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  3. Paper 1: multiple choice and data-based questions

Paper 1: multiple choice and data-based questions

How to approach Paper 1A multiple-choice questions and Paper 1B data-based questions, with worked questions and markschemes.

  • IB Biology
  • Exam skills
  • SL and HL

14 min readLast reviewed 14 Oct 2026

Paper 1 has two parts taken in one sitting. Paper 1A is multiple choice: each question has four options and one correct answer. Paper 1B is a set of data-based questions on experimental work, built on results from investigations you have not seen before. The two parts test different skills, so they need different approaches.

At SL, Paper 1 lasts 1 hour 30 minutes: Paper 1A has 30 multiple-choice questions (30 marks) and Paper 1B has data-based questions worth 25 marks, 55 marks in all. At HL, Paper 1 lasts 2 hours: Paper 1A has 40 questions (40 marks) and Paper 1B is worth 35 marks, 75 marks in all. Paper 1 carries 36% of the final grade at both levels.

Because the two parts share one sitting with no break, time saved on Paper 1A can be spent on Paper 1B, where questions are longer and carry more marks each.

Paper 1A: multiple-choice questions

Paper 1A covers the whole course. Most questions test recall and understanding of a single idea; some ask you to read a diagram or micrograph, interpret a short data set or carry out a quick calculation such as magnification or percentage change. There is no penalty for a wrong answer, so every question should be answered.

A method for every multiple-choice question

  1. Read the stem and cover the options. Decide what the answer should be before looking.

  2. Read all four options, even if the first one matches your answer. A later option may be more complete or more precise.

  3. Eliminate the options you know are wrong, and say to yourself why each is wrong.

  4. Check the stem again for words that change its meaning: NOT, LEAST, MOST, ALWAYS, ONLY.

  5. For a calculation, work it out on paper first; the distractors are usually the results of common slips.

  6. If you cannot decide, choose between the remaining options and mark the question to return to; never leave it blank.

Definition

Distractor: a wrong option in a multiple-choice question, written to be attractive to a candidate who holds a common misconception or makes a common error.

Each distractor is built on a specific error. Recognising the type helps you eliminate it quickly.

Type of distractor

How it works

How to spot it

A common misconception

States an idea many students believe, such as "mitochondria produce energy" or "plants respire only at night".

Ask whether the statement would survive a careful textbook definition.

True but irrelevant

A correct statement that does not answer the question asked.

Reread the stem: does the option answer this question, or a different one?

Partly correct

Right process but wrong detail: the right enzyme with the wrong bond, the right organelle with the wrong membrane number.

Check every word of the option, not just the key term.

A calculation slip

The number you get by dividing the wrong way, forgetting a unit conversion or using the wrong denominator.

Work the calculation yourself and check units at each step.

Reversed cause and effect

Water is polar because it forms hydrogen bonds, instead of the other way round.

Ask which comes first in the mechanism.

Worked multiple-choice questions

Worked example

Which property of water results from hydrogen bonding between water molecules?

  • A. Water dissolves non-polar substances such as lipids.

  • B. Water can be pulled up xylem vessels in a continuous column.

  • C. Water has a low specific heat capacity.

  • D. Ice is denser than liquid water.

Show the solution

Answer: B. Hydrogen bonds between water molecules cause cohesion, which allows a column of water to be pulled up the xylem under tension (A1.1 Water).

  • A is a misconception: water dissolves polar and charged substances, not non-polar ones.

  • C reverses a fact: hydrogen bonds give water a high specific heat capacity.

  • D reverses a fact: hydrogen bonds hold ice in an open lattice, so ice is less dense and floats.

Each wrong option in that question names a real property of water but states it incorrectly, which is why every word of an option needs checking.

Worked example

A cell in a micrograph is 30 mm long. The magnification of the micrograph is ×1500. What is the actual length of the cell?

