Exam technique · Biology · 9 min read
IB Biology Paper 1B: how to read unfamiliar data fast and score on every question
What the data-based half of Paper 1 asks, the five calculations that come up, and two original worked examples with the marks shown.
On this page
- What is Paper 1B, and how does it fit into Paper 1?
- How do you read an unfamiliar graph or table quickly?
- Which calculations come up, and how do you avoid the slips?
- How do you describe a trend, and when do you explain it?
- How should you use the marks?
- What do experimental-design questions ask for?
- How should you split your time across 1A and 1B?
- Worked example 1: seed germination in salt solutions
- Worked example 2: reading rate from an amount graph
IB Biology Paper 1B is the data-based half of Paper 1: four questions built around experimental data you have usually never seen, worth 25 marks at SL and 35 marks at HL. It is sat in the same sitting as Paper 1A (multiple choice), so the time is shared between both. The marks go to reading graphs and tables quickly, doing a handful of standard calculations without slips, and using biology you already know to explain what the data show.
What is Paper 1B, and how does it fit into Paper 1?
On the current course (first exams 2025), Paper 1 comes in two booklets, 1A and 1B, sat together with no break.
| Item | SL | HL |
|---|---|---|
| Paper 1 total | 1 h 30 min, 55 marks, 36% of the grade | 2 h, 75 marks, 36% of the grade |
| Paper 1A | 30 multiple-choice questions, 30 marks | 40 multiple-choice questions, 40 marks |
| Paper 1B | 4 data-based questions, 25 marks | 4 data-based questions, 35 marks |
The 1B questions are about experimental work and can draw on any theme of the syllabus, so a question can start with data on stomata (B3.1) and end by asking about water potential (D2.3). Calculators are allowed. There is no Biology data booklet for the 2026 and 2027 exams; one is introduced from the 2028 exams. For the full picture of both papers, see what the Biology exams look like now.
How do you read an unfamiliar graph or table quickly?
Spend the first 30 seconds on each question reading, not writing. In that time, find:
- The independent variable (usually the x-axis or first column) and its units.
- The dependent variable and its units. Is it an amount (total product, mass, number) or a rate (per minute, per day)? This one check prevents the most expensive 1B error.
- The key: how many groups or conditions are being compared.
- What the bars or ranges mean: standard deviation, range, or 95% confidence interval. The caption tells you.
- Sample size and repeats: "mean of 3 dishes of 20 seeds" tells you how far to trust a difference.
- The question parts, before you study the data in detail. Then you read the graph for what is asked.
Then annotate the figure: circle the units, mark the values you will quote, and pencil in a ruler line where you need to read off a value. Examiners usually accept a small range around a value read from a graph, but only if you read it carefully.
Which calculations come up, and how do you avoid the slips?
| Calculation | How | The usual slip |
|---|---|---|
| Percentage change | (final − initial) ÷ initial × 100 | Dividing by the final value; dropping the minus sign for a decrease; confusing percentage points with percent |
| Rate | change in quantity ÷ time taken; for a curve, the gradient of a tangent | Missing units (cm3 s−1); reading the rate off the y-axis of an amount graph |
| Mean | sum ÷ number of values | Leaving out a value without saying why; giving more decimal places than the raw data |
| Ratio | divide both numbers by the smaller one, e.g. 12 : 180 = 1 : 15 | Writing it the wrong way round from how the question names the two items |
| Magnification / actual size | image size ÷ actual size, both in the same unit | mm vs µm (1 mm = 1000 µm) |
Show the working even for one-mark calculations. A correct method with an arithmetic slip may still earn something; a wrong number on its own earns nothing. Give units unless the question supplies them in the answer line. Error bars and statistical tests have their own guide: which statistical test to use in Biology, and graphs and error bars covers how to read overlapping bars.
How do you describe a trend, and when do you explain it?
"Describe" asks what the data show. A description that earns full marks has three parts:
- Direction: increases, decreases, no clear relationship.
- Shape: steady, steepest at first, peaks at, levels off from.
- Values with units: at least one pair of quoted data points, and any anomaly or large spread.
"Explain" asks why, using biology: enzymes, water potential, limiting factors, selection. "Suggest" means a sensible biological reason for something you are not expected to know for certain; any well-reasoned answer within biology can score. When two conditions are compared, refer to both in every sentence ("whereas", "both"), and be careful with "compare": in IB terms it is about similarities. The command terms guide has the full list.
Loses marks (Describe, 2 marks.) As the salt goes up, germination goes down because of osmosis.
Earns marks (Describe, 2 marks.) Germination decreases as NaCl concentration increases, from 95% at 0 to 10% at 200 mmol dm−3. The decrease is small up to 50 mmol dm−3 and steepest between 100 and 150.
How should you use the marks?
Each mark usually corresponds to one distinct, correct point. Before writing, count the marks and plan that many points:
- 1 mark, calculate: the answer with units, and working shown.
- 2 marks, describe: the trend plus quoted values, or two features of the shape.
- 2–3 marks, explain or suggest: two or three linked reasons, each a separate idea, not one reason reworded.
- Evaluate a conclusion or hypothesis: what in the data supports it, what does not (overlapping error bars, small sample, a single species), and a judgement.
If you have written five lines for a one-mark question, stop: the time belongs to a later question.
What do experimental-design questions ask for?
1B rewards the same thinking as the IA. Expect to identify variables, say how and why a variable should be controlled, judge whether the method is valid or reliable, and suggest a realistic improvement. Specific beats generic every time.
Loses marks Control the temperature and repeat the experiment to make it more accurate.
Earns marks Keep the dishes in an incubator at 20 °C, because temperature affects enzyme activity in the germinating seeds. Use five dishes per concentration instead of three so that a mean is less affected by one unusual dish.
