IB DP Biology: Organelles, compartmentalization and cell specialization (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sections B2.2 and B2.3 of the IB Diploma Programme Biology guide, first assessment 2025. Questions marked HL only cover additional higher level statements. Read the study guide and revision notes first, and see the IB Biology hub and checklist.

Questions

1. From the list cytoplasm, ribosome, cell wall, vesicle, identify the two that are organelles. [2]

2. Explain two advantages to a cell of keeping hydrolytic enzymes inside lysosomes. [3]

3. Explain why post-transcriptional modification of mRNA can occur in a eukaryotic cell but not in a prokaryotic cell. [3]

4. Two cube-shaped model cells have sides of 10 µm and 40 µm. Calculate the surface area to volume ratio of each, and explain what the result suggests about the largest size a real cell can reach. [4]

5. An electron micrograph of a mitochondrion has a magnification of ×21 000. The mitochondrion measures 63 mm in the image. Calculate its actual length in µm. [3]

6. Haematopoietic stem cells were cultured for seven days, with or without bone marrow niche cells. In each condition 240 cells were counted. With niche cells, 150 were still undifferentiated stem cells. Without niche cells, 36 were. (a) Calculate the percentage of undifferentiated cells in each condition. [2] (b) State the difference between the conditions in percentage points. [1] (c) Deduce one conclusion about the stem cell niche. [1] (d) Suggest one limitation of this experiment. [2]

7. (a) State the type of potency of a cell from a four-cell human embryo. [1] (b) State the type of potency of a stem cell in adult bone marrow. [1] (c) Distinguish between pluripotent and multipotent cells. [2]

8. (HL only) (a) Explain three ways in which the structure of a mitochondrion is adapted for ATP production in aerobic cell respiration. [3] (b) State two adaptations of the chloroplast for photosynthesis. [2]

9. (HL only) Cells were exposed to radioactive amino acids for a short time. The table shows the percentage of the cells’ radioactivity in each location over time.

Time / min	rough ER	Golgi apparatus	vesicles	outside the cell
10	60	25	5	0
30	20	50	20	2
60	5	10	55	25

(a) Deduce the route taken by the labelled protein. [2] (b) State whether the protein was made by free ribosomes or ribosomes on the rough ER, with a reason. [1] (c) Outline the role of the Golgi apparatus. [2] (d) State the role of clathrin in vesicle formation. [1]

10. (HL only) Explain two advantages that the double membrane of the nucleus gives the cell. [4]

11. (HL only) A flat patch of membrane on a kidney tubule cell measures 20 µm × 20 µm. Microvilli are then modelled on it as 2500 cylinders, each 0.1 µm wide and 0.8 µm tall. Assume each tip replaces the flat area beneath it. (a) Calculate the surface area after microvilli are added, and the factor of increase. [4] (b) Explain how this helps a proximal convoluted tubule cell. [2] (c) Explain how type I and type II pneumocytes are each adapted to their function. [3]

12. (HL only) A human egg is modelled as a sphere of radius 55 µm. For a sphere, surface area to volume ratio equals 3/r. (a) Calculate the ratio, to 3 significant figures. [2] (b) Suggest why a low ratio does not harm the egg. [1] (c) Describe two adaptations of a sperm cell. [2] (d) Discuss whether a striated muscle fibre is a single cell. [2]

Answers

1. Ribosome [1] and vesicle [1]. [2] Examiner insight: The question asks for two, so name exactly two rather than a longer list.

2. Hydrolytic enzymes would digest proteins and other structures if free in the cytoplasm [1]. A lysosome keeps them apart, so the cell is protected [1]. The lysosome can also hold an acidic pH where the enzymes work best, and keep the enzymes concentrated [1]. [3] Examiner insight: Name the hazard and the benefit separately. “Keeps things separate” alone is too vague to be credited.

3. In a eukaryote the nucleus separates transcription from translation [1]. The primary transcript can be modified, for example by removing introns, before it leaves through pores to ribosomes [1]. A prokaryote has no nucleus, so ribosomes can bind to the mRNA straight away while it is still being made [1]. [3] Examiner insight: Compare both cell types explicitly. Describing only the eukaryote does not explain why the prokaryote cannot do it.

4. 10 µm cube: area = 6 × 10² = 600 µm², volume = 1000 µm³, ratio = 0.6 µm⁻¹ [1]. 40 µm cube: area = 9600 µm², volume = 64 000 µm³, ratio = 0.15 µm⁻¹ [1]. Quadrupling the side cuts the ratio to a quarter [1]. As a cell grows, exchange across the surface cannot meet the needs of the volume, which limits cell size [1]. [4] Examiner insight: Give units for the ratio. Link the trend to exchange area and need rather than commenting on the trend alone.

