1
Cell Biology Lab
Cell Membrane Permeability and Transport Properties
Cell Membranes
“Just as chemistry could not have developed without test tubes to hold reacting
substances, so organisms could not have evolved without relatively impermeable
membranes to surround the cell constituents.”
E.N. Harven
DuPraw Cell and Molecular Biology
The cell is a dynamic system which interacts with its surroundings, The structure which
controls this interaction is the plasma membrane that is freely permeable to water yet
selectively permeable to most other substances. The major mechanisms by which
materials cross the membrane are: 1. Passive Transport (diffusion) — the movement of
molecules from regions of high concentration to regions of low concentration; 2.
Assisted Transport — the movement of molecules with the assistance of carrier
substances in the membranes either along the concentration gradient (facilitated
transport) or against it with the expenditure of cellular energy (active transport); 3.
Endocytosis — the bulk movement of molecules or particles across the membrane by
the formation of vesicles with an expenditure of energy.
Experiment 1. Passive transport
There is considerable evidence that water diffuses freely through the membrane. The
diffusion of water across the cell membrane is called osmosis. The direction of the water
flow is dependent on the concentration of water molecules. There are three terms used
to describe solutions. Solutions that contain the same concentration of solutes as the
cytoplasm are said to be iso-osmotic and the water diffuses at equal rates in both
directions. When the solution has a higher concentration of solutes than the cytoplasm
the water flows from the cytoplasm into the solution. Under these conditions the solution
is hyper-osmotic. When the solution has a lower concentration of solutes than the
cytoplasm the solution is hypo-osmotic. To visualize the effects that these three types of
solutions have on animal cells, set up the following demonstrations using red blood cells
and the epidermal cells of red onion.
Procedure
1.
2.
3.
Prepare 50 ml of a 0.15 M NaCl solution from the 1 N NaC1 stock solution.
Cut off a small section of a red onion. With a sharp razor blade make a thin slice
parallel to the epidermis. Section should be at least 2-3 mm square.
Mount on microscope slide in 0.15 M NaCl solution and note the appearance of of
both the pigmented and nonpigmented cells.
2
4.
5.
6.
7.
8.
Add 2-3 drops 1 M NaCl on the slide near one edge of the coverglass.
Place a pad of paper toweling along the other edge and draw the NaCl across the
slide.
Continue to add NaCl until the cells are bathed in the fluid.
Observe the appearance of the cells. Note the decrease in volume of the
protoplasts,
Using new toweling, add 0.15 M NaCI in the manner described above and observe.
Define plasmolysis and deplasmolysis.
Experiment 2. Differential Permeability of Cell Membranes
One of the basic characteristics of all cells is that they possess plasma membranes by
which the cells are able to control the movement of materials in and out of their
cytoplasms. A fundamental problem in biology is to understand the methods by which
materials are able to move through such membranes. One of these methods is
osmosis. Plasma membranes are known to be differentially permeable; that is, under
certain conditions they allow solvent molecules to more through them in greater
numbers in one direction than in the other. Osmosis, then, can be generally defined as
the diffusion of a solvent through a differentially permeable membrane. In biological
systems the solvent is water. As a result of the diffusion of water into cells, a turgor
pressure develops in the cell. The maximum potential turgor pressure that could be
developed by a solution if it were separated from pure water by a rigid membrane
permeable only to water is called the osmotic pressure. However, since plasma
membranes are not completely rigid and not permeable only to water, the turgor
pressure of cells is never as high as the osmotic pressure. The more solute particles
there are in a solution, the higher the osmotic pressure and the lower the water
concentration. Thus, osmotic pressure is merely a convenient method of expressing the
water concentration of a solution. If a cell is placed in a solution of osmotic pressure
lower than the cell contents, water will pass into the cell, causing it to swell. In a solution
with higher osmotic pressure than the cell contents, water will pass out, causing
shrinkage of the cell contents.
Factors Affecting the Permeability of Cell
1. The effect of concentration of electrolytes and nonelectrolytes on the
permeability.
Red blood cells (erythrocytes) can be conveniently used to illustrate the osmotic
relations of the living cell. The plasma membrane of red cells is freely permeable to
water, but relatively impermeable to salts. If placed in an isotonic saline solution ( a
solution containing the same salt concentration as found in normal plasma 0.9% NaCl),
it retains its normal shape and size. If placed in a hypotonic solution (a solution
containing a solute concentration lower than that of the cell), water enters the cell more
rapidly than it leaves. The red blood cell increases in volume as substances such as
water and dissolved solutes enter it from the outside, but it does not increase
3
indefinitely. Instead, it reaches a limit of size, and then the plasma membrane ruptures
and hemoglobin diffuses out of the cell, leaving only the cell membrane (ghost) behind.
This phenomenon is called hemolysis. If red blood cells are places in a hypertonic
solution, more water leaves the cell than enters it, and consequently it shrinks, showing
a wavy and irregular outline (crenation).
