, while the Core route caps the grade at C. If your mocks are comfortably at grade C or above, Extended entry is usually the right call.
How should I revise for the long 5 and 6 mark questions?
Practise writing them as separate short statements, one biological point per sentence, then mark yourself against real mark schemes. The scoring points are usually syllabus phrases, so revising from syllabus wording beats revising from paraphrased notes.
Do I need to memorise diagrams for Paper 4?
You need to interpret and label them more than draw them from scratch. The heart, the eye, the kidney and nephron, the villus, a leaf section and the reflex arc are the ones I would know cold.