IB Biology 2025 · Paper 1B · Standard level · 45 minutes

IB Biology Paper 1B practice paper. Standard level, 2025 format.

Four data-based questions, 25 marks, with the full markscheme under the paper. Paper 1B is the part of Paper 1 that hands you data you have not seen before (results from an experiment or a study, a graph, a diagram) and asks what it shows, then ends each question on content you were taught. Sit this one on screen against a 45-minute clock and mark it yourself, or print it for a class.

What Paper 1B is

Paper 1 at standard level: 1 hour 30 minutes, 55 marks, 36% of the final grade. Its two parts are sat together, in one sitting.

Paper 1A30 marks 30 multiple-choice questions. One mark each.
Paper 1B25 marks 4 data-based questions of about 5 to 7 marks each. Each is built on a data set, an experiment or a micrograph, and ends with a part that leaves the data and tests content you were taught.

On the front of the paper

A calculator is required for this paper.
Answers must be written within the answer boxes provided.

On the day the two parts share the sitting, so how you split the time is up to you. Sat on its own, this paper is timed at 45 minutes, half of Paper 1, which leaves the other half for the 30 multiple-choice questions.

How this paper was built

Four of the 204 data-based questions the bank can set at standard level, one from each theme, worth 7, 7, 5 and 6 marks: 25 in all, the real paper’s total.

Question 1: C1.2 Cell respiration 7 marks
Question 2: B1.1 Carbohydrates and lipids 7 marks
Question 3: D3.1 Reproduction 5 marks
Question 4: A2.2 Cell structure 6 marks

The code on the paper rebuilds this exact paper and its markscheme. Build a fresh Paper 1B draws a new set from the same questions on any topics you tick; a fresh one comes out close to 25 marks, but not always as four questions.

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The paper

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Biology · Paper 1B

Standard level · data-based questions · practice paper

45 minutes25 marks

Paper code: this paper was generated, so the code is its recipe. Enter it at biologybybradford.com/exam-maker to rebuild this exact paper and its markscheme.

Covers Whole syllabus: data-based questions from all four themes

A calculator is required for this paper.

Name:

A student investigated the effect of glucose concentration on the rate of anaerobic respiration in a yeast suspension by counting the number of carbon dioxide bubbles released per minute through a delivery tube. The results are shown in the table.

Glucose concentration / % w/v	Mean bubbles per minute
0	0
2	9
4	17
6	24
8	25
10	25

(a)Describe the relationship between glucose concentration and the rate of carbon dioxide production shown in the table.[2]

(b)Explain the shape of the relationship you described in (a).[2]

(c)Identify one variable that should be controlled in this investigation.[1]

(d)Calculate the percentage increase in the rate of carbon dioxide production when the glucose concentration is raised from 2 % to 6 %.[2]

A researcher measured the melting points of four fatty acids, each with an 18-carbon chain but differing in the number of carbon–carbon double bonds. Each measurement was repeated three times; the table shows the means with their standard errors (± SE).

Fatty acid	Number of C=C double bonds	Melting point / °C (mean ± SE)
stearic acid	0	69.4 ± 0.3
oleic acid	1	13.2 ± 0.4
linoleic acid	2	−5.6 ± 0.5
linolenic acid	3	−11.3 ± 0.6

(a)Analyse the relationship between the number of double bonds and the melting point of these fatty acids.[2]

(b)Explain, in terms of molecular structure, why increasing the number of double bonds lowers the melting point.[2]

(c)The researcher claims the melting points of oleic and linoleic acid are significantly different. Evaluate whether the data support this claim, naming an appropriate statistical test and stating the null hypothesis it would test.[2]

(d)Suggest why the researcher used fatty acids that all had 18 carbon atoms.[1]

The table shows data from one country for in vitro fertilization (IVF) treatment cycles started in one year, grouped by the age of the patient. The middle column shows the percentage of cycles started that led to a live birth. The last column shows the percentage of those live births that were twins or more.

Patient age / years	Live births per cycle started / %	Live births that were twins or more / %
under 35	32	18
35–37	25	15
38–40	16	12
41–42	8	9
over 42	3	6

(a)Describe how the live-birth rate per cycle changes with patient age.[1]

(b)Transferring several embryos at once raises the chance of a birth but also of a multiple pregnancy. Evaluate this practice.[3]

(c)Suggest why some couples still choose IVF even when the success rate at their age is low.[1]

A student calibrated the eyepiece graticule of a light microscope. The diagram shows the field of view with the eyepiece graticule, which carries 100 arbitrary divisions, lying above a stage micrometer ruled in millimetres. No magnification is stated on the diagram.