Know the difference between repeatability (more repeats, a more reliable mean) and accuracy (closeness to the true value, which repeats do not fix), and what a control does (shows the effect is caused by the independent variable). The annotated science IA example shows the same ideas in a full investigation.
How should you split your time across 1A and 1B?
Paper 1 gives about 1.6 minutes per mark at both levels (90 minutes for 55 marks at SL, 120 for 75 at HL). Split strictly by marks, that is roughly:
- SL: 1A about 50 minutes, 1B about 40 minutes (about 10 minutes per 1B question).
- HL: 1A about 65 minutes, 1B about 55 minutes (about 14 minutes per 1B question).
Most multiple-choice questions take less than their share, and any time saved on 1A is worth more on 1B, where reading the data takes time. Decide your plan before the exam, write the clock times you want to reach on scrap paper, and don't leave 1B until the last 20 minutes. Our guide to time management in IB exams covers what to do when you fall behind.
Worked example 1: seed germination in salt solutions
Original example written for this guide; it is not an IB question.
Radish seeds were placed in dishes containing NaCl solutions. Each value is the mean of 3 dishes of 20 seeds, counted after 5 days at 20 °C.
| NaCl concentration / mmol dm−3 | 0 | 50 | 100 | 150 | 200 |
|---|---|---|---|---|---|
| Mean germination / % | 95 | 90 | 70 | 35 | 10 |
(a) Calculate the percentage decrease in germination between 0 and 150 mmol dm−3. [1]
(95 − 35) ÷ 95 × 100 = 63% (63.2%). The trap is answering 60%: that is the difference in percentage points, not the percentage change.
(b) Describe the effect of NaCl concentration on germination. [2]
Germination decreases as NaCl concentration increases, from 95% to 10%. The decrease is small up to 50 mmol dm−3 and largest between 100 and 150 mmol dm−3 (70% to 35%).
(c) Suggest a reason for the effect. [2]
A more concentrated NaCl solution (lower water potential, in HL terms) means less water enters the seed by osmosis. Without enough water, enzymes that break down the seed's food stores are not activated, so the embryo cannot grow. (A high concentration of Na+ ions may also be toxic to the cells: an alternative point.)
(d) Identify one variable that should be controlled, and explain why. [2]
Volume of solution per dish, because a dish with less solution could dry out, raising the salt concentration and reducing the water available to the seeds. (Temperature and seed age are also good answers, each with its own reason.)
Worked example 2: reading rate from an amount graph
Original example written for this guide; it is not an IB question.
Catalase solution at two concentrations was added to equal volumes of the same hydrogen peroxide solution, and the oxygen released was collected.
| Time / s | 0 | 30 | 60 | 90 | 120 | 150 | 180 |
|---|---|---|---|---|---|---|---|
| High catalase / cm3 | 0 | 9.0 | 14.0 | 16.0 | 16.0 | 16.0 | 16.0 |
| Low catalase / cm3 | 0 | 4.0 | 8.0 | 11.5 | 14.0 | 15.5 | 16.0 |
(a) Calculate the mean rate of oxygen production at high catalase concentration in the first 30 s. [1]
9.0 cm3 ÷ 30 s = 0.30 cm3 s−1. (The low-concentration rate over the same interval is 4.0 ÷ 30 = 0.13 cm3 s−1.)
(b) Explain why both curves level off at the same volume. [2]
Both tubes contained the same amount of hydrogen peroxide, so the same maximum volume of oxygen can be produced. Once all the substrate has been broken down, no more oxygen is released. The enzyme is a catalyst, is not used up and does not change the final amount of product, only how quickly it is reached.
Loses marks The high-concentration enzyme stopped working after 90 s.
Earns marks After 90 s the substrate had all been converted, so the rate fell to zero; the enzyme was still active.
(c) Explain why the rate at low catalase concentration decreases over time. [2]
As hydrogen peroxide is broken down, its concentration falls, so there are fewer collisions between substrate molecules and active sites per second. Fewer enzyme–substrate complexes form per second, so the gradient of the curve decreases.
The same rate-versus-amount reading appears in our list of common IB Biology mistakes, alongside the other data slips worth checking before the exam.
Questions students ask
How many questions are in IB Biology Paper 1B?
The subject guide describes Paper 1B as four data-based questions at both levels: 25 marks at SL and 35 marks at HL. It is sat in the same sitting as Paper 1A, the multiple-choice section.
Can I use a calculator in Biology Paper 1B?
Yes. On the current course, calculators are allowed on Paper 1 (both 1A and 1B) and on Paper 2, at SL and HL. Check the IB's calculator rules for your session with your exams coordinator.
Is there a data booklet for IB Biology Paper 1?
Not for the 2026 and 2027 exams: the current Biology guide provides no data booklet, so learn the formulas for percentage change, rate, mean and magnification. A Biology data booklet is introduced from the 2028 exams, for students who started the course in August 2026.
How can I practise Paper 1B when there are few past papers?
Use the data question in Paper 2, Section A of the old Paper 3 (the closest older format), and the results from your own practicals: for each graph, write a two-mark description, a two-mark explanation and one improvement to the method.
Try one: Biology
In an osmosis experiment, a potato cylinder had a mass of 2.50 g before and 2.20 g after 30 minutes in a sucrose solution. What is the percentage change in mass?
- −12.0%
- −13.6%
- +12.0%
- −88.0%
Show the answer and why
Answer: A
Percentage change = (final − initial) ÷ initial × 100 = (2.20 − 2.50) ÷ 2.50 × 100 = −12.0%. The negative sign matters: the cylinder lost water by osmosis because the solution was more concentrated than the potato cells (at HL: it had a lower water potential). B divides by the final mass instead of the initial mass. D gives the final mass as a percentage of the initial (88.0%), which is not a change.