IB MYP Sciences – Cells and Organisms 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.

This practice set is for IB MYP Sciences and follows the International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014. It covers cells and organisms, two topics the brief lists for the on-screen examinations at the end of MYP year 5, and suits MYP years 4 and 5 (there is no SL/HL split). MYP has no prescribed content list – schools design their own units – so your teacher will share the task-specific clarifications.

Each question is labelled with the criterion it trains. Real MYP work is judged against criterion level descriptors (1–8, in four bands), so the mark points here are a revision aid, not IB marks.

See also the study guide, MYP Sciences hub and printable checklist.

Useful facts: 1 mm = 1000 µm; 1 µm = 1000 nm; magnification = image size ÷ actual size.

Questions

1. (Criterion A) State the function of each of these cell structures: (i) mitochondrion, (ii) ribosome, (iii) cell membrane. [3]

2. (Criterion A) Describe two differences between the structure of a bacterial cell and the structure of a plant cell. [4]

3. (Criterion A) A student views cheek cells.

(a) The eyepiece is ×10 and the objective ×40. State the total magnification. [1]

(b) At this magnification, the image of one cell is 18 mm long. Calculate the actual length of the cell in µm. [2]

4. (Criterion A) An electron micrograph has a magnification of ×24 000. A mitochondrion in the image is 12 mm long. Calculate the actual length of the mitochondrion in µm, and give it in nm. [3]

5. (Criterion A) A micrograph’s scale bar is labelled 10 µm and measures 20 mm with a ruler. A leaf cell in the image is 58 mm long.

(a) Calculate the magnification of the micrograph. [2]

(b) Calculate the actual length of the cell in µm. [2]

6. (Criterion A) Explain how each cell is adapted to its function.

(a) A red blood cell, which carries oxygen. [2]

(b) A palisade cell in a leaf, which carries out photosynthesis. [2]

7. (Criterion A)

(a) Put these in order, from smallest to largest: organ system, cell, organ, tissue. [1]

(b) State the level of organisation of the heart. [1]

(c) State the level of organisation of xylem in a plant stem. [1]

8. (Criterion A) Three members of the dog family (Canidae) are the grey wolf, Canis lupus, the coyote, Canis latrans, and the red fox, Vulpes vulpes.

(a) State the genus of the coyote. [1]

(b) Which two of these animals are most closely related? Explain your answer. [2]

(c) Name the kingdom that all three belong to. [1]

9. (Criterion A) Here is a dichotomous key.

1. Legs present → go to 2. Legs absent → earthworm
2. Eight legs → spider. Six legs → go to 3
3. Front wings are hard cases covering the back wings → beetle. Wings covered in coloured scales → butterfly

(a) Animal X has six legs and two pairs of wings covered in coloured scales. Use the key to identify it. [1]

(b) A woodlouse has fourteen legs. Explain why this key cannot identify it, and suggest a change to the key that would. [2]

10. (Criterion B) A student asks whether leaves from the sunny side of a tree have more stomata on their lower surface than shaded leaves. She has clear nail varnish, sticky tape, slides and a light microscope.

(a) Write a testable hypothesis, with a scientific reason. [2]

(b) State the independent variable and the dependent variable. [2]

(c) Describe a method she could use, including how she would count stomata and how she would make her results reliable. [4]

11. (Criterion C) The student’s results are shown below. Each value is the number of stomata counted in one field of view at ×400. All data are fictional.

Field	1	2	3	4	5
Sun leaf	22	25	19	24	20
Shade leaf	14	17	12	16	16

(a) Calculate the mean number of stomata per field of view for each type of leaf. [2]

(b) The field of view is a circle 0.40 mm in diameter. Calculate its area, then the mean number of stomata per mm² on the sun leaf. Give your answer to 3 significant figures. [3]

(c) State a conclusion, and evaluate how well the data support it. [3]

12. (Criterion D) A fictional hospital in Easthaven can buy either one electron microscope or eight light microscopes. Its laboratory checks blood samples and identifies bacteria from infections. Evaluate the two options and give a justified recommendation, using scientific knowledge. [8]

Answers

1. (i) Mitochondrion: site of aerobic respiration, releasing energy for the cell [1]. (ii) Ribosome: makes (synthesises) proteins [1]. (iii) Cell membrane: controls which substances enter and leave the cell [1]. [3] Examiner insight: “Makes energy” for a mitochondrion is not credited; respiration releases energy, it does not make it.

2. Difference 1: a bacterial cell has no nucleus [1]; a plant cell has its DNA inside a nucleus [1]. Difference 2: a bacterial cell has plasmids / a loop of DNA free in the cytoplasm, or has no chloroplasts, mitochondria or permanent vacuole, or has a cell wall not made of cellulose [1]; the plant cell has no plasmids / has those organelles / has a cellulose wall [1]. [4] Examiner insight: A “difference” needs both cells described; “bacteria have plasmids” alone earns half, because it does not say what the plant cell has instead.

3. (a) 10 × 40 = ×400 [1]

(b) 18 mm = 18 000 µm [1]; actual length = 18 000 ÷ 400 = 45 µm [1] Examiner insight: The conversion step earns its own point, so show “18 mm = 18 000 µm” rather than jumping straight to 45; a lone final answer that is wrong earns nothing.