Very dilute suspension of intact red blood cells transmit very little light, because each
cell acts as an opaque disc. The opacity is a property not only of the hemoglobin
content of the cell but also of the size and shape of the cells. When cells in dilute
solutions are hemolyzed, the solution becomes transparent.
In this experiment you will determine the effect of the concentration of electrolytes and
nonelectrolytes on the permeability of red blood cells.
Procedure
1. Set up two rows of five test tubes each, one row for glucose solutions and the
second row for NaCl solutions. To each appropriate tube add 2 ml of the various molar
solutions of glucose or NaCl
glucose: 0.1 M, 0.2 M, 0.3 M, 0.4 M, 0.5 M;
NaCl: 0.065 M, 0.1 M, 0.l5 H, 0.25 M, 0.3 M).
2. Then to each tube add 2 drops of a sheep red blood cell suspension provided by the
instructor. Mix immediately by holding parafilm over the tube and inverting it once or
twice.
3. Set the tubes aside for 15 or 20 minutes, and then indicate the tubes in which
hemolysis has occurred. This is simply done by holding the tube against a printed page
(this sheet). If you can plainly see the printing, hemolysis has occurred.
4. Which concentrations of NaCl and glucose are closest to being isotonic to the red
blood cells?
Although the percentage concentrations of the solutions of nonelectrolytes which just
fail to hemolyze may be quite different, in actuality they are found to have approximately
the same molecular concentration (that is, they have the same number of molecules per
liter). Thus, solutions having the same number of molecules per liter show, by
hemolysis, the same osmotic pressured If, however, NaCl or any other salt (electrolyte)
to which the cell is comparatively impermeable is used, hemolysis occurs at much lower
concentrations because -- investigation has revealed -- every ion in a solution exerts the
same osmotic pressure that would be produced by a molecule. For example, if 5.5 M
sugar is isotonic with 1O.03 M KC1, we find that the sugar is 1.87 times as concentrated
as the KC1 (5.5÷10.3 =1.87). In other words, 100 molecules of KC1 exert as much
osmotic pressure as 187 molecules of sugar. The ratio computed above between the
concentration of a nonelectrolyte and that of an electrolyte having the same osmotic
pressure is called the isotonic coefficient for the given solution of the electrolyte. Thus, if
4
the isotonic coefficient of KC1 solution is 1.87, it behaves as if 87% of its molecules
were dissociated. In other words, out of every 100 molecules, 87 appear to have formed
two ions, so that there is a total of 187 particles capable of exerting osmotic pressure;
this is equal to 187 sugar molecules in a 5.5 M sugar solution.
The isotonic coefficient (i) of NaCl is given by the formula:
=
isotonic molar concentration of glucose
isotonic molar concentration of NaC1
Using your hemolysis data, calculate i and account for the difference in isotonic
concentrations of glucose and NaC1.
With the same data it is possible to calculate the proportion of molecules of NaCl which
are dissociated:
i= 1 + (k—l) a
where i = isotonic coefficient of NaC1
k = number of ions from each molecule of the salt
a = degree of dissociation
Calculate the degree of dissociation, a, for NaC1 in your experiment. What is the
difference between hemolysis and crenation?
Experiment 3. Measurement of the Relative Rates of Penetration in Erythrocytes:
The effect of molecular size.
Major Hypotheses Tested in this Experiment
1. Erythrocytes burst, or hemolyze, when they are suspended in hypotonic saline.
2. Erythrocytes maintain normal, biconcave shape in isotonic saline.
3. Erythrocytes crenate (dehydrate, shrink, and form projections along their margins)
when they are suspended in hypertonic saline.
4. The rate of penetration of a series of alcohols into erythrocytes depends, in part, on
the lipid solubilities of the alcohols. The higher the lipid solubility of an alcohol, the
faster the rate of penetration into the erythrocyte.
5. Detergents disrupt membrane structure by solubilizing membrane components
(lipids and proteins) and bringing about the hemolysis of erythrocytes.
6. An organic lipid solvent dissolves the lipid bilayer and thereby brings about the
hemolysis of erythrocytes.
7. Within a series of polar, organic compounds, the rate of penetration into erythrocytes
5
partially depends on the molecular weight (size of the compound). The lower the
molecular weight (the smaller the size) of the compound, the faster the rate of
penetration.
Procedure
1. Obtain 13 clean glass test tube, ranging in size from approximately 13 X100 mm to
16 X 150 mm, one for each of the thirteen treatments in the study. It is important
that all of the test tubes for all thirteen treatments are the same size. Obtain a
color slide for viewing the effects of the treatments.
2. Obtain a tube containing approximately 1 ml of defibrinated whole sheep blood. This
should be enough blood to complete all thirteen treatments.
3. Using a 10-ml pipet, transfer 10 ml of each of the following thirteen test solutions into
separate test tubes. A separate pipet must be used for each individual reagent.