  • A. 0.02 µm

  • B. 2 µm

  • C. 20 µm

  • D. 45 µm

Show the solution

Answer: C. Convert the image size to micrometres first, then divide by the magnification:

actual size=image sizemagnification=30 000 μm1500=20 μmactual size=magnificationimage size​=150030000 μm​=20 μm
  • A is 30 mm ÷ 1500 = 0.02 mm, with the unit wrongly changed to µm.

  • B is a slip of a factor of 10 in the conversion.

  • D is 30 × 1500 with the unit lost: multiplying instead of dividing.

Magnification calculations are taught in A2.2 Cell structure.

Calculation questions in Paper 1A are short, but the distractors are chosen to match the most common slips, so a careless answer is almost always among the options.

Worked example

Which row correctly identifies features of RNA?

  • A. Sugar: deoxyribose; base found only in this nucleic acid: thymine; usually double-stranded

  • B. Sugar: ribose; base found only in this nucleic acid: uracil; usually single-stranded

  • C. Sugar: ribose; base found only in this nucleic acid: thymine; usually single-stranded

  • D. Sugar: deoxyribose; base found only in this nucleic acid: uracil; usually double-stranded

Show the solution

Answer: B (A1.2 Nucleic acids). In a "which row" question, check one column at a time. The sugar in RNA is ribose, which removes A and D. Uracil replaces thymine in RNA, which removes C. Only one row survives both checks, so the third column does not need to be checked, although it confirms the answer.

Tip

For a "which row" question, take one column, cross out every row that is wrong in it, then move to the next column. Two or three checks are usually enough to leave a single row.

Worked example

What is the role of helicase in DNA replication?

  • A. It joins fragments of the new strand together.

  • B. It separates the two strands of DNA by breaking hydrogen bonds between bases.

  • C. It adds free nucleotides to the 3′ end of the new strand.

  • D. It separates the two strands of DNA by breaking covalent bonds between bases.

Show the solution

Answer: B (D1.1 DNA replication). A describes DNA ligase and C describes DNA polymerase: true statements about replication that answer a different question. D is partly correct: helicase does separate the strands, but the bonds between complementary bases are hydrogen bonds, not covalent bonds.

Options B and D differ by a single word. Questions like this test the detail of a process as well as recognition of the enzyme involved.

Worked example

A population of 40 rabbits increased to 70 rabbits in one year. What was the percentage increase?

  • A. 30%

  • B. 43%

  • C. 57%

  • D. 75%

Show the solution

Answer: D. The change is 70 − 40 = 30 rabbits, divided by the initial value:

70−4040×100=75%4070−40​×100=75%

B divides the change by the final value (30 ÷ 70). C is 40 ÷ 70. A is the change itself, not a percentage.

Common mistake

Changing an answer without a reason. A first answer based on a clear method is more often right than a later change made from doubt. Change an answer only when you find a definite error, such as a misread NOT in the stem or a slip in a calculation.

Exam tip

Answer every Paper 1A question, including guesses: a blank scores nothing and a guess between two options has a 50% chance. If a question is taking more than about a minute, choose the best remaining option, mark it and move on.

Paper 1B: data-based questions on experimental skills

Each Paper 1B question describes an investigation and gives its results as a table, graph, micrograph, gel or statistical output. The questions then test the skills of experimental work: identifying variables, processing data, reading graphs, evaluating the method and drawing conclusions. The biology is from the course, but the context is usually new, so answers must refer to the experiment given.

Typical stem

What it asks you to do

Identify the independent variable.

Name the variable the investigator changed, as precisely as possible ("concentration of sucrose solution", not "sucrose").

State the trend in … between day 0 and day 10.

Direction of change, with values at the start and end and units.

Calculate the mean rate / percentage change. Show your working.

Formula, substitution, answer with unit.

Deduce whether the data support the hypothesis that …

Yes, no or partly, with the specific data that show it.

Suggest one reason for the large error bars at 40 °C.

A cause specific to this method or organism, not "human error".

Evaluate the method used to measure …

At least one strength and one limitation, each tied to the method, and a judgement.

Outline how the experiment could be modified to …

A specific, practical change, with what would be measured or controlled.

