Lab #2: Permeability of Cell Membranes

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Lab #2: Permeability of Cell Membranes

Introduction

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

In general, biological membranes are far less permeable than an equivalent thickness of pure solvent. 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.

Here we will investigate the effects of lipid solubility and molecular weight (i.e. size) on the permeability of cell membranes. Our model system will be red blood cells. Since a membrane is largely phospholipid, the permeability of a substance (in the absence of active transport mechanisms) is largely determined by the lipid solubility and physical size of the molecule. Non-polar molecules readily dissolve in a membrane and readily pass; small molecules (particularly small uncharged molecules) readily pass.

There are mechanisms in place to increase the rate at which some molecules pass into cells. Membrane channels of various kinds help specific molecules pass through a membrane that would normally not pass. 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 through which polar molecules readily pass. Thus, polar solutes can diffuse through the membrane in an aqueous solution. Membrane carriers, responsible for passive and active 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. Passive transport (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. Active transport moves target molecules against a concentration gradient. It 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 diffusion rates decrease with increasing molecule 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 molecule dissolve more readily and are able to pass more quickly through the lipid membrane.

Because cell membranes are selectively permeable, we need to distinguish between the concentration of all solutes across a membrane (described by "-osmotic") and the effects of solute movement water movement on cell shape (described by "-tonic"). Imagine a cell placed in a solution having a total solute concentration equivalent to the concentration of solutes within the cell. The cell is in an isoosmotic solution. (Hyperosmotic and hypoosmotic describe the concentration of total solutes across the membrane in prior to the movement of solutes by diffusion or water by osmosis.)

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 and there will usually be no net movement of water. If the membrane is not permeable to the solute, there will be no movement of solute; instead water will move across the membrane by osmosis. Such a cell is in a hypertonic or hypotonic solution since the cell will shrink or swell by osmosis. For example, suppose that a cell is in a isoosmotic 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 (that is, 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 isoosmotic solution (equal initial total solute concentration) that is hypotonic (ultimately causes the cell to swell).

The concentration of solutes can be described in a number of different ways:

a) The simplest expression is the weight of solute (grams) per 100 ml of solution. This is a percent solution. For example, 15 g in 100 ml solution may be expressed as 15 %.

It is frequently more useful to express concentration in terms of molarity or molality. 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 (C6H12O6). The common measure to both solutions is the mole (6.02 * 1023 molecules).

b) A one-molar (1 M) solution contains one mole of solute in 1 l (litre) of solution.

c) A one-molal (1 m) solution contains one mole of solute in 1,000 g of solvent. If water is the solvent, 1,000 g of water (at 4o C) is 1 litre and the final volume of solution slightly exceeds 1 litre. (A 1 M aqueous solution is very close (but not identical) in concentration to a 1 m 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 (water)- 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 1m solution of glucose has one mole of glucose in solution. This would have the same osmotic pressure as 1m solutions of urea, sucrose, etc Some molecules dissociate when they are dissolved in solution. Sodium chloride (NaCl), for example, completely dissolves in water and dissociates into Na+ and Cl. Therefore, a 1m 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.d) Osmolality (Osm) is a measure of the number of particles in solution (which is related to the osmotic pressure generated by the solution). Thus, a 1m solution of NaCl has an osmolality of 2Osm. It is frequently more convenient to express concentration in terms of milliosmolality (mOsm; 1/1000Osm). For example, a 0.1m solution of NaCl has a osmolal concentration of 200 mOsm, whereas a 0.10m glucose solution has a osmolal concentration of 100mOsm.

How do we measure permeability?

One of the best model systems for studying membrane permeability is the mammalian erythrocyte. Their membranes are sufficiently flexible and weak that placement in solutions of tonicity that significantly diverges from that of the cell itself results in rapid and visible changes in cell form. For example, when hemolysis (blood + burst) occurs, the internal solutes, including hemoglobin, escape and the solution turns clearer (i.e. it absorbs less light). The rate at which a solute diffuses into the cell is related to the rate at which it hemolyses. The opposite effect is "crenation" in which sufficient water leaves an erythrocyte that the membrane shrinks and appears knobby. The effects of this process cannot be seen by visual inspection of a test tube, but cells do take a distinctive shape when viewed under a microscope.

We will monitor hemolysis using the spectrophotometer. As hemolysis proceeds, the absorbance of the solution decreases. Therefore,

membrane permeability(the proportion of cells hemolyzed(1/absorbance

Materials and Methods:

Part I- osmosis across a semi-permeable membrane and the generation of osmotic pressure

1. Cut a 6 cm piece of dialysis tubing. Soak in distilled water until the layers separate. Slide one blade of a scissors between the two sides and cut along the margin to produce a sheet of semi-permeable plastic polymer.

Note: dialysis tubing is a porous polymer used to separate molecules based merely on their molecular size. Molecules that are larger than the pore size do not pass, whereas smaller molecules (including water) move through the pores by diffusion.

