Diffusion and Osmosis
Atoms and molecules are in constant motion. This kinetic energy causes the molecules to collide and move in different directions. This movement is the fuel for diffusion. Diffusion is the random movement of molecules from an area of higher concentration to an area of lower concentration. This happens until the two areas reach dynamic equilibrium. When this dynamic equilibrium is reached, the concentration of the molecules is approximately equal, and after this point there is no net movement of the molecules. The molecules are still moving, but the concentrations remain the same.
Osmosis is a special type of diffusion in which water moves across a selectively permeable membrane. A selectively permeable membrane allows diffusion of only certain solutes (the substance that dissolves) and water, the most common solvent (a substance that dissolves). The most common selectively permeable membrane is the cell membrane. Water moves from an area of high water potential to an area of low water potential. Water potential is a measure of the free energy of water in a solution and is represented by the symbol ψ (psi). The water potential is affected by two physical factors: the addition of a solute (ψs) and the pressure potential (ψp). Adding solutes to a concentration lowers the water potential of that solute, causing water to enter the area. The movement of water is directly proportional to the pressure potential. The water potential can be determined by the equation:
ψ = ψp + ψs
Pure water has zero water potential. Adding solutes results in a negative water potential reading, while increasing the pressure potential results in a more positive water potential reading.
There are three relationships that can occur between two solutions. When two solutions have equal solute concentrations, they are isotonic and there is no net movement of solutes. There is also no net movement of water. If the two solutions differ in solute concentrations, they are hypertonic or hypotonic. The hypertonic solution has a lower concentration of solutes. Water will move out of a hypertonic solution as solutes flow out (increasing concentration gradient is similar to water potential). This depends on the selective properties of the membrane. In cell terms, this causes the cell to shrink or become flaccid. The hypotonic solution has a higher concentration of solutes and therefore less water. This solution gains water while losing solute. This movement between hypotonic and hypertonic solutions continues until the point of dynamic equilibrium is reached. A hypertonic cell can also undergo plasmolysis. Plasmolysis is the contraction of the cytoplasm in a plant cell in response to the diffusion of water out of the cell. When a cell is hypotonic, it can lyse. In plant cells, it creates a turgor pressure against the cell walls that prevents the plant from wilting.
Besideosmosis and diffusionMolecules and ions can move by active transport. This process involves the use of ATP to drive molecules into or out of a cell. Active transport is generally used to move molecules against a concentration gradient from a region of lower concentration to a region of higher molecular concentration.
In this experiment, diffusion and osmosis occur until dynamic equilibrium is reached. This experiment is carried out under theoretical conditions, with no other variables than the water potential that influence the movement of solutes.
For this exercise, you will need 30 cm of 2.5 cm dialysis tubing, a 250 ml beaker, distilled water, 2 dialysis tubing clamps, 15 ml of 15% glucose/1% starch solution, 4 pieces of glucose tape, 4 ml of Lugol's solution (Iodine Potassium Iodide or IKI) and watch or stopwatch.
This experiment requires six 30-cm strips of dialysis tubing, a 250-mL beaker, 12 dialysis tubing clamps, six cups of distilled water, a balance, a stopwatch or clock, paper towels, and approximately 25 mL of each of these solutions: distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 M glucose, and 1.0 M glucose.
This experiment requires a large potato, a potato corer (about 3 cm long), a 250 mL beaker, a paper towel, a scale, six cups, a knife, and approximately 100 mL of each of these. solutions: distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 M glucose, and 1.0 M glucose.
For this experiment you will need a calculator, paper, pencil, and graph paper.
This experiment requires onion skins, dye, microscope, slides, coverslips, salt water (15%), and tap water.
First, soak the dialysis tubing in distilled water for 24 hours. Remove the hose and tie one end with the clamp (twist the end of the hose and fold it back on itself about 7 times, slide it into the clamp). Next, open the other end of the tube (rub the end between your fingers) and fill it with the glucose/starch solution. Use the glucose band and observe the change in color of the band and the color of the sac. Tie off end with hose clamp (leave slack but no air). Fill the beaker with distilled water and add the 4 mL of Lugol's solution, noting the color change. Use the glucose tap to test the glucose in the water (recording). Place the dialysis tubing in the beaker and let it sit for about 30 minutes. Remove the bag and observe the change in the water and the color of the bag. Use the last two pieces of glucose tape to measure the glucose in the water and the bag and record the results.
First, soak the dialysis tubing for about 24 hours. Tie one end of each hose with the clamps. Next, fill each tube with a different solution (distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 M glucose, and 1.0 M glucose) and tie off the end (leave space, but not air). Weigh each tube individually and record the masses. Soak the tubes in separate beakers filled with distilled water for about 30 min. Remove the tube, dry it, weigh it again, and record the mass.
First cut the potato into 3 cm slices. Use the potato corer and core 24 seeds (you don't get any). Weigh 4 nuclei together and record the mass. Fill each beaker with a different solution (distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 M glucose, and 1.0 M glucose). Put 4 potato seeds in each cup and leave them overnight. Remove the seeds and dry. Note the change in mass. Calculate the information and compare.