(a)Determine the distance, in µm, represented by one division of the eyepiece graticule.[2]

(b)A cell measured with the same objective spans 26 eyepiece graticule divisions. Calculate its length in µm.[1]

(c)The student could read the graticule only to the nearest half division. State the uncertainty of the length in (b), in µm.[1]

(d)Explain why the eyepiece graticule alone cannot give a measurement in µm.[2]

Original practice questions © Biology by Bradford · CC BY-NC-SA 4.0 · Not affiliated with or endorsed by the International Baccalaureate Organization.
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Markscheme BbB-EgD______wAAK2bC

One mark per point; / separates alternative wording within a point, OR separates alternative answers, words in brackets are not required, underlined words are essential. OWTTE = or words to that effect.

1. C1.2.6

(a) [2]
- rate of CO2 production increases with glucose concentration up to about 6–8 %;
- above about 8 % glucose concentration, the rate levels off (no further increase), OWTTE;
(b) [2 max]
- at low glucose concentrations, glucose (substrate) availability limits the rate of respiration, so more glucose increases the rate;
- at higher concentrations another factor becomes limiting, e.g. the amount/activity of respiratory enzymes (or oxygen for the aerobic component), OWTTE;
(c) [1]
- temperature (accept: yeast concentration/mass, volume of suspension, time allowed);
(d) [2]
- (24 − 9) ÷ 9 × 100;
- = 167 % (accept 166.7 %);
Award ECF for a correct method using misread values.

2. B1.1.10

(a) [2]
- melting point decreases as the number of C=C double bonds increases (a negative relationship);
- the fall is largest for the first double bond (0→1: ~56 °C) and smaller thereafter, so the relationship is not linear, OWTTE;
(b) [2]
- each C=C double bond (cis) introduces a kink/bend in the chain;
- kinked chains pack together less closely, so intermolecular attractions are weaker and less energy/lower temperature is needed to melt them, OWTTE;
(c) [2 max]
- the means (13.2 vs −5.6 °C) differ by far more than the sum of their standard errors / the ±SE ranges do not overlap, so a real difference is very likely;
- a t-test would test whether the two means differ significantly, with the null hypothesis that there is no difference between the mean melting points of the two fatty acids (any difference being due to chance), OWTTE;
Accept 'error bars do not overlap' reasoning. Null hypothesis must state no difference.
(d) [1]
- to control chain length, so that any difference in melting point is due to the number of double bonds (the independent variable) and not to chain length, OWTTE;

3. D3.1.7

(a) [1]
- the live-birth rate per cycle falls with age (from 32 % under 35 to 3 % over 42);
(b) [3]
- benefit: transferring more embryos raises the chance that at least one implants, so a higher live-birth rate, valuable when success per embryo is low (e.g. older patients);
- risk: it also raises the chance of twins or more, which carries greater dangers (premature birth, low birth mass, risk to the mother);
- reasoned judgement: single-embryo transfer is safer and may be preferred where success is reasonable, while multiple transfer may be justified only when the chance per embryo is very low, OWTTE;
(c) [1]
- it may be their only realistic chance of a biological child / the emotional value of even a small chance is high, OWTTE;

4. A2.2.2

(a) [2 max]
- reads from the diagram that 40 eyepiece divisions align with 0.5 mm of the stage micrometer;
- converts to 500 µm and divides by 40;
- 12.5 (µm);
Award full marks for a correct answer with no working. Accept 12.5 µm per division. ECF from a misread alignment.
(b) [1]
- 325 (µm);
ECF from part (a). Units not required.
(c) [1]
- ± 3.1 (µm) / ± 3.13 (µm) / ± 3.125 (µm);
ECF from part (a): accept ± a quarter of the value found in (a), since a reading to the nearest half division is uncertain by ± 0.25 division. Accept ± 6.25 µm only where the answer explains that both ends of the cell are read, each to ± 0.25 division.
(d) [2 max]
- the graticule divisions are arbitrary / have no fixed size of their own;
- the size a division represents depends on the magnification (of the objective in use), so it changes between objectives;
- so it must be compared with a scale of known length, the stage micrometer, before any reading means anything, OWTTE;
Do not accept 'the graticule is not accurate'. Accept 'it is not to scale' only with a reason.

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