2. Each group will choose one of the following treatments. (fill in the blanks)

TreatmentSolution *Molarity (M)approximate osmolality (in mOsm)

#130 % sucrose

#230 % glucose

#360 % glucose

#4Distilled water

* glucose FW = 180.16; sucrose FW = 342.30

3. Hold a thistle tube vertically with the mouth upward. Have one person hold a finger over the lower opening while another pours the solution into the mouth of the tube and down the inside surface of the tube until it overflows.

4. Remove as many air bubbles as possible.

5. Place the rectangular piece of dialysis tubing tightly over the mouth of the thistle tube so that no air is trapped between it and the solution. Keeping the dialysis tubing taut, secure it to the thistle tube with several wrappings of a rubber band.

6. Invert the thistle tube in a beaker of distilled water until about 5 mm of the straight tubing is submerged and the rest is in the air. Clamp in place and check for leaks.

7. Using the syringe with the attached length of tubing, withdraw solution until the level approximates that of the surface of the water in the beaker.

8. Note the time and measure the distance the fluid moves up the thistle tube every 10 minutes until the end of the laboratory period or until the fluid stops moving.

Prior to class, the University Nurse will collect a blood sample from one member of each group. This will provide sufficient erythrocytes for remaining experiments.

Caution: exercise due caution when working with all blood and blood products. Never allow blood to contact exposed skin. Handle only your own blood or use gloves or use a clean transfer pipette once, then discard in the pan full of bleach. Place all instruments and glassware that has contacted blood in the containers provided by the instructor.

Part II- determining the solute concentration that is isotonic to erythrocytes

1. Start the spectrophotometer, computer and spectrophotometer software.

2. Obtain a supply of clean, scratch-free cuvettes.

3. Create 3 ml of the appropriate solution to serve as a spectrophotometer blank. Set the spectrophotometer to 640nm and zero the spectrophotometer (0.000 A or 100.000% T) using the blank.

4. You will be adding blood to 3 ml of each of the following NaCl solutions in turn:

(Fill in the blanks)

solution numbersolution molarity (M)percent solutionapproximate molality (m)approximate osmolality (in mOsm)

#11.71 M NaCl

#20.60 M NaCl *

#30.15 M NaCl

#40.08 M NaCl

#50.06 M NaCl

#60.03 M NaCl

* approximate concentration of sea water

NaCl FW = 58.45

5. Ensure that the computer interface, software and printer are working properly. Set the software to record absorbance vs. time for at least 120 seconds (2 minutes).

6. Prepare the cuvette with solution #1.

7. One team member should add 2 drops of blood (using a pipette), mix QUICKLY, place it in the spectrophotometer and close the lid.

8. A second team member should start recording as soon as the blood contacts the solution.

9. Monitor absorbance until it plateaus. Note the final absorbance.

10. Other team members should replicate the procedure in a plain test tube. Taking note of the time of addition, adding 2 drops of blood to the contents of the tube and mixing.

11. The tube should be held up to window and observed over the course of several minutes. Notes should include the elapsed time since blood addition of any significant changes. NOTE: if there is a change, it will usually occur in less than 60 seconds.

12. Repeat steps 7.- 11. with solution #6.

13. When finished, note the differences in appearances and the final absorbance of the two solutions (#1 & #6). These two tubes represent the two extremes of response by erythrocytes to solutions of differing NaCl concentrations. The appearance of the one of the two tubes that underwent a major change is your hemolysis endpoint (which tube?). TAKE CAREFUL NOTE of the visual appearance of hemolysis and equate that with the final absorbance provided by the spectrophotometer.

14. With these two end points in mind, repeat steps 7.-11. Using solutions 2-5.

15. Discard the test tubes in the dishpan with bleach.

Part III- permeability to water soluble compounds

1. Repeat the protocol of part II using the following solutions. You will be adding 1 drop of blood to 3 ml of each of the following solvents in turn. NOTE: Since some of these solutions are barely miscible in water, shake the stock solutions well prior to withdrawing the volume for the cuvette:

solution numbertest solutionmolecular weight (g/mole); relative size

#10.3 M methanol32 (smallest)

#20.3 M ethylene glycol62

#30.3 M glycerol92

#40.3 M erythritol122 (largest)

2. Finally, repeat the protocol with the following solution. BEFORE YOU DO- what is the appropriate blank? Create this blank and re-zero the spectrophotometer before proceeding.

Solution numbertest solutionmolecular weight (g/mole)

#50.3 M glycerol + 10-3M Cu++92

Part IV- permeability to nonpolar compounds

1. Repeat the protocol for Part III with the following solvents:

solution numbertest solution molarity (M)lipid solubilitymolecular weight (g/mole); size

#10.3 M glycerol + 10-3M Cu++least92 (smallest)

#20.3 M monoacetin134

#30.3 M diacetin176

#40.3 M triacetinmost218 (largest)

Glycerol forms the backbone for each of these molecules. Mono- through tri-acetin adds progressively larger esters to this backbone.

2. Make sure you use the appropriate spectrophotometer blank for each test solution.

Clean the bench top with 95% Ethanol (EtOH).