5. 63 mm = 63 000 µm [1]. Actual length = 63 000 ÷ 21 000 [1] = 3 µm [1]. [3] Examiner insight: Show the unit conversion. A correct number with the wrong unit is a common slip.

6. (a) With niche cells: 150 ÷ 240 × 100 = 62.5% [1]. Without: 36 ÷ 240 × 100 = 15% [1]. (b) 62.5 − 15 = 47.5 percentage points [1]. (c) The niche cells help to maintain stem cells (or reduce differentiation) [1]. (d) Two from: culture conditions differ from the body [1]; one count per condition, with no repeats or statistics [1]. Other valid limits include a single cell type or one time period. [6] Examiner insight: Percentage points and percentage change are different, so use the term that matches what you calculate.

7. (a) Totipotent [1]. (b) Multipotent [1]. (c) Pluripotent cells can differentiate into most cell types of the body [1], while multipotent cells form only a limited range of related types [1]. [4] Examiner insight: A distinction needs both sides, so define both terms rather than only one.

8. (a) Three adaptations: a double membrane with a small intermembrane space, so protons accumulate quickly and form a gradient [1]; cristae provide extensive membrane carrying electron transport chains and ATP synthase [1]; the matrix concentrates Krebs cycle enzymes and substrates [1]. (b) Two from: large area of thylakoid membranes carrying photosystems [1]; small thylakoid volume so a proton gradient builds quickly [1]. The stroma concentrating Calvin cycle enzymes and substrates is also valid. [5] Examiner insight: Each adaptation needs its effect. “It has cristae” is a feature only, so add what the cristae allow.

9. (a) The peak moves from rough ER at 10 min [1] to Golgi at 30 min, then vesicles at 60 min, and then outside the cell [1]. (b) On the rough ER, because it is secreted from the cell [1]. (c) The Golgi apparatus processes the protein, for example by modifying it [1], and sorts and packages it into vesicles for secretion [1]. (d) Clathrin forms a cage that helps the membrane curve and pinch off as a vesicle [1]. [6] Examiner insight: Quote the times when you read a data table. The order of peaks should be tied to the figures.

10. Pores allow mRNA and ribosomal subunits out, and proteins such as enzymes for replication and transcription in [1], so the passage is controlled [1]. In mitosis and meiosis the envelope breaks into vesicles, which lets spindle microtubules reach the chromosomes [1], and the vesicles later fuse to reform nuclei [1]. [4] Examiner insight: Linking the structure (pores, fragmenting membrane) to its effect is what makes an answer complete.

11. (a) Side area of one microvillus = π × 0.1 × 0.8 = 0.251 µm² [1]. Total added = 2500 × 0.251 = 628 µm² [1]. New area = 400 + 628 = 1030 µm² (1028) [1]. Factor = 1028 ÷ 400 = 2.57 [1]. (b) Larger surface area gives room for more transport proteins [1], so more solutes are reabsorbed per unit time [1]. (c) Type I pneumocytes are extremely thin, so the diffusion distance between air and blood is short [1]. Type II pneumocytes contain lamellar bodies, which release surfactant into the alveolar lumen [1], reducing surface tension [1]. [9] Examiner insight: In a modelling calculation, state your assumption (tip replaces the base) because it shows how you reached your area.

12. (a) 3 ÷ 55 [1] = 0.0545 µm⁻¹ [1]. (b) The egg stores food, so it does not need fast exchange with its surroundings [1]. (c) Two from: flagellum for swimming [1]; many mitochondria for ATP [1]. An acrosome with enzymes to penetrate the egg’s outer layers is also valid. (d) For: it has one plasma membrane and acts as one functional unit [1]. Against: it is multinucleate, much larger than typical cells and formed by many cells fusing [1]. [7] Examiner insight: “Discuss” needs both sides. Stating only one view leaves the answer incomplete.

Where marks are usually lost

• Naming the feature of a cell structure but not its effect on function.
• Confusing cytoplasm with an organelle.
• Missing units on a ratio or a size.
• Skipping the conversion between mm and µm in magnification.
• Saying a bigger cell has a bigger ratio.
• Treating differentiation as a change to the genome, when it is a change to which genes are expressed.
• Confusing potency terms or giving only one half of a distinction.
• Answering “discuss” with one side only.

Next steps

• Review the revision notes for the points you missed.
• Reread the study guide.
• Open the IB Biology hub and tick off the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

International Baccalaureate Organization, Diploma Programme Biology guide, first assessment 2025, published February 2023, sections B2.2 and B2.3.