Test Solutions:
1) Distilled deionized water (DDH20)
2) 0.065 M sodium chloride in DDH20
3) 0.145 M sodium chloride in DDH20
4) 0.450 M sodium chloride in DDH20
5) 0.8M ethanol in DDH2O
6) 0.8 M ethylene glycol in DDH20
7) 0.8 M n-propanol in DDH2O
8) 0.8 M propylene glycol in DDH2O
9) 0.8 M glycerol in DDH2O)
10) 2% (v/v) triton X-100 in DDH2O
11) 1% (w/v) sodium dodecyl sulfate (SDS) and 10 mM dithiothreitol (DTT) in DDH2O
12) 0.3 M D-glucose (dextrose) in DDH2O
13) 0.3 M urea in DDH2O
All test solutions should be at room temperature.
Solutions 1 through 4 form the hypotonic/isotonic/hypertonic saline series; pure DDH20
is used as a control for all treatments and is, of course, hypotonic.
Solutions 5 through 9 contain the series of alcohols differing in lipid solubility. In order of
decreasing lipid solubility, the alcohols are: n-propanol, ethanol, propylene glycol,
ethylene glycol, and glycerol. See Table 1 for partition coefficients and chemical
formulas.
Solutions 10 and 11 contain chemicals that act directly on the erythrocyte membranes.
Triton X-100 and SDS are, respectively, nonionic and ionic detergents that solubilize
membrane lipids and proteins. DTT is a strong reducing agent that, along with SDS,
6
denatures membrane proteins.
In the final series, solutions 12 and 13 contain polar, organic compounds of varying
sizes. Urea and glucose have molecular weights of approximately 60 and 180 g/mole,
respectively.
4. Perform the permeability test one solution at a time, as follows. You will need a
timer and one small square of parafilm to seal each test tube. The permeability test is
based on the time that it takes for hemolysis to progress to the point where a film image
can be seen through a suspension of erythrocytes in test solution. The test must be run
quickly and efficiently. While one person observes the tube and calls out time-zero and
stop times, a second person should operate the timer. A given sample must be run
through all six of the following steps (a through f) before you begin testing the next
sample. For step f, use a separate Pasteur pipette for each test sample.
a. Using a micropipette, transfer exactly 50 l (0.05 ml) of defibrinated whole blood into
a tube filled with the test solution.
b. Immediately seal the tube with parafilm. Agitate it quickly by tilting it upside down and
back. This is time-zero. Start timing.
c. Immediately hold the tube up to a light source with a slide held against the tube,
between the light and the tube. The light source can be a ceiling fixture in the lab, the
incandescent or fluorescent bulb of a lamp, or a microscope illuminator. You will obtain
the best results by conducting all thirteen treatments within a given study using the
same or a very similar light source.
d. Looking at the slide through the tube filled with the suspension of erythrocytes in test
solution, watch for the point when the image on the film can first be clearly resolved.
This is the stop time.
e. Record the time, in seconds, which it took for the image to resolve through the
erythrocyte suspension. Then, take the reciprocal of that time and record that as the
relative rate of penetration in sec-1. If the image cannot be discerned within ten minutes
of time-zero, then record the hemolysis time as >600 sec and the relative penetration
rate as <0.0017 sec-1. As soon as this step is completed, move on to the next.
f. Using a Pasteur pipet, transfer a drop of the test mixture to a microscope slide. Apply
a cover glass and immediately examine the preparation with your microscope. Focus in
on erythrocytes, first with the 10X objective, then with the 40X objective. Describe and
sketch the appearance of the erythrocytes. Normal erythrocytes, viewed from the top,
are round with smooth margins; viewed from the side, they are biconcave. Since they
are thinner in the center than along the margins, these two regions appear slightly
different in color and/or brightness. Crenated erythrocytes are shrunken and have
spiked projections along their margins. All that remains of hemolyzed erythrocytes are
7
membrane ghosts, the empty "shells" of plasma membranes devoid of cytoplasmic
contents, which are not easily observed. Measure the diameters of representative
normal and crenated erythrocytes. Include these measurements with your descriptions.
Results and Discussion
Present your results in tables and drawings. Make sure they are labeled properly.
Drawings should have plate magnifications.
The discussion should address each of the hypotheses at the beginning of the lab. Do
your results support the hypotheses? Include an explanation for your statements. For
example: The discussion relating to the alcohols should include the relationships
between penetration rates, molecular structures and partition coefficients. Is there a
pattern?
Table 1. Ether: water partition coefficients, (solubility in diethyl ether/ solubility in water),
for a series of alcohols. The alcohols are grouped according to the number of carbon
atoms and arranged within each group by increasing number of hydroxyl groups.
Alcohol
Methanol
Ethanol
Ethylene glycol
n-Propanol
Propylene glycol
Glycerol
Condensed
Structural formula
CH3OH
CH3CH2OH
HOCH2CH2OH
CH3CH2CH2OH
CH3CH(OH)CH2OH
HOCH2CH(OH)CH2OH
Partition coefficient
0.14
0.26
0.0053
1.9
0.018
0.00066