C



Diffusion and Osmosis (lab 2.6)

Background information on diffusion and osmosis

The background information for understanding this lab is in lab 2.6. Although we will use different procedures than the ones described in lab 2.6, you should read lab 2.6 and answer the lab report questions in the lab report section.

Procedure A) Osmosis across an artificial semipermeable membrane

Read section B of lab 2.6 for background information on this activity, but use the following procedure instead:

You will make 3 “cells” using dialysis tubes. One cell will be filled with a solution of 15% sucrose (sucrose is a small sugar molecule), one will be filled with a solution of 30% sucrose, and one cell will be filled with a solution of 15% albumin (albumin is a large protein molecule). Percent concentration means grams of the solute per 100 ml of solution. After filling each cell with a solution, the cells will be placed in beakers of water. Each cell will gain weight as water moves into the cell by osmosis.

1) Obtain three 5.5 inch sections of dialysis tubing, six plastic clamps for the

tubes, and three 300 ml beakers.

Using a wax pencil, label the beakers: 15% sucrose, 15% albumin, and

30% sucrose. Fill each beaker with 200 ml water.

2) Into each beaker of water place one dialysis tube. Let the tubes soak in the

water for 5 minutes.

3) After 5 minutes, remove the dialysis tube from your 30% sucrose beaker. By gently massaging the top end of the tube with your fingers, open that end.

Clamp the bottom end of the tube shut by first folding that end in half and then applying a plastic clamp to the fold. Your instructor has prepared a demonstration clamped dialysis tube to help you with this step.

Carefully fill the tube with 25 ml of 30% sucrose solution. After adding the sucrose to the tube, fold the open end of the dialysis tube in half (expelling as much air as possible) then clamp it shut. This dialysis tube represents a “cell” with a solute concentration of 30 grams per 100 ml. If done properly, your tube should not leak and should look the same as the instructor’s demonstration tube.

Using paper towels, gently pat dry the “cell”. Try to remove all drops of water from the surface of the tube and from the plastic clamps. Take the cell to the

electronic balance, tare (zero) the balance, then place the cell on the balance and record its mass in the results table on page 3.

Place the cell into the “30% Sucrose” beaker of water. Be sure that the cell is fully

submerged under the water.

4) Repeat step 3 with the dialysis tube soaking in your 15% sucrose beaker, but fill it with the 15% sucrose solution and return it to the 15% sucrose beaker.

5) Repeat step 3 with the dialysis tube soaking in your 15% albumin beaker, but fill it with the 15% albumin solution and return it to the 15% albumin beaker.

6) Let the three dialysis tube “cells” remain in the water for 30 minutes.

7) After 30 minutes, remove the 30% sucrose cell. Use paper towels to gently pat dry the cell. Again, try to remove all drops of water from the surface of

the tube and from the plastic clamps.

Tare (zero) an electronic balance. Place the cell on the balance and record its

mass in results table A.

8) Repeat step 7 with the 15% sucrose cell and the 15% albumin cell.

9) In the space below, answer the following questions:

a) Which tube had the highest weight gain by osmosis? ______

b) Explain why that tube had the highest weight gain:

c) Of the two 15% tubes, which had the highest weight gain by osmosis?

______

d) Explain why the two 15% tubes did not have the same amount of osmosis,

even though they were both have the same solute concentration (15% solute):

To answer questions b and d, you will need to convert the % concentration units of the three solutions into osmolarity (OsM) concentration units. This conversion has been done for you. The 30% sucrose has an osmolarity of 0.88 OsM, the 15% Sucrose has an osmolarity of 0.44 OsM, and the 15% Albumin has an osmolarity of 0.015 OsM.

Results Table A:

Conc. Conc. Mass (grams) Mass (grams)

(%): (OsM) before osmosis after osmosis Mass increase

30% sucrose

15% sucrose

15% albumin

Procedure B) Solubility of compounds in Polar and Non-polar solvents

Read section A of lab 2.6 for background information on this activity, but use the following procedure instead:

1) Obtain a test tube from your physiology lab basket.

2) Add 10 ml water and 2 ml corn oil to the test tube. They do not mix because

one is hydrophilic and one is hydrophobic.

Which substance forms the top layer? ___ Which forms the bottom layer? ___

The molecular formula for water is H2O. A typical molecular formula for corn oil

triglyceride is C57H105O6. Judging from the molecular formulas, which substance

is hydrophilic? _______ Which is hydrophobic? _______

3) Add 2 drops of Methylene blue dye and one rice-grain size amount of Sudan red

dye. One of the dyes is hydrophobic and one is hydrophilic. Put a rubber stopper

into the test tube and then mix the tube well by shaking it vigorously for a minute.

At first the tube will be maroon color (red + blue), but as the oil and water slowly separate, you will see the two colored dyes separate also. This is because

of the “like mixes with like” principle: The hydrophobic dye moves into the

hydrophobic solvent and the hydrophilic dye moves into the hydrophilic solvent.

4) Let the test tube stand until end of period. This will allow the dyes to fully

separate. When they have fully separated, answer these questions:

Which dye is hydrophobic? ________ Which is hydrophilic? ________

5) When done, wash the test tube using test tube brush in your basket and the

Dawn™ soap on the counter. Remember, it’s always clear and bright when it’s

Dawn™

Procedure C) Concentration and tonicity

Read section C of lab 2.6 for background information on this activity.

1) On the countertop are three test tubes (A, B, and C). Each tube contains a drop of

sheep’s blood in salt solution. The salt solution in one tube is hypotonic, the

salt solution in one tube is isotonic, and the salt solution in one tube is hypertonic.

2) Before the lab started, a sample from each of the three test tubes was placed on a microscope slide for viewing. The samples from each of the test tubes can be

viewed under microscopes A, B, and C.

3) View each slide under the microscopes. Because the cells were in different osmotic solutions, their shapes will be different (due to water loss or water gain), as described below:

Normal shape = isotonic solution

Crenated (shriveled shape) = hypertonic

No cells = hypotonic (due to hemolysis (blood cell tearing)).

4) Judging from the shapes of the cells, fill in the following table with either A, B, or

C. Show your instructor your results when done.

Hypotonic solution = _______

Isotonic solution = _______

Hypertonic solution = _______

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