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Writing a Laboratory Report

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Writing a Laboratory Report Powered By Docstoc
					                          Billy muse
                          AP Bio 4th hour
                          Group number 1
                          9/17/02




Osmosis/Diffusion Lab 1
      Lab 1A - Diffusion
   I.   Background - Diffusion is defined as the random mixing of molecules or atoms due
        to there constant motion and impact.

   II.        Purpose - To investigate diffusion through a semi-permeable cellulose dialysis
              membrane and the effect of size on this diffusion.


   III.       Hypothesis – Large polymers, such as starch, will not be able to diffuse (effuse)
              through the finite size holes in the membrane. Smaller molecules may be slowed
              down, but not prevented from diffusion through.

   IV.        Procedure - Imbibe a dialysis membrane and tie off one end. Add 15-20 mls of a
              15% glucose/1% starch solution to the bag and tie of the end leaving some air space
              for the tube to expand due to osmosis. Soak the dialysis bag in distilled water
              containg 3-4 mls of iodine/iodide solution.

   V.         Results – The iodine diffuses into the bag and interacts with the starch to create a blue
              complex. After 30 minutes no glucose is detected outside the bag nor is their any
              blue starch complex outside the bag.

Data tables              Outside bag             Inside bag        Benedicts assay (out) Starch assay I2
Initial                  Water, iodine        Glucose, starch      negative              negative
30 mins                  water                Glucose, starch,     negative              Blue complex in
                                            iodine                                       bag
24 hours                  Some glucose      Glucose, starch        Slightly positive     +
48 hours                  glucose           Glucose, starch        positive              +

   VI.        Observations - Starch stayed in the bag as no blue complex was noticed outside the
              bag. Glucose was slowed by the bag, but the benedict’s test showed that some of the
              glucose left the bag over time.

   VII. Conclusion (What does all this mean?) – The cellulose dialysis membrane mimics the
        typical cell wall and lipid bilayer. It selectively lets small molecules out but retains
        large molecules such as starch.
   Questions from this section
   1) Iodine and water enter the bag. The bag swells slightly and the starch complexes with iodine to form a
      blue precipitate. Glucose eventually leaves the bag as the benedic’s assay after 24 hours shows a
      positive test for sugar.
   2) The pore size of the membrane must be slightly larger than the glucose.
   3) Yes, weigh the bag periodically
   4) Rank: small to large water, iodine (IKI), glucose, membrane pores, starch.
   5) I would expect the outside water to turn blue as the IKI left the bag.
   Exercise 1 B: Osmosis

   Background - Osmosis is defined as the movement of water down a concentration gradient
           across a semi-permiable membrane

   Purpose - To observe osmosis in a controlled setting

   Hypothesis - The bags will shrink in hypertonic solution, but there will be no net change in
           the isotonic distilled water.

    Procedure - Add solutions of different concentrations sucrose to 6 dialysis membranes and
              put the bags into distilled water.

    Results. The initial mass increased in all bags except the one filled with distilled water.

Contents in bag        Initial mass           Final mass             difference             % change
ddH2O                  5g                     5g                     0                      0
0.2 M sucrose          5g                     5.35 g                 .35 g                  5%
0.4                    5                      5.6 g                  .6 g                   10%
0.6                    6                      6.85 g                 .85 g                  15%
0.8                    6                      8g                     2g                     25%
1.0 M                  6                      10 g                   4g                     50%

                      Include graph of data here (recommend excel)

Observations: All the concentrations above 0 M sucrose were in hypotonic medium when in
              the beaker of distilled water, thus water rushed in to equalize the water
              concentrations.

Conclusions - Osmosis is a powerful force and can be quantitativally determined by measuring
            the mass of water change over a given time.

Questions from lab handout 1 b
           1) The higher the molarity sucrose, the greater the mass change. More water would be needed to
              equalize the concentrations
           2) If the bags were placed in 0.4 molar sucrose instead of distilled water,. Then the distilled water
              and 0.2 molar bags would shrink as water left them ((hypertonic) , the 0.4 molar bag would stay
              the same, and the 0.6 – 1.0 molar will still swell.
           3) Because all the bags start out at slightly different masses, we looked at the % change to
              normalize the data to the starting weight.
           4) 20-18/20 x 100 = 10%
Exercise 1c Water potential

Background - The driving force behind osmosis is water potential. Water tends to go
     from high concentration to low concentration.

