Cells in Hypotonic and Hypertonic Media
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should understand the concept of water potential and be able to calculate the net direction of water movement based on solute concentration gradients.
Why this matters
Cell survival depends on maintaining internal volume and integrity against external osmotic pressure.
Cells are not passive bags; they are dynamic systems that must regulate their internal environment. When placed in solutions with different solute concentrations, water moves across the plasma membrane to equalise chemical potentials. This movement causes the cell to swell or shrink, which can lead to lysis or plasmolysis if unchecked. Understanding these responses is fundamental to physiology, from plant agriculture to medical IV therapy.
The Core Concept: Volume and Turgor
Imagine a cell as a water balloon with a semi-permeable skin. If you place this balloon in pure water, water rushes in because the inside has more dissolved particles than the outside. The balloon expands. If you place it in a very salty solution, water rushes out, and the balloon shrivels. In biology, we describe these states using specific terms. A hypotonic solution has a lower solute concentration than the cell's cytoplasm. Consequently, water enters the cell. In animal cells, this causes swelling and potentially bursting, a process called lysis. In plant cells, which have a rigid cell wall, the cell swells until the wall pushes back. This internal pressure is called turgor pressure, and it keeps the plant upright. A hypertonic solution has a higher solute concentration than the cytoplasm. Water leaves the cell. Animal cells shrivel in a process called crenation. Plant cells lose water, and the plasma membrane pulls away from the cell wall, a state known as plasmolysis. An isotonic solution has the same solute concentration as the cytoplasm. There is no net water movement, and the cell maintains its normal shape. These principles explain why red blood cells burst in distilled water but shrink in saline, and why wilted plants perk up when watered. The key driver is always the gradient in water potential, which is determined by the concentration of solutes.
Animal Cell Responses
Animal cells lack a cell wall, making them highly susceptible to osmotic changes. In a hypotonic environment, the osmotic pressure drives water into the cell. As the volume increases, the plasma membrane stretches. If the influx is too rapid or the solution too dilute, the membrane cannot withstand the internal pressure and ruptures. This is haemolysis in red blood cells. In a hypertonic environment, water leaves the cell. The cell volume decreases, and the membrane folds inward, creating a spiky appearance known as crenation. This loss of volume can disrupt cellular functions and lead to cell death. To survive, animal cells have evolved mechanisms to regulate their volume. For example, kidney cells use ion pumps to adjust their internal solute concentration, balancing the osmotic pressure. In clinical settings, intravenous fluids must be isotonic to prevent red blood cells from lysing or crenating. A common isotonic solution is 0.9% sodium chloride, which matches the osmolarity of human plasma. Deviations from this concentration can cause severe physiological stress, highlighting the critical importance of osmotic balance in multicellular organisms.
Plant Cell Responses and Turgor
Plant cells possess a rigid cell wall composed of cellulose, which provides structural support and prevents bursting. In a hypotonic solution, water enters the vacuole, increasing the hydrostatic pressure against the cell wall. This pressure is termed turgor pressure. When turgor pressure is high, the cell is turgid, and the plant tissue is firm and upright. This is the normal state for most plant cells in a healthy environment. If the external solution becomes hypertonic, water leaves the vacuole and cytoplasm. The protoplast shrinks, and the plasma membrane detaches from the cell wall. This separation is called plasmolysis. While mild plasmolysis is reversible if the cell is returned to a hypotonic or isotonic environment, prolonged plasmolysis can cause permanent damage to cellular structures. The ability of plant cells to generate turgor pressure is essential for growth, as it drives cell expansion. The cell wall limits the extent of swelling, allowing the cell to maintain a stable volume even in very hypotonic conditions. This structural difference is why plant cells do not lyse in pure water, unlike animal cells.
