Introduction: 
 Cells have kinetic energy.  This causes the molecules of the cell to move around and bump into each other.  Diffusion is one result of this molecular movement.  Diffusion is the random movement of molecules from an area of higher concentration to areas of lower concentration.  Osmosis is a special kind of diffusion where water moves through a selectively permeable membrane (a membrane that only allows certain molecules to diffuse though).  Diffusion or osmosis occurs until dynamic equilibrium has been reached.  This is the point where the concentrations in both areas are equal and no net movement will occur from one area to another. 
If two solutions have the same solute concentration, the solutions are said to be isotonic.  If the solutions differ in concentration, the area with the higher solute concentration is hypertonic and the area with the lower solute concentration is hypotonic.  Since a hypotonic solution contains a higher level of solute, it has a high solute potential and low water potential.  This is because water potential and solute potential are inversely proportional.  A hypotonic solution would have a high water potential and a low solute potential.  An isotonic solution would have equal solute and water potentials.  Water potential (y) is composed of two main things, a physical pressure component, pressure potential (yp), and the effects of solutes, solute potential (ys).  A formula to show this relationship is y = yp + ys.  Water will always move from areas of high water potential to areas of low water potential.
The force of water in a cell against its plasma membrane causes the cell to have turgor pressure, which helps maintain the shape of the cell.  When water moves out of a cell, the cell will loose turgor pressure along with water potential.  Turgor pressure of a plant cell is usually attained while in a hypotonic solution.  The loss of water and turgor pressure while a cell is in a hypertonic solution is called plasmolysis.
Hypothesis: 
 During these experiments, it will be proven that diffusion and osmosis occur between solutions of different concentrations until dynamic equilibrium is reached, affecting the cell by causing plasmolysis or increased turgor pressure during the process.
Materials: 
Lab 1A - To begin Lab 1A, first collect the desired equipment.  The materials needed are dialysis tubing, Iodine Potassium Iodide (IKI) solution, 15% glucose/ 1% starch solution, glucose Testape or Lugol’s solution, distilled water, and a 250-mL beaker.
Lab 1B - For Lab 1B you will need to collect six presoaked dialysis tubing strips, distilled water; 0.2M, 0.4M, 0.6M, 0.8M, and a 1.0M sucrose solution; six 250-mL beakers or cups, and a scale.
Lab 1C - Lab 1C these items are needed: a potato, knife, potato core borers, six different solutions, and a scale.
Lab 1D - During Lab 1D, only paper, pencil, and a calculator will be needed to make the calculations.
Lab 1E - n Lab 1E these items are needed: a microscope slide, cover slip, onion cells, light microscope, and a 15% NaCl solution. 
 Procedures:
Lab 1A - After gathering the materials, pour glucose/starch solution into dialysis tubing and close the bag.  Test the solution for presence of glucose.  Test the beaker of distilled water and IKI for presence of glucose.  Put the dialysis bag into the beaker and let stand for 30 minutes.  When time is up test both the bag and the beaker for presence of glucose.  Record all data in table.
Lab 1B - Obtain the six strips of dialysis tubing and fill each with a solution of a different molarity.  Mass each bag.  Put each bag into a beaker of distilled water and let stand for half and hour.  After 30 minutes is up, remove each bag and determine its mass.  Record all data in its appropriate table.
Lab 1C - sing the potato core borer, obtain 24 cylindrical slices of potato, four for each cup.  Determine the mass of the four cylinders.  Immerse four cylinders into each of the six beakers or cups.  Let stand overnight.  After time is up, remove the cores from the sucrose solutions and mass them.  Record all data in its appropriate table.
Lab 1D - Using the paper, pencil, and calculator collected, determine solute potentials of the solutions and answer the questions asked to better understand this particular part of the lab.
Lab 1E - Using the materials gathers, prepare a wet mount slide of the epidermis of an onion.  Draw what you see of the onion cell under the microscope.  Add several drops of the NaCl solution to the slide.  Now draw the appearance of the cell.  
Data:
Lab 1A - Table 1.1

 Lab 1A Questions
1)     Glucose is leaving the bag and Iodine-Potassium-Iodide is entering the bag.  The change in color of the contents of the bag and the presence of glucose in the bag prove this.
2)     In the results, the IKI moved from the beaker to the bag, this caused the change in the color of the bag.  The IKI moved into the bag to make the concentrations outside the bag equal to inside the bag.  The glucose solution moved out of the bag making glucose present in the beaker.  The glucose moved to make the solute concentration inside and out of the bag equal.
3)     If the initial and final percent concentration of glucose and IKI for in the bag and the beaker were given, they would show the differences and prove the movement of these substances to reach dynamic equilibrium.
4)     Based on my observations, the smallest substance was the IKI molecule, then the glucose molecules, water molecules, membrane pore, and then the starch molecules being the largest.
5)     If the experiment started with glucose and IKI inside the bag and starch in the beaker, the glucose and IKI would move out of the bag to make the concentrations equal, but the starch could not move into the bag because its molecules are too big to pass through the semipermeable membrane.

Lab 1B -  Table 1.2   Dialysis Bag Results

Table 1.3     Dialysis Bag Results: Class Data

Lab 1B Questions:
1)     The molarity of the sucrose in the bag determines the amount of water that either moves into or out of the bag, which changes the mass.  For example, when the bag contained a 0.2M solution, water entered the bag to make the concentrations inside and outside of the bag more equal.  As this happened, the mass rose 1.2g
2)     If each of the bags were placed into a 0.4M solution instead of distilled water, the masses of the bags would have changed in different ways.  The mass of the bags filled with distilled water and 0.2M sucrose would have gone down because water would have left the bag.  The mass of the 0.4M bag would have stayed the same because the concentrations are now equal.  The masses of the 0.6, 0.8, and 1.0M bags would have increased because water would have moved into the bag to equalize the concentrations.
3)     In the data collected, the percent change in mass was calculated to show how greatly the mass increased or decreased.  The difference in mass is not enough to go by because the initial masses of the dialysis bags were not all the same.
4)     If a dialysis bag’s initial mass was 20g and it’s final mass was 18g, the percent change in mass is 20%.
5)     The sucrose solution in the beaker would have been hypotonic to the distilled water in the bag.
Lab 1C        Table 1.4

Lab 1C        Table 1.5     Class Results