Your Assignment (Part I):

Prepare a narrative text for the materials and methods and results sections of a laboratory report. Note: by narrative text, I mean complete English sentences and no outlines. In other words, using the guides outlined in the document describing the preparation of laboratory reports in Human Physiology laboratory.The results sections should include the following analyses. Include i) a narrative, ii) figures (in the standard format) and iii) reference to, and explanation of, all figures presented in the narrative text.

Results-Part I1. Collect the data for each treatment from the other groups.

2. The goal of the experiment is to determine the relationship between solute concentration and the osmotic pressure it generates.a. Calculate the milliosmolal concentration of each of the test solutions. This is your independent variable to be graphed on the x-axis of a final summary figure.

b. Determine the approximate osmotic pressure each solution generates (our dependent variable) to be graphed on the y-axis of that final summary figure:a. For each solution plot time (min) vs. vertical distance traveled (mm) in column.b. Use each of these plots to estimate the osmotic potential generated by the solution by either a) choosing a section of the graph that is most nearly linear and draw a straight line that approximates the slope (mm/min) through that range and/or b) determining the final height of the fluid in the column (mm).c. Draw your summary figure using approximate milliosmolality vs. approximate osmotic pressure (determined by either means).Part IINote: Red blood cells are essentially impermeable to NaCl. (Think about why, but there is no need to address that here) Therefore, only water is moving (by osmosis) in this experiment.1. For each form of data analysis (visual inspection and spectrophometer) present your data.

a. In the case of the visual inspection, prepare a table of your observations.

b. In the case of the spectrophotometer data, prepare figures of time vs. absorbance.

i. Lay the figures out in order and note that there is a pattern to the data- [NaCl] and the form of change of absorbance over time.

ii. Create a means to provide a numerical value to the rate at which cells are lysing in these solutions. (There are at least three appropriate means that Dr. Baube can think of.)

iii. Apply this method to each of the figures to arrive at a single number so that you can relate the concentration of a solution to the membrane response (no lyse, slow lysing, rapid lysing, etc) of RBCs to that solution.2. Create a summary figure of approximate solution concentration (mOsm) vs. RBC membrane response.

3. Identify any patterns common to both graphs.

Note: In parts III & IV, cells are permeable to the solutes in varying degrees. If cells hemolyse, it is by osmosis after the solute has crossed the cells' membrane by diffusion.

Part III1. Create figures of absorbance vs. time for each solution.

2. As before, create and apply a means to generate a numerical value for the rate of cell hemolysis. This rate of hemolysis is directly related to the rate at which the solute penetrates the cells. (That is, fast penetration = fast hemolysis, and vice versa.)

3. Prepare a summary figure relating the values of the independent variable of the experiment to the values of the dependent variable determined in the previous step.

4. In initially describing the results, ignore the data generated for glycerol (without Cu++). What do the rest of the data say with respect to the relationship between molecular size and cell permeability for polar compounds? Support with data.

5. Now, include the data from glycerol with Cu++ previously omitted data. Are there any systematic departures from a monotonic (i.e. more-or-less linear) sequence of increasing or decreasing permeabilities across treatments?

(Note: only glycerol has molecule-specific active transport proteins- no other molecules used in this experiment have such a mechanism. That is, all information collected describes the passive diffusion of these molecules across the membrane.)

Part IV-1. Use the same approach as the previous section to analyzing these data.2. However, as your describe your results, note that each molecule differs both in molecular size AND lipid solubility. Describe the effects of each factor, independent of other (or at least as much as you can separate them). Make reference to specific data as much as possible.Your Assignment (Part II):

The following are the kinds of questions that would be considered in a discussion section, if you were to be writing a full research report on this exercise. However, you are NOT writing a full research report. Rather, answer the following questions in paragraph format. That is, no need to produce a discussion section as such.1. Generalize the relationship between cell permeability and molecular i) polarity and ii) size. Describe why this makes sense in light of what you know about membrane structures and properties.2. Cu++ interacts with a mechanism for the active transport of glycerol across membranes. Using the data generated, propose and support a hypothesis for the effect of Cu++ on this mechanism. It can either be as a facilitator (working as a enzyme helper) or inhibitor (as a channel blocker or enzyme inhibitor).3. What is the function of glycerol in living cells? Why does the pattern identified make sense in light of the critical importance of glycerol to cell structure and function?4. Many organic solvents such as benzene are highly toxic to cells. Apply what you have learned to explain the basis of this toxicity.

Answer the following questions as an appendix to your report:

a) Suppose a salt solution and a glucose solution are separated by a membrane that is permeable to water but to neither solute. The NaCl solution has a concentration of 1.95 g per 250 ml solution (FW = 58.45). The glucose solution has a concentration of 9.0 g per 250 ml solution (FW = 180.16). Calculate the initial molarity, approximate molality and approximate milliosmolality of both solutions. State whether osmosis will occur and, if it will, in which direction. Explain.

b) Now, suppose the membrane is permeable to NaCl, but not to glucose. Describe the movement of NaCl across the membrane. Estimate the theoretical concentration of NaCl on each side after its diffusion, but before any osmosis. Then, describe the resulting direction of water flow by osmosis.