6 Chapter 6 – Diffusion and Osmosis

Permeability of Cell Membranes: Osmosis


The passive movement of materials across cell membranes is the result of the of the membrane to the diffusing substance and a . For uncharged compounds, the driving force is the between the inside and outside of the cell. In the case of ions, the driving force is a balance between the concentration gradient and the between the inside and outside of the cell.

In general, biological membranes are far less permeable than an equivalent thickness of pure . They are completely impermeable to many compounds. The combination of membrane permeability, driving force, and the density and identity of active transport mechanisms determine the rate of movement of nutrients into and waste products out of cells. The rates of penetration through the membrane by various molecules are important in determining the chemical composition of the cell. A substance to which the membrane is impermeable can only be found in the cell if it is synthesized there or actively transported into the cytoplasm.

There are mechanisms in place to increase the rate at which some molecules pass into cells. of various kinds help specific molecules pass through a membrane that would normally be impermeable to that molecule. A membrane channel is a tube through the membrane formed by one or more intrinsic proteins. The outer surface of this protein channel is formed by amino acids with nonpolar groups that readily dissolve in the phospholipid of the membrane. The lining of the channel is composed of amino acids with polar groups that allow the channel to fill with water and polar molecules to readily pass. Thus, polar can diffuse through the membrane in an aqueous solution. , responsible for active and passive transport, are also intrinsic membrane proteins. These undergo a configurational change when in contact with a target molecule or ion. This configurational change forces the target through the membrane. (facilitated diffusion) is the movement of molecules by membrane carrier proteins along the concentration gradient of the target molecule. It involves no input of energy. moves the target molecule against a concentration gradient and requires the input of energy in the form of ATP.

Although the permeability of membranes is highly variable among and within organisms, two generalizations can be made. First, for chemically related polar molecules, membrane rates decrease with increasing molecular size. Chemically related molecules are those groups of molecules that are variations of a basic structure. This is a direct byproduct of the physical structure of membranes. Smaller molecules move between phospholipid subunits and through membrane channels more quickly than larger molecules. Second, for chemically related nonpolar compounds, membrane diffusion rates increase with increasing lipid solubility and are more or less independent of molecular size. More lipid soluble molecules dissolve more readily and are able to pass more quickly through the lipid membrane.

Because cell membranes are , we need to distinguish between the concentration of all solutes across a membrane (“-osmotic) and the effects of solute movement or water movement on cell shape (“-tonic”). Imagine a cell placed in a solution having a total solute concentration equivalent to the concentration of solutes within the cell. This cell is in an solution. and describe the concentration of total solutes across the membrane prior to the movement of solutes by diffusion or water by .

If a particular solute molecule is in differing concentrations across the membrane, the effect on the membrane will depend on that membrane’s permeability to the solute. If the solute can cross the membrane, it will cross the membrane to equalize a concentration gradient, usually with no net movement of water. If the membrane is not permeable to the solute, there will be no movement of solute; water will move across the membrane by osmosis instead. Such a cell is in a or solution since the ceil wall will shrink or swell by osmosis. For example, suppose that a cell is in an iso-osmotic solution but there is a higher concentration of a permeable (or penetrating) solute outside the cell than inside. The solute enters the cell by diffusion, the osmotic pressure inside the cell increases (i.e. there is now a higher concentration of solutes inside the cell than outside) and water enters the cell. This influx of water causes the cell to swell. In this case, the cell is in an iso-osmotic solution (i.e. equal initial total solute concentration) that is hypotonic (causes the cell to swell).

The concentration of solutes can be described in a number of different ways. The simplest expression is the weight of solute (grams) per 100 ml of solution. This is a . For example, 15 g in 100 ml solution may be expressed as 15%. It is frequently more useful to express concentration in terms of or . These measurements take into account the differing molecular weights of the solutes. The production of a 1 Molar (1 M) or 1 molal (1 m) solution of sodium chloride (NaCl) would require weighing out a different amount of solute than a 1 M or 1 m solution of glucose (C6H1206). The common measure to both solutions is the (6.02 x 10 23 molecules). One molar (1 M) solution contains one mole of solute in 1 liter of solution. One molal (1m) solution contains one mole of solute in 1,000 grams of solvent.

If water is the solvent, 1000g of water at 4°C is 1 liter and the final volume of solution slightly exceeds 1 liter. A one molar aqueous solution is very close, but not identical, in concentration to a one molal aqueous solution. This is the measure of concentration that has the most relevance to physiological systems because it more accurately describes the ratio of solute to solvent – the important driving force in osmosis and the generation of osmotic potential.

The osmotic pressure of a solution is proportional to the number of solute molecules in solution. For example, a 1 m solution of glucose has one mole of glucose in solution. This would have the same osmotic pressure as 1 m solutions of urea, sucrose, etc. Some molecules when they are dissolved in solution. Sodium chloride (NaCl), for example, completely dissolves in water and dissociates into Na+ and Cl-. Therefore, a 1 m aqueous solution of NaCl contains 1 mole of Na+ and 1 mole of Cl-for a total of 2 moles in solution. Therefore, the NaCl solution would generate twice the osmotic potential of the glucose solution because it has twice the particles in the same volume (mass) of water.

(Osm) is a measure of the number of paråcles in solution, which is related to the osmotic pressure generated by the solution. Thus, a 1 m solution of NaCl has an osmolality of 2 Osm. It is frequently more convenient to express concentration in terms of milliosmolality (mOsm; 1/1000 Osm). For example, a 0.1 m solution of NaCl has an osmolal concentration of 200 mOsm, whereas a 0.10 m glucose solution has an osmolal concentration of 100 mOsm.

Have you ever had a plant that you just never remembered to water? In that case, you have deprived the plant of liquid that it uses to swell its cells and produce turgor pressure. As the local environment becomes drier, water moves from an area of higher concentration (in this case, remember that the cell is mostly water) into an area of lower concentration (in this case, the soil). As a result, we see , or water exiting the plant cell. The result is a wilted, droopy vegetative appendage.

Conversely, we often see water moving from an area of higher external environmental concentrations into the cell, causing the cell to swell and lyse in a process called . For example, a red blood cell, lacking the cell wall of a plant, has only a thin membrane preventing it from swelling to epic proportions before lysing as water enters the cell. Remember, compared to distilled water, cells are hypertonic (since we have ribosomes, DNA, RNA, etc.) and thus water will flow into the cell.


Key Terms

  • Concentration gradient
  • Hypotonic
  • Diffusion
  • Facilitated diffusion
  • Hypertonic
  • Solutes
  • Passive transport
  • Isotonic
  • Solvent
  • Active transport
  • Plasmolysis
  • Osmosis
  • Selective permeability
  • Hemolysis



  • Discuss the flow of water across various types of membranes.
  • Relate how factors such as charge and size influence transportation.
  • Differentiate between active and passive transport, and why each is employed in cells.
  • Learn how to define and correctly use the terms hypertonic, hypotonic and isotonic.




  •  10% Sucrose solution                                      
  • Iodine
  • 20% Glucose solution        
  • Benedict’s reagent
  • 50% Sucrose solution          
  • Dialysis tubing (5)       
  • 2% Starch solution                                   
  • String
  • Deionized water                    
  • p-1000 micropipette and tips
  • 100 ml beakers (5)   
  • 60°C water bath
  • Test tubes (2)        
  • Test tube rack
  • Scale


1. What factors determine the permeability of cell membranes?
2. How does the concentration gradient influence the passive movement of materials across cell membranes?
3. Explain the driving force for the movement of uncharged compounds across cell membranes.
4. What is the role of electrical potential difference in the movement of ions across cell membranes?
5. Describe the structure and function of membrane channels in facilitating the movement of specific molecules.
6. How do membrane carriers participate in active and passive transport across cell membranes?
7. What is the difference between passive transport and active transport?
8. Define iso-osmotic, hyperosmotic, and hypo-osmotic solutions and their effects on cells.
9. Explain the concept of percent solution and how it is used to express the concentration of solutes.
10. Differentiate between molarity and molality as measures of solute concentration and their relevance in physiological systems.


Exercise 1: Testing Permeability


  1. Take one piece of dialysis tubing, twist and fold one end over, and tie it tightly with cotton string.
  2. Fill the tube with 1 ml glucose solution and 1 ml starch solution.
  3. Fold over the top of the tube, squeeze the air out, and tie it tightly with string. The tube should be limp.
  4. Place the tied-off tube in a 100 ml beaker half filled with distilled water. Let sit for one hour.
  5. At the end of the hour, test the water in the beaker for glucose (a simple sugar). Remove 1 ml of the water from the beaker and place in a test tube. Add 1 ml Benedict’s reagent to the test tube and place in the water bath for 5 minutes.
  6. Test the water in the beaker for starch by removing 1 ml of water from the beaker and placing it into a second test tube. Add 1 ml of iodine as an indicator.
Table 1: Testing Permeability
Test Result


Exercise 2: The Effect of Concentration Gradients on Diffusion Rates


  1. Obtain four pieces of dialysis tubing. Fold and tie one end of each tube as in Exercise 1. Fill two of the tubes with 1.5 ml distilled water. Fill the third tube with 1.5 ml of 10% sucrose solution, and the fourth tube with 1.5 ml of 50% sucrose solution.
  2. Fold the tops of the tubes, squeeze out the air, and tie them tightly with string.
  3. Weigh each tube to determine its mass and record it under time zero on the data table.
  4. Place the tube containing 10% sucrose solution in a beaker half filled with distilled water. Label this beaker A.
  5. Place the tube containing 50% sucrose solution in a beaker half filled with distilled water and label this beaker B.
  6. Place one of the tubes containing water in a beaker filled one-third with 50% sucrose solution and label this beaker C. Place the remaining tube in a beaker filled one-third with water and label this beaker D.
  7. Weigh each tube every 15 minutes for one hour. Record the masses in the data table (Table 2).
Table 2: Effect of Concentration Gradient on Diffusion


Beaker Contents Mass (g) O min. Mass (g) 15 min. Mass (g) 30 min. Mass (g)

45 min

Mass (g) 60 min.

10% Sucrose


50% Sucrose

Water 50% Sucrose





  1. In the Testing Permeability exercise, do you think any of the water from the beaker moved into the bag? Explain.
  2. Does the water in the beaker now contain glucose? Explain why or why not.
  3. Does the water in the beaker now contain starch? Explain.
  4. Why did some of the bags lose weight and some gain weight?
  5. How does the concentration gradient affect the rate of diffusion?
  6. Explain why sucrose is used instead of glucose in this experiment to determine the effect of concentration gradient on diffusion rate.

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Biology I Cellular Processes Laboratory Manual by The authors & Hillsborough Community College is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.

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