State the null hypothesis for this test. / Explain what the p-value indicates.

The "no difference / no association" statement; whether p is below 0.05 and what that means.

Statistics, error bars and the full set of calculations are worked through in Data, graphs and statistics. Experimental methods such as colorimetry, chromatography and sampling are in Practical techniques and biological drawing.

Reading graphs and tables

Before answering any question on a graph, read its title, both axis labels with their units, the scale of each axis and any key. Many lost marks come from reading the wrong curve or missing that an axis does not start at zero, is logarithmic, or that a second y-axis is on the right.

  • Quote values with units: "the rate rose from 2.0 to 8.5 cm³ min⁻¹".

  • Refer to the independent variable when you give a trend: "as temperature increased from 10 to 40 °C".

  • Name turning points, plateaus and the range over which a trend holds.

  • Distinguish a rate (per unit time) from a total.

  • Read error bars: wide bars mean variable repeats; overlapping standard deviation bars suggest the difference may not be significant.

The graph of germination against salt concentration is typical of a Paper 1B figure: a line through a small number of means, with the independent variable on the x-axis.

020406080100120140160180200220020406080100NaCl / mmol dm⁻³Germination / %
  • Germination
Practice data: a student's results for the percentage of radish seeds germinating after 5 days at five sodium chloride concentrations (mean of three dishes of 50 seeds), with a curve of best fit.

Worked example

Using the graph of germination against sodium chloride concentration:

(a) Describe the effect of sodium chloride concentration on germination. [2]

(b) Calculate the percentage decrease in germination between 0 and 150 mmol dm⁻³. [2]

(c) Suggest a reason for the effect of high sodium chloride concentration on germination. [2]

Show the markscheme
  • (a) Germination decreases as NaCl concentration increases [1]; the decrease is small up to 50 mmol dm⁻³ and becomes steeper at higher concentrations, from 96% at 0 to 14% at 200 mmol dm⁻³ [1].

  • (b) 50−9696×1009650−96​×100 [1] =−47.9%=−47.9% / a decrease of 47.9% [1]. Accept 48%.

  • (c) The salt solution has a lower water potential than the seed [1]; less water enters the seed by osmosis / water is not taken up for germination [1]; Na⁺ ions may be toxic / inhibit enzymes [1]. [max 2]

Part (a) is a description, so it needs the trend and values but no reason. Part (c) asks for a reason: the link to water potential (D2.3 Water potential) is the biology the question is testing.

Common mistake

Reading a value from the curve of best fit when the question asks about the measured data, or the other way round. "According to the data" means the plotted points; "from the graph" usually allows the curve. When in doubt, use the plotted points and say which value you used.

A full Paper 1B question with a data table

A student investigated the effect of temperature on the activity of catalase in potato extract. Equal volumes of extract and hydrogen peroxide solution were mixed in a water bath at each temperature, and the volume of oxygen produced in 60 s was collected in a gas syringe. The results, a practice data set, are shown in the table.

Temperature / °C

Trial 1 / cm³

Trial 2 / cm³

Trial 3 / cm³

Mean / cm³

10

4.0

4.4

4.2

4.2

20

7.8

8.2

8.3

8.1

30

12.1

11.6

12.4

12.0

40

14.0

9.2

14.6

?

50

6.1

5.8

6.4

6.1

Worked example

Using the table of oxygen volumes produced by catalase:

(a) Identify the independent and dependent variables. [2]

(b) Identify the anomalous result and calculate the mean at 40 °C without it. [2]

(c) Calculate the mean rate of oxygen production at 30 °C, giving the unit. [2]

(d) Suggest one cause of the anomalous result. [1]

(e) State two variables that should be controlled. [2]

(f) Evaluate the use of a gas syringe to measure the volume of oxygen. [2]

Show the markscheme
  • (a) Independent: temperature [1]. Dependent: volume of oxygen produced in 60 s [1].

  • (b) 9.2 cm³ (trial 2 at 40 °C) [1]; mean of 14.0 and 14.6 = 14.3 cm³ [1].