Purpose - to determine the water potential in a potato cell. In other words to determine
             the iso-osmolarity within the cell to outside conditions.

Hypothesis - The water potential in the cell will be greater than a 0.2 M sucrose solution,
             but less than a 1 M sucrose solution. In hypertonic medium the potato cells
     will shrink as water leaves them. In hypotonic medium the cells will swell. In
     isotonic media there will be no net flow and this will be the concentration to
     calculate the water potential from.

Procedure - Each lab group makes up a slightly different concentration of sucrose water.
            Each group gets a chunk of potato containing many thousands of cells to
            soak in their concentration of sucrose water.

Results - Class data looked pretty rocky - (see Muse’s handout). This is probably due to
            Class only soaking for ½ hour or less.

             Use and plot Muse’s data from the handout.




Observations – The point at which the plot of solution concentration vs mass change
            crosses the 0 line (y intercept) is 0.37 M

Conclusion - The concentration of water in the potato cell is equal to the concentration of
            water in 0.37 M sucrose solution. This is the iso-osmolar point.
Exercise 1D
      Background – The formula to calculate solute potential is s = - iCRT
                     Water potential and solute potential will be equal.
      Purpose – To use the data from 1C to calculate water potential from a number of
               different cells.

      Hypothesis – I believe that different types of cells will have similar water
               potentials.

      Results - for potato the iso-osmolar concentration was solute to 0.37 molar

                   s = - iCRT       ( - 1) (0.37 M) (0.0831 liter bar/mole K) (296 K)

                   s = - 9.1 bars

Question 1 from 1d: Would water potential increase or decrease if potato was
dehydrated? It would decrease because the concentration of solute rises as water
decreases

          2. Hypotonic to its environment, it will swell.

          3. p = 0

          4. beaker with 0.4% sucrose has lower water potential

          5. Water will diffuse out of the bag

Zucchini is 0.37 iso-osmolar. Molar concentration of solutes = 0.37 M

Question 8 - same as calculation of potato shown above
s = - iCRT     ( - 1) (0.37 M) (0.0831 liter bar/mole K) (296 K)
                                      s = - 9.1 bars
Question 9 Adding more solute to the Zucchini cells will decrease the water potential of
            the Zucchini plugs.

Question 10 a. ddH2O has a higher concentration of water molecules
         10 b. ddH2O has a higher water potential
          10 c. The RBC would swell and burst if put in distilled water. Water would
          rush into it.
Exercise 1E Onion Cell Plasmolysis

      Background – Onion cells can be easily seen under 20 -100 fold magnification.
                 The cellulose cell wall will remain constant sized but the internal
                 membrane can shrink.

      Purpose – To visibly watch osmosis in a live cell

      Hypothesis - Onion cells will shrink in 15% salt solution (hypertonic)

      Procedure – Take a thin slice of onion and put a cover slip over it. Focus a cell at
            100 x magnification and watch membrane. Put a droip of 15% salt on the
            edge of slide and draw it in with capillary action with a paper towel.

      Results:
            1. Cell wall and internal membrane very close to each other
            2. The internal membrane has peeled away from cell wall and cell has
               shrunk.
            3. The membrane returns to close to wall when fresh water is added.
               The cell has swelled. Over time, the cell and wall bursts.

      Analysis of results questions:
      1. What is plasmolysis? It is the pulling away from the wall of the plasma
         membrane due to the osmotic shrinking of the cell.

      2. The onion cell when added to ddH2O was forced to swell because the water
         potential inside was less than that outside. Water followed an osmotic gradient
         and burst the cell.

      3. Grass that is in a salty environment cannot overcome the tendency of the water
         to leave the roots by going from lower water potential to higher water potential.
         The salt effectively makes the soil hypertonic compared to the inside of the
         grass roots. The cells shrink and the plants die.



What have I learned from Lab exercise 1a – 1e.
       Water tends to move from areas of high concentration (potential) to low
concentration. Plasma membranes are semi-permeable. They let water cross easily, but
not large solutes such as starch polymers. Small molecules such as glucose or salt ions
can cross, but not as fast as water. Some plant cells have internal solute concentrations
equal to 0.37 molar. Therefore, their water concentrations are less than that of ddH2O
and will swell if put into fresh water.

				
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