AP® Biology review sheet from Aim for Five (aimforfive.com/bio/units/2/2-7)
Unit 2 · Topic 2.7
2.7 Tonicity and Osmoregulation
Osmosis is the passive movement of water across a membrane toward the side with more dissolved solute, or more precisely, from higher to lower water potential. Comparing solute concentrations (hypotonic, hypertonic, isotonic) tells you whether a cell will gain or lose water, and the water potential equations let you calculate it. Organisms use structures like contractile vacuoles and central vacuoles to keep their water balance, which is called osmoregulation.
Key terms
- osmosis
- water potential
- hypertonic
- hypotonic
- isotonic
- osmoregulation
Tonicity: comparing two solutions
Tonicity compares the solute concentration of two solutions separated by a membrane. The words always describe one solution relative to another, usually the outside solution relative to the cell.
Water moves toward the side with the higher solute concentration, which is the side with less 'free' water. You can also say water moves from the hypotonic side to the hypertonic side.
| Outside solution is… | Meaning | Net water movement | Animal cell | Plant cell |
|---|---|---|---|---|
| Hypotonic | Lower solute concentration than the cell | Into the cell | Swells; may burst (lyse) | Becomes turgid (firm), which is healthy |
| Isotonic | Same solute concentration | No net movement | Normal | Flaccid (limp) |
| Hypertonic | Higher solute concentration than the cell | Out of the cell | Shrivels | Plasmolyzes: membrane pulls away from the wall |
Water potential
Water potential (Ψ, the Greek letter psi) measures water's tendency to move. Water always moves from higher Ψ to lower Ψ. It's usually measured in bars (or megapascals: 1 MPa = 10 bars). Pure water in an open container has Ψ = 0, and adding solute makes Ψ negative.
Ψ = Ψp + Ψs. Pressure potential (Ψp) is the physical pressure on the water. In an open container, Ψp = 0. Inside a plant cell pushing against its wall, Ψp is positive. Solute potential (Ψs) comes from dissolved solute. It's zero for pure water and becomes more negative as you add solute.
Ψs = −iCRT. Here i is the ionization constant (the number of particles each unit of solute forms in water: 1 for sucrose or glucose, about 2 for NaCl, which splits into Na⁺ and Cl⁻), C is molar concentration (mol/L), R is the pressure constant 0.0831 L·bar/(mol·K), and T is temperature in kelvins (°C + 273). All of these formulas are on the formula sheet.
Osmoregulation
Cells grow and stay in balance only because molecules, including water, keep moving across their membranes. Osmoregulation is how an organism controls its water balance and the concentration of solutes inside it.
A freshwater protist such as Paramecium lives in hypotonic water, so water constantly enters by osmosis. It has a contractile vacuole that collects the extra water and pumps it out, which takes energy. Without it, the cell would burst.
A plant cell's large central vacuole holds water and solutes. When the plant has enough water, the full vacuole presses the cell against its wall, creating turgor pressure (positive Ψp) that keeps leaves and stems firm. When the plant loses water, the cells become flaccid and the plant wilts.
Animals osmoregulate too. Freshwater fish produce large amounts of dilute urine, while ocean-dwelling bony fish drink seawater and get rid of extra salt through their gills.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Calculating solute potential
Calculate the solute potential of a 0.2 M sucrose solution at 20 °C. Then calculate it for a 0.15 M NaCl solution at 25 °C. Assume NaCl fully ionizes.
Show the solutionHide the solution
- Step 1: Use Ψs = −iCRT with R = 0.0831 L·bar/(mol·K).
- Step 2: Sucrose doesn't ionize, so i = 1. Convert temperature: T = 20 + 273 = 293 K.
- Step 3: Ψs = −(1)(0.2 mol/L)(0.0831 L·bar/(mol·K))(293 K) = −4.87 bars.
- Step 4: NaCl splits into Na⁺ and Cl⁻, so i = 2. T = 25 + 273 = 298 K.
- Step 5: Ψs = −(2)(0.15)(0.0831)(298) = −7.43 bars.
Answer: Sucrose: Ψs ≈ −4.87 bars. NaCl: Ψs ≈ −7.43 bars.
- Example 2Calculator allowed
Finding the water potential of potato cells
Potato cores are placed in sucrose solutions of different molarity in open beakers at 22 °C. A graph of percent change in mass against sucrose molarity crosses zero (no mass change) at 0.30 M. Find the water potential of the potato cells.
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- Step 1: Interpret the graph: where mass doesn't change, there's no net water movement, so the potato cells and that solution have equal water potential.
- Step 2: Find the solution's Ψ. In an open beaker, Ψp = 0, so Ψ = Ψs.
- Step 3: Sucrose: i = 1. T = 22 + 273 = 295 K.
- Step 4: Ψs = −(1)(0.30)(0.0831)(295) = −7.35 bars.
- Step 5: So the potato cells' water potential is also about −7.35 bars.
Answer: Ψ of the potato cells ≈ −7.35 bars.
- Example 3Calculator allowed
Which way does water move? (classic trap)
A plant cell has a solute potential of −8.0 bars and a pressure potential of +3.0 bars. It's placed in an open beaker of 0.25 M sucrose at 27 °C. Does water move into or out of the cell?
Show the solutionHide the solution
- Step 1: Cell: Ψ = Ψp + Ψs = 3.0 + (−8.0) = −5.0 bars.
- Step 2: Solution: open beaker, so Ψp = 0. T = 27 + 273 = 300 K. Ψ = Ψs = −(1)(0.25)(0.0831)(300) = −6.23 bars.
- Step 3: The trap: −6.23 is a 'bigger number' than −5.0, but it's more negative, so it's lower. Compare on a number line: −5.0 > −6.23.
- Step 4: Water moves from higher Ψ (the cell, −5.0 bars) to lower Ψ (the solution, −6.23 bars).
Answer: Water moves out of the cell, from Ψ = −5.0 bars to Ψ ≈ −6.23 bars.
Common mistakes
- Saying water moves toward higher water potential. It moves from higher Ψ to lower (more negative) Ψ.
- Using °C instead of kelvins in Ψs = −iCRT. Always add 273.
- Using i = 1 for salts like NaCl. NaCl forms two ions, so i ≈ 2.
- Describing a cell as hypertonic without saying compared to what. Tonicity always compares two solutions.
On the exam
- Expect a calculation with Ψs = −iCRT and Ψ = Ψp + Ψs followed by a prediction of water movement. Show the formula, the substituted values with units, and a clear direction statement.
- For potato or dialysis-tubing lab data, the concentration where mass change is zero is where water potentials are equal. Questions often ask you to find it from a graph.
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Check yourself
4 questions on 2.7 Tonicity and Osmoregulation. Pick an answer to see if you got it, and why.
What is the solute potential of a 0.2 M sucrose solution at 27 °C? (Ψs = −iCRT; R = 0.0831 L·bar/(mol·K); sucrose does not ionize in water.)
| Sucrose concentration (M) | Average percent change in mass |
|---|---|
| 0.0 | +18 |
| 0.2 | +6 |
| 0.4 | −4 |
| 0.6 | −13 |
| 0.8 | −20 |
Experimental data: potato cores were weighed, soaked in sucrose solutions for 24 hours, and weighed again. Each value is the mean of five cores.
Based on the data, the sucrose concentration that is isotonic to the potato cells is closest to which of the following?
Which of the following best explains the result in the 0.8 M solution?
Which change would most improve confidence in the estimate of the potato cells' solute concentration?
0 of 4 answered