  • (c) 12.0÷6012.0÷60 [1] =0.20 cm3 s−1=0.20 cm3 s−1 [1]. Accept 12.0 cm³ min⁻¹.

  • (d) Syringe plunger stuck / gas leaked from a joint / timing started late / the mixture had not reached 40 °C. [1]

  • (e) Two of: volume and concentration of hydrogen peroxide; volume of extract / mass of potato used; pH; time of collection. [2]

  • (f) Strength: measures volume directly quantitatively, so results can be compared between temperatures [1]. Limitation: the plunger may stick / oxygen produced before the bung is fitted is lost / some oxygen dissolves in the solution [1].

Including the anomaly would give a mean of 12.6 cm³, which hides the true peak. Always identify anomalies before calculating means, and say that you excluded them.

Exam tip

In Paper 1B, give every calculated value to the same number of decimal places as the raw data (or one more) and include the unit. "0.2" for a rate of 0.20 cm³ s⁻¹ with no unit can lose the answer mark even if the working is right.

Evaluating methods and suggesting improvements

Evaluation questions reward limitations that are specific to the method described, with their effect on the results, and improvements that address them. Generic answers such as "human error", "be more careful" or "do more repeats" score nothing on their own.

Specific limitation

Effect on the results

Realistic improvement

Oxygen escapes before the bung is fitted

Volumes are too low; the error is larger at high rates

Add the enzyme through a sealed port / use a pressure sensor

Temperature of the mixture not checked

True temperature may differ from the bath, especially at 10 and 50 °C

Pre-incubate both solutions to the bath temperature and check with a thermometer

Only three trials with one large anomaly

Mean at 40 °C uncertain; optimum cannot be located precisely

Five trials per temperature; extra temperatures between 35 and 45 °C

Potato extract from different tubers

Catalase concentration may differ between trials

Prepare one batch of extract and use it for all trials

A strong evaluation states which limitation matters most. Here, the lost oxygen is a systematic error that affects every result in the same direction, while the anomaly is a random error that affects one value.

Common mistake

Writing "repeat the experiment more times" as the only improvement. More repeats reduce the effect of random variation but do not fix a systematic error such as lost gas. Name the limitation first, then give an improvement that fixes it.

Hypotheses, conclusions and statistics

Paper 1B often gives the output of a statistical test and asks what it shows. A null hypothesis states that there is no difference or no association. If the p-value is less than 0.05, the result is statistically significant and the null hypothesis is rejected; if it is greater than 0.05, the null hypothesis is not rejected.

Worked example

A t-test comparing the mean height of plants grown with and without a fertiliser gave p = 0.003. State the null hypothesis and deduce what the result shows. [2]

Show the markscheme
  • Null hypothesis: there is no difference in mean height between plants grown with and without fertiliser [1].

  • p = 0.003 is less than 0.05, so the difference is significant / the null hypothesis is rejected; the fertiliser is associated with a difference in height [1].

Do not write that the test "proves" the fertiliser works: a significant result means the difference is unlikely to be due to chance alone.

YouTube
An overview of Paper 1B: the structure of the data-based questions and how to approach variables, data processing and evaluation.

Managing the sitting

Paper 1A rewards speed and Paper 1B rewards care. A sensible plan is to work through Paper 1A at a steady pace, mark questions to come back to, and move on to Paper 1B with time in hand, returning to the marked multiple-choice questions at the end. Paper 1B questions are multi-part and each part is short, so leaving one part blank to finish the others is better than running out of time with a whole question untouched.

  • Read the introduction to each Paper 1B question fully before answering: it says what was measured and how.

  • Annotate the data: circle anomalies, mark the range of a trend, underline units.

  • Answer in the context given: name the organism, the variable and the values.

  • Check that each calculated answer has a unit and sensible precision.

Exam tip

In Paper 1B, use the language of the experiment. "The volume of oxygen collected in 60 s was highest at 40 °C (14.3 cm³)" scores; "the enzyme worked best at 40" may not, because it does not refer to the data or the dependent variable.

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