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IB Biology: Cells and Membrane Transport – Practice Paper 3 (10 Questions)

IB Biology: Cells and Membrane Transport – Practice Paper 3 (10 Questions)

  • 2026-10-30

IB Biology: Cells and Membrane Transport – Practice Paper 3 (10 Questions)

Paper 3 concentrates on calculation and judgement: converting between magnification and actual size, surface area to volume ratios, summing water potential, and the effects of carrier saturation and inhibitors on active transport.

Quick answer Ten times forty gives four hundred times magnification, and an eight millimetre image at that power is twenty micrometres in reality. A cube of side two centimetres has a ratio of three, and a cell at minus five hundred kilopascals in a solution of minus seven hundred loses water and plasmolyses.

01 Paper details (check answers afterwards)

Item Detail
Questions 10 questions
Suggested time 25 minutes
Difficulty mix Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter Chapter: IB Biology: Cells and Membrane Transport – Tutor Lesson Plan
Level IB SL

02 Questions (10 items)

  1. FoundationThe eyepiece is ten times and the objective is forty times. What is the total magnification?
  2. FoundationAn image is eight millimetres long at a magnification of four hundred. What is the actual length?
  3. FoundationA cell is fifty micrometres long and is drawn at four hundred times magnification. How long is the image?
  4. FoundationA cube has sides of two centimetres. What is its surface area to volume ratio?
  5. IntermediateWhat are the surface area to volume ratios of cubes with sides of three centimetres and one centimetre?
  6. IntermediateA cell has a solute potential of minus six hundred kilopascals and a pressure potential of two hundred. What is its water potential?
  7. IntermediateCell A has a water potential of minus two hundred and fifty kilopascals and cell B minus four hundred and fifty. Which way does water move?
  8. ChallengeA cell with solute potential minus eight hundred and pressure potential three hundred is placed in a solution of minus seven hundred kilopascals. What happens?
  9. ChallengeThe rate of active transport levels off as sugar concentration rises further. Why, and what does a respiratory inhibitor do?
  10. Error redoCommon error: pressure potential is treated as negative, giving minus eight hundred kilopascals. What is the correct value?

03 Answers and one-line notes

# Answer One-line note
1 400 times Total magnification is the eyepiece multiplied by the objective.
2 20 micrometres Eight millimetres divided by four hundred is zero point zero two millimetres, or twenty micrometres.
3 20 millimetres Fifty micrometres times four hundred is twenty thousand micrometres, or twenty millimetres.
4 3, from 24 square centimetres over 8 cubic centimetres Surface area is six times four, and volume is two cubed.
5 2 and 6 Fifty-four over twenty-seven is two, and six over one is six, so smaller cubes have the larger ratio.
6 minus 400 kilopascals Water potential is the sum of solute and pressure potentials.
7 From A to B Water moves from the higher potential of minus two hundred and fifty to the lower one.
8 The cell reaches minus five hundred kilopascals; water leaves and plasmolysis follows Minus eight hundred plus three hundred is minus five hundred, which is above the outside, so water leaves.
9 The carrier proteins are saturated; a respiratory inhibitor stops active transport Active transport needs carrier proteins and ATP, and both are limited.
10 minus 400 kilopascals Pressure potential is positive in a living cell, so the terms add.

04 What to do after marking

  • After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
  • Repeat the same item two days later; move to the next paper only when both attempts are correct.
  • Parents only need the answer and note columns; there is no need to explain every item.

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