Transport Across the Cell Membrane
Senior 2 · Biology
Transport Across the Cell Membrane
Lesson 3 of 5
20 min
Let's begin
Leave slices of raw potato in salty water for an hour and they become soft and floppy. Leave them in tap water and they become firm and crisp. The salt did not enter the potato to soften it. Water moved out of the cells in one case and into them in the other.
This movement of water across a membrane is osmosis. It decides whether plants stand up or wilt, why hospital drips must be exactly the right strength, and how isambaza can be preserved with salt.
Osmosis is the net movement of water molecules from a region of higher water potential (a dilute solution) to a region of lower water potential (a concentrated solution), through a partially permeable membrane.
Water potential (symbol ψ, the Greek letter "psi", measured in kilopascals, kPa) describes how freely water molecules can move out of a solution.
Question: Cell X has a water potential of −300 kPa and the cell next to it, Y, has −700 kPa. Which way does water move?
Solution: −300 kPa is higher (less negative) than −700 kPa. Water moves by osmosis from X to Y, from the higher to the lower water potential.
| The outside solution is… | Compared with the cell, it is… | Net water movement |
|---|---|---|
| hypotonic | more dilute (higher ψ) | into the cell |
| isotonic | the same concentration (equal ψ) | no net movement |
| hypertonic | more concentrated (lower ψ) | out of the cell |
Question: Five potato strips were weighed, left in sucrose solutions for 2 hours, blotted dry and reweighed. Calculate the missing value, then find the concentration that is isotonic with the potato cells.
| Sucrose (mol/dm³) | Starting mass (g) | Final mass (g) | Change in mass (%) |
|---|---|---|---|
| 0.0 | 2.50 | 2.80 | ? |
| 0.2 | 2.50 | 2.65 | +6 |
| 0.4 | 2.50 | 2.45 | −2 |
| 0.6 | 2.50 | 2.30 | −8 |
| 0.8 | 2.50 | 2.15 | −14 |
Solution: percentage change = (final − starting) ÷ starting × 100 = (2.80 − 2.50) ÷ 2.50 × 100 = +12%.
The strips gained mass in 0.0 and 0.2 mol/dm³, because water entered by osmosis: the solution had a higher water potential than the cells. They lost mass from 0.4 mol/dm³ upwards, because water left. The isotonic point is where the change is 0%, between 0.2 (+6%) and 0.4 (−2%). It lies 68 of the way along that interval: 0.2 + (6 ÷ 8) × 0.2 = 0.35 mol/dm³. The cell sap of the potato has about the same water potential as a 0.35 mol/dm³ sucrose solution.
We use percentage change, not change in grams, so that strips of different starting masses can be compared fairly. Blotting removes surface water, which would otherwise add false mass.
Tip
Memory hook: water runs downhill in water potential: from 0 towards more negative, from dilute towards concentrated.
Exam technique: say "water molecules move", never "the solution moves" or "the salt moves". Always mention the partially permeable membrane.
Define osmosis.
Solution: The net movement of water molecules from a region of higher water potential (a dilute solution) to a region of lower water potential (a concentrated solution), through a partially permeable membrane.
Which has the higher water potential: pure water, or a 5% sugar solution?
Solution: Pure water, at 0 kPa. Any dissolved solute lowers the water potential below zero.
Cell A has ψ = −450 kPa and cell B has ψ = −250 kPa. In which direction does water move?
Solution: From B to A. −250 kPa is higher (less negative) than −450 kPa, and water moves from the higher to the lower water potential.
A strip of cassava weighing 4.0 g weighs 3.4 g after an hour in strong salt solution. Calculate the percentage change in mass and explain it.
Solution: (3.4 − 4.0) ÷ 4.0 × 100 = −15%. The salt solution had a lower water potential than the cassava cells (it was hypertonic), so water left the cells by osmosis.
Summary
Sources and credits
The explanation of water potential, the potato-strip data and worked examples, and the practice questions are written by Ganzaa.
