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Investigation of Rate of Diffusion

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Investigation of Rate of Diffusion

1. Define the following terms: solute, solvent, concentration
gradient, osmotic pressure, and selectively permeable.

2. Define the following processes and identify the characteristics
that distinguish them from one another: diffusion, osmosis,
facilitated diffusion, and active transport.

3. Give three factors that affect the rate of diffusion and state
whether they increase or decrease the rate.

Introduction

Cells maintain a constant internal environment, a process called
homeostasis. In a constant environment enzymes and other cellular
components can operate at optimum efficiency. The ability to
selectively exchange materials with the environment is one component
of the cell’s homeostatic mechanism. Ions and molecules, such as
sugars, amino acids and nucleotides, must enter the cell, and the
waste products of cellular processes must leave the cell. Regardless
of the direction of movement, the common mediator of these processes
is the cell membrane, or plasma membrane.

The plasma membrane is a fluid, or mobile, mosaic of lipids and
proteins. Ions and molecules cross the plasma membrane by a number of
processes. Large particles are engulfed by the membrane, forming a
vesicle or vacuole that can pass into (endocytosis) or out
(exocytosis) of the cell. Some small, electrically neutral molecules
diffuse through the spaces between the lipid molecules of the plasma
membrane. Others bind to transport proteins embedded in the plasma
membrane and transported into or out of the cell.

Atoms, ions and molecules in solution are in constant motion and
continuously collide with each other because of their kinetic energy.
As the temperature is raised, the speed of movement of the molecules
increases and they collide more frequently and with greater force. An
observable consequence of this motion is Brownian motion, an erratic,
vibratory motion of small particles suspended in water, which is
caused by the collisions of water molecules with the particles.

Diffusion also results from the kinetic energy of molecules. If a
small crystal of a soluble substance is added to water, molecules of
the substance break away from the crystal surface and enter solution.
As a consequence of the collisions with water molecules, molecules of
the substance move in a random pattern in the solution, but always
away from the crystal, with some moving to the farthest reaches of the
solution. This process continues until the molecules of the substance
are evenly distributed throughout the water (the solvent). In a
general sense, in any localized region of high concentration, the
movement of molecules is, on average, away from the region of highest
concentration and towards the region of lowest concentration. The
gradual difference in concentration over the distance between the
regions of high and low concentration is called the concentration
gradient. The steeper the concentration gradient, the greater the rate
of diffusion. The rate of diffusion is also directly proportional to
the temperature and inversely proportional to the molecular weight of
the diffusing molecules. In other words, the higher the temperature
the faster the rate of diffusion, but larger molecules move slower
than smaller molecules at the same temperature.

Small, electrically neutral substances diffuse into and out of cells
by passing through the spaces between the lipids of the plasma
membrane or by dissolving in the lipids or proteins of the membrane.
Substances that are large or electrically charged cannot pass through
membranes. Membranes that block or inhibit the movement of molecules
are called differentially permeable, or selectively permeable.
Selective permeability explains the phenomenon of osmosis, the
diffusion of water across a membrane under certain conditions.

If two solutions containing different concentrations of a solute are
separated by a selectively permeable membrane (permeable to water but
not to the solute), water will move from the solution with low solute
concentration to the solution of high solute concentration. The water
flows in this direction because the solution with low solute
concentration has a high water concentration and the solution with
high solute concentration has a low water concentration. Thus, the
water diffuses from a region of high water concentration to a region
of low water concentration.

Many ions and molecules important to cells are taken into cells by
specific transport proteins found in cell membranes. Facilitated
diffusion occurs when such a protein functions as a binding and entry
port for the substrate. In essence, the protein functions as a
pipeline for the specific substance. The direction of flow is always
from high concentration to low concentration. The gradients are
maintained because frequently the molecules are metabolically
converted to other types of molecules once they enter the cell.

For many other molecules and ions, favorable diffusion gradients do
not exist. For example, sodium ions are present at higher
concentrations outside mammalian cells than inside the cells, yet the
net movement of sodium ions is from the inside to the outside of the
cell. Likewise, potassium ions are found inside mammalian cells at
significantly higher concentrations than outside the cell, but the net
movement of potassium ions is from the outside to the inside of the
cell. For such molecules and ions, cellular energy must be used to
transport the molecules across the plasma membrane. Active transport
occurs when transport proteins in the cell membrane bind with the
substrate and use cellular energy to drive the “pumping” of the
molecules into or out of the cell, against the concentration gradient.

In today’s lab you will observe Brownian motion, osmosis, and
diffusion in the solid, liquid and gaseous state, and investigate the
parameters that affect the rate of diffusion of molecules. In next
week’s lab, which is a continuation of your investigation of diffusion
and the properties of cell membranes, you will model a semi-permeable
membrane and investigate the behavior of different types of cells in
hypotonic, hypertonic and isotonic solutions.

Observing Brownian Motion

The vibratory movement exhibited by small particles in suspension in a
fluid was first observed by the Scottish botanist Robert Brown in
1827. Brown incorrectly concluded that living activity was the cause
of this movement, but we now know that Brownian movement results from
the collisions between water molecules and small particles (less than
10 micrometers in diameter) suspended in the water.

To illustrate Brownian movement, place a drop of water on a microscope
slide. Dip a dissecting needle into India ink and then touch the tip
of the needle into the water drop. (India ink consists of small
particles of carbon suspended in a fluid.) Add a coverslip and observe
the slide with a high-power objective.

Briefly record your impressions of the movement of the particles. If
you gently warm the slide over a light bulb, what effect does this
have on the movement of the particles? How do you account for any
changes in motion you observe after heating the slide?

Osmosis

The rate of water movement in osmosis can be observed with an
osmometer (see figure below). A starch solution in the thistle funnel
is separated from the water in the beaker by a dialysis membrane that
allows water to pass through but is impermeable to starch. (The starch
solution has had food coloring added to it so that you can track any
movement of the solution in the thistle funnel.) What do you expect
will happen over time in this type of setup?

Figure 1. A Simple Osmometer

Early in the lab, measure the height of the column of fluid in the
thistle funnel. At intervals of about 20 minutes during the lab,
repeat the measurement. Record the time and height of the fluid column
in the table below.

Time

Elapse Time

Height of Fluid Column

Describe what is happening to both the starch and water molecules in
the osmometer.

Over time do you expect that the rate of water movement will increase,
decrease, or remain the same? Why?

Diffusion in a Solid

The solid we will use is agar, which forms a colloid (a gel-like
matrix) when mixed with water, and is clear so you can see into it.
Molecules can diffuse through the water-filled channels in the agar
matrix. Your instructor may assign you to a group to carryout this
experiment. Alternatively, your instructor may do this experiment as a
demonstration and provide you with the data at the end of the lab
period.

1. Obtain 6 agar plates for your group. Label one pair of plates 4oC,
the second pair RT (for room temperature), and the third pair 37oC.

2. Take off the lids of the first pair of plates. Using a toothpick,
place a small crystal of potassium permanganate (KMnO4) on the agar
surface of one plate, and a similar amount of methyl orange on the
agar surface of the second plate. Be careful not to poke a hole in the
agar surface. Replace the lid on each plate.

3. Repeat step 2 for the remaining pairs of plates, taking care to use
similar sized KMnO4 crystals and methyl orange on each pair of plates
as used in step 2.

4. Place the 4oC plates in a refrigerator, leave the RT plates in a