4. 12 mm = 12 000 µm [1]; actual length = 12 000 ÷ 24 000 = 0.5 µm [1]; 0.5 × 1000 = 500 nm [1]. [3] Examiner insight: An answer of 0.0005 means the millimetres were not converted; a mitochondrion is about 1 µm, so 0.5 µm is sensible.

5. (a) Scale bar image = 20 mm = 20 000 µm [1]; magnification = 20 000 ÷ 10 = ×2000 [1]

(b) Cell image = 58 mm = 58 000 µm [1]; actual length = 58 000 ÷ 2000 = 29 µm [1] Examiner insight: Part (b) can earn follow-through: a wrong magnification from (a) used correctly in (b) still earns the method point, so show the division.

6. (a) It has no nucleus / is packed with haemoglobin [1], so there is more space to carry oxygen [1]. Also accept: biconcave shape, giving a large surface area for diffusion.

(b) It contains many chloroplasts [1], so it absorbs more light for photosynthesis [1]. Also accept: near the upper surface, where light is brightest. Examiner insight: The second point in each part is for the link to function; “it has lots of chloroplasts” with no mention of absorbing light gains only the first point.

7. (a) cell → tissue → organ → organ system [1]

(b) Organ [1]

(c) Tissue [1] Examiner insight: Xylem is often wrongly called an organ; it is one tissue made of similar cells, while a stem, which contains xylem, phloem and other tissues, is the organ.

8. (a) Canis [1]

(b) The grey wolf and the coyote [1], because they share the same genus, Canis, while the fox is in a different genus [1].

(c) Animals (Animalia) [1] Examiner insight: In (b) the naming point and the reason are separate, so naming the wolf and coyote without mentioning the shared genus earns only one of the two points.

9. (a) Butterfly (1 → 2 → 3 → butterfly) [1]

(b) The key only offers six or eight legs at step 2, so fourteen legs matches neither option [1]. Suggested change: add a step, for example “More than eight legs → woodlouse; eight or fewer legs → go to 2” [1]. Examiner insight: Each new step must still offer exactly two options; a step with three choices is not dichotomous and is not credited.

10. (a) Leaves from the sunny side will have more stomata per mm² on their lower surface than leaves from the shaded side [1], because sun leaves photosynthesise faster and need more carbon dioxide to diffuse in through stomata [1].

(b) Independent variable: side of the tree (sun or shade) that the leaf comes from [1]. Dependent variable: number of stomata per mm² (or per field of view) on the lower surface [1].

(c) Paint nail varnish on the lower surface, let it dry, lift it with sticky tape onto a slide [1]. Count stomata in a field of view at a fixed magnification, such as ×400 [1]. Control: same tree, leaves of similar age and size, same magnification [1]. Count several fields on several leaves from each side and find a mean [1]. Examiner insight: “Keep everything the same” earns nothing; name the specific control variables to gain that point.

11. (a) Sun: (22 + 25 + 19 + 24 + 20) ÷ 5 = 22 [1]. Shade: (14 + 17 + 12 + 16 + 16) ÷ 5 = 15 [1].

(b) Area = π × 0.20² = 0.126 mm² [1]. Density = 22 ÷ 0.1257 [1] = 175 stomata per mm² [1] (3 s.f.).

(c) The sun leaf had more stomata than the shade leaf (mean 22 against 15 per field) [1]. The ranges do not overlap (19–25 and 12–17), so the difference is likely to be real [1]. But only five fields on one leaf of each type were counted, so more leaves and trees are needed to generalise [1]. Examiner insight: Use the radius, not the diameter, in πr²; using 0.40 makes the area four times too big and loses the accuracy point.

12. Indicative points, one mark each, up to 8:

• Light microscopes are cheaper, so eight can be bought and several staff can work at once [1].
• A light microscope’s resolution (about 200 nm) is enough to see blood cells and bacteria, which are several µm long [1].
• Light microscopes can use coloured stains, which help staff tell types of bacteria and blood cells apart [1].
• Samples are prepared quickly and living cells can be viewed, so results come sooner [1].
• An electron microscope has far higher resolution (under 1 nm), so it can show viruses and structures inside cells that a light microscope cannot [1].
• But specimens must be dead and in a vacuum, preparation is slow, and trained staff are needed [1].
• The electron microscope’s high cost would also leave less money for other patient care [1].
• Justified recommendation: buy the light microscopes, because blood checks and bacteria need speed, colour and capacity; send rare samples to a specialist centre with an electron microscope [1]. [8] Examiner insight: An evaluation needs strengths and limitations of both options and a final judgement; describing only the electron microscope, however accurately, stays in the lower bands.

Where marks are usually lost

• Dividing a length in mm by a length in µm without converting first.
• Writing “×400 µm” or “500 times µm” – magnification has no unit.
• Using the diameter instead of the radius when calculating the area of a field of view.
• Giving a specialised-cell feature with no explanation of how it helps.
• “Keep everything the same” in place of named control variables.
• Conclusions that ignore sample size and the spread of the data.
• In evaluations, covering only one option or giving no final judgement.

Next steps

• Cells and organisms revision notes
• Cells and organisms study guide
• Investigation skills exam preparation
• Criteria in practice questions
• MYP Sciences hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

This practice set is aligned to the International Baccalaureate Organization, Middle Years Programme Subject Brief – Sciences, from 2014, which lists “cells” and “organisms” among the topics explored in MYP sciences on-screen examinations.