Quantifying Osmotic Pressure
The driving force for water movement can be quantified using the van 't Hoff equation for osmotic pressure. Osmotic pressure (\(\Pi\)) is directly proportional to the molar concentration of solute particles (\(C\)), the absolute temperature (\(T\)), and the ideal gas constant (\(R\)). The relationship is expressed as \(\Pi = iCRT\), where \(i\) is the van 't Hoff factor, representing the number of particles produced per formula unit. For non-dissociating solutes like glucose, \(i = 1\). For dissociating salts like sodium chloride, \(i \approx 2\) because it splits into Na+ and Cl-. This equation allows us to predict the direction and magnitude of water flow. A higher osmotic pressure indicates a greater tendency for water to enter the solution. By calculating \(\Pi\) for both the intracellular and extracellular environments, we can determine the net osmotic gradient. This quantitative approach is crucial in physiology for understanding how cells maintain homeostasis and for designing appropriate medical and agricultural interventions. The use of the ideal gas constant in this biological context highlights the physical basis of osmotic phenomena.
The concept
Diffusion and osmosis
Passive movement of molecules down a gradient.
A common misconception is that osmosis is driven by the concentration of water molecules. In reality, water molecules are present in vast excess in both the cell and the solution. The driving force is the difference in chemical potential, which is determined by the solute concentration. Water moves to dilute the higher solute concentration, not because there are "fewer water molecules" outside, but because the entropy of mixing favours the state with lower solute concentration. This distinction is critical for understanding why non-penetrating solutes create osmotic pressure while penetrating solutes do not, once equilibrium is reached. C
Worked example
Problem. A plant cell has a cytoplasmic solute concentration of \(0.30\ \mathrm{mol\,L^{-1}}\) (non-dissociating). It is placed in a solution with a solute concentration of \(0.10\ \mathrm{mol\,L^{-1}}\) (non-dissociating). Calculate the osmotic pressure difference (\(\Delta \Pi\)) at \(298\ \mathrm{K}\). Use \(R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
- Identify the concentrations: \(C_{\text{in}} = 0.30\ \mathrm{mol\,L^{-1}}\), \(C_{\text{out}} = 0.10\ \mathrm{mol\,L^{-1}}\).
- Determine the van 't Hoff factor: Since the solutes are non-dissociating, \(i = 1\) for both.
- Calculate the osmotic pressure inside the cell: \(\Pi_{\text{in}} = i C_{\text{in}} R T = 1 \times 0.30 \times 8.314 \times 298\).
- Calculate the osmotic pressure outside the cell: \(\Pi_{\text{out}} = i C_{\text{out}} R T = 1 \times 0.10 \times 8.314 \times 298\).
- Find the difference: \(\Delta \Pi = \Pi_{\text{in}} - \Pi_{\text{out}} = (0.30 - 0.10) \times 8.314 \times 298\).
- Compute the value: \(\Delta \Pi = 0.20 \times 2477.572 = 495.5144\ \mathrm{J\,L^{-1}}\). Convert to kPa: \(495.5144\ \mathrm{J\,L^{-1}} = 495.5144\ \mathrm{kPa}\).
Units check. \(\mathrm{J\,L^{-1}}\) is equivalent to \(\mathrm{kPa}\), which is the standard unit for pressure in this context.
Three quick tests
- What happens to a red blood cell placed in a hypertonic solution?
Answer
It loses water, shrinks, and undergoes crenation. - Why do plant cells not burst in hypotonic solutions?
Answer
The rigid cell wall provides counter-pressure (turgor) that prevents lysis. - Calculate the osmotic pressure of a \(0.50\ \mathrm{mol\,L^{-1}}\) NaCl solution at \(300\ \mathrm{K}\) (assume \(i=2\)).
Answer
\(\Pi = 2 \times 0.50 \times 8.314 \times 300 = 2494.2\ \mathrm{J\,L^{-1}} = 2494.2\ \mathrm{kPa}\).
Back to BU-106 →
Every lecture and every concept in this unit, with prerequisites and threads.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.
Concepts cited: T-028 · Threads: regulation, systems