First, determine the solute potential of the glucose solution, the pressure potential, and the water potential. Then record the information about the zucchini seeds graphically.
First, prepare a wet slide of stained onion peel. Look under a light microscope and draw what the cells are. Add a few drops of the saline solution, observe and draw the change.
tabla 1Dialysis tube and cup color change
Presence of glucose (glucose band)
|15% glucose/1% starch||light white||Indigo||dark brown||mahogany|
|Water + IKI||Bernstein||Bernstein||teal yellow||tan green|
Which substances go into the bag and which come out of the bag? What experimental evidence supports your answer?
The glucose slowly comes out of the bag; This is indicated by the glucose band on the cup. iodine penetrates the dialysis bag; this is indicated by the color change in the pocket. the water enters the bag; the bag is getting thicker, which shows this.
Explain the results you got. Include differences in concentration and membrane pore size in your discussion.
The substances moved out of the bag according to the gradient. Some were small substances and went in and out of the pocket quickly and easily. Larger substances, such as starch and glucose, were slow or did not go in or out of the bag.
Quantitative data uses numbers to measure observed changes. How could this experiment be modified so that quantitative data could be collected to show that water diffuses into the dialysis bag?
The mass of the bag could be recorded before and after soaking.
Based on your observations, rank the following by relative size, starting with the smallest: glucose molecules, water, IKI, membrane pores, and starch molecules.
Water→ IKI→ glucose molecules→ membrane pores→ starch molecules
What results would you expect if the experiment started with a solution of glucose and IKI in the bag and just starch and water outside? Why?
The IKI would have left the bag and changed the color of the solution in the beaker. Eventually dynamic equilibrium would be reached and there would be no net motion.
Tabla 2Dialysis bag results: individual data
Dialysis bag content
Percent mass change*
a) Distilled water
b) 0.2 million
c) 0.4 million
d) 0.6 million
e) 0.8 million
f) 1.0 million
* To calculate: percentage change in weight = final mass – initial mass
Initial mass * 100
Dialysis Bag Outcomes: Class Data
a) Distilled water
|0,77 %||1,53 %||0,83 %||1,04 %|
b) 0.2 million
|1,86 %||5,30 %||1,9 %||3,02 %|
c) 0.4 million
|2,4 %||2,22 %||2,2 %||2,27 %|
d) 0.6 million
|12,54 %||9,75 %||11,8 %||11,36 %|
e) 0.8 million
|13,07 %||9,64 %||12,3 %||11,67 %|
f) 1.0 million
|16,55 %||18,98 %||16,9 %||17,48 %|
|tripp y stephanie||Hudgens and Kris||Elizabeth and Julia|
Percent change in dialysis tubing mass in glucose solutions of different molarities
Explain the relationship between the change in mass and the molarity of sucrose in dialysis bags.
They are directly proportional. The percentage change in mass increases with increasing molarity of the sugar.
Predict what would happen to the mass of each bag in this experiment if all the bags were placed in 0.4 M sucrose solution instead of distilled water. Justify your answer.
There is no net movement when there is 0.4 M in the dialysis bag. If the concentration is higher than 0.4, the bag will lose water. If the concentration is less than 0.4, the bag will absorb water.
Why did you calculate the percent mass change and not just use the mass change?
The volumes of the solutions were not exactly equal.
A dialysis bag is filled with distilled water and then placed in a sucrose solution. The initial mass of the bag is 20 g and the final mass is 18 g. Calculate the percent mass change and show your calculations in the chart below.
Weight Change Percentage = Final Mass - Initial Mass x 100 = 18-20 x 100 = 10%
See fair 20
The sucrose solution in the beaker would have been hypertonic.to distilled water in the bag.
Tabla 4Potato core: individual data
initial time (g)
Final time (g)
% change in mass
a) Distilled water
b) 0.2 million
c) 0.4 million
d) 0.6 million
e) 0.8 million
f) 1.0 million
Tabla 5Potato seed results: class data
|16,7 %||28,5 %||45,2 %||22,6 %|
0,2 M Sacarosa
|13,3 %||21,4 %||34,7 %||17,35 %|
0,4 M Sacarosa
|20,0 %||14,28 %||34,28 %||17,14 %|
0,6 M Sacarosa
|-18,75 %||-20,0 %||-38,75 %||-19,38 %|
0,8 M Sacarosa
|-21,4 %||-26,66 %||-48,06 %||-24,03 %|
1,0 M Sacarosa
|-18,75 %||-21,42 %||-40,17 %||-20,09 %|
Percent change in mass of potato kernels at different glucose molarities
If a potato were left to dry in the open air, would the water potential of the potato cells decrease or increase? Why?
The water potential of the cells decreases, the osmotic potential decreases, and the solute increases. This happens because the cells are dehydrated.
If a plant cell has a lower water potential than its surroundings and the pressure is zero, is the cell hypertonic or hypotonic with respect to its surroundings? Does the cell absorb or lose water? Justify your answer.
The environment is hypotonic, so it will gain water. This is because it contains less water than the surrounding area.
In Figure 1.5 the cup is open to the atmosphere. What pressure potential does the system have?
The pressure potential is zero.
Where is the greatest water potential in Figure 1.5?
The dialysis bag.
the water diffusesthe pocket. Why?
The water diffuses because there is a greater potential for water inside the bag.
Zucchini seeds immersed in sucrose solutions at 27 °C showed the following percentage changes after 23 hours:
% change in molarity of sucrose by mass
20% distilled water
-3% 0.4 Million
-17% 0.6 Million
-25% 0.8 million
-30 % 1,0 Mio
Percent change in the mass of zucchini kernels from sucrose solutions of different molarities
b)What is the molar concentration of solutes in the zucchini in the cells?
about 0.36 million
See Procedure for Calculating Water Potential from Experimental Data.
Calculate the solute potential (ψs) of the sucrose solution in which the mass of the zucchini seeds does not change. Show work.
ψs = -8,580075
ψ =0+ ψs
Calculate the water potential (ψ) of the solutes in the zucchini seeds. show work here.
ψ = ψs + ψp
-8,580075 = ψs + 0
-8,580075 = ψs
What effect does the addition of solutes have on the potential component of this solution? Why?
Adding more solute increases the solute potential and decreases the water potential making it more negative.
Consider what would happen to a red blood cell placed in distilled water:
a)Which would have the highest concentration of water molecules?
b)Which would have greater water potential?
C)What would happen to the red blood cells? Why?
The red blood cells would absorb water and lyse.
Prepare a wet slide from a small piece of epidermis from an onion. Observe at 100x magnification. Draw and describe the appearance of the onion cells.
Describe the appearance of the onion cells after adding NaCl.
The plasma membrane contracted from the cell wall causing plasmolysis.
Remove the coverslip and flood the onion with fresh water. Observe and describe what happened.
Onion cells absorbed water and increased turgor pressure.
What is plasmolysis?
Plasmolysis is the contraction of the cytoplasm of a plant cell in response to the diffusion of water out of the cell into a hypertonic solution surrounding the cell.
Why does the onion cell plasmolyze?
The environment became hypertonic for the cell, and water was forced out of the cell by the salt with its concentration gradient. With all the water flowing out of the cell, it shrank, leaving behind its cell wall.
In winter, grass often dies off near roads that have been salted to remove ice. What does that do?
Salt causes plant cells to plasmolyze.
One error that may have occurred in this experiment is that some of the glucose/starch solution may have leaked into the beaker before inserting the dialysis bag but after the glucose test.
An error that may have occurred in this experiment is that the sugar did not fully mix with the water, or some sugar was lost during mixing. Another mistake could be that the bags were not inflated enough.
An error that may have occurred in this experiment is that the sugar did not fully mix with the water, or some sugar was lost during mixing. Another mistake could be that the potato seeds did not puff up enough. Also, the measurement of liquid volumes may not have been accurate.
Mistakes that may have been made in this exercise could be entering the numbers incorrectly into the calculator.
One mistake that could have been made in this exercise was that the skin of the onion could have dried out before adding the salt water, which affected the results.
Discussion and conclusion:
In this experiment, the ability of substances to move across a selectively permeable membrane was examined. A glucose/starch solution was placed in the dialysis bag. Glucose molecules leaked out of the bag (learned before and after the glucose band test). Using IKI to test for starch, the color change of the bag only indicates that the starch molecule was too large to escape the dialysis bag, but the IKI molecule was small enough to enter the bag.
In this experiment, the study of hypotonic and tested hypertonic solution. When the bags were placed in a hypotonic solution, they absorbed water. This could be observed by massaging the bags before and after soaking.
In this experiment it was found that potato seeds contain some sugar. When placed in low-sugar or no-sugar environments, they ingested water. In high-sugar environments, the beans lost water.
In this experiment all these conclusions from previous experiments were reinforced with scientific equations.
In this experiment, the turgor pressure of the plasma membrane of a plant cell was observed. The plasma membrane gives shape and form to the cell. When salt is added, the cytoplasm loses water and shrinks the plasma membrane. This causes the plant to wilt.
The concentration gradient and water potential affected all aspects of this experiment. Many scientists use water potential to study the effects of various substances on plants, either for good or for ill. Pressure potential and dissolved potential are the two main components. Water moves to different areas depending on the water potential. Water moves from high water potential to low water potential or low to high water potential. This change is called a gradient. Ψs = -iCRT is the formula used for solute potential. The water potential and the dissolved potential are inversely proportional. When the water potential is increased, the solute potential is low. All of these factors influenced the results of these experiments.
Plant and animal cells respond differently to different environments. A hypertonic solution causes an animal cell to shrink in size. A hypotonic solution for an animal cell causes its lysis. Isotonic solutions are ideal for animal cells. The plant cell plasma membrane shrinks when in a hypertonic solution, causing the cell wall to lose shape or become plasmolyzed. While an isotonic solution is good, it does not provide enough support for the cell wall. Hypotonic solutions are better for plant cells. The plasma membrane presses against the inside of the cell wall and gives it a lot of support. Without this pressure, turgor pressure, the plant will wither and die. These experiments allow us to better understand and care for living things, including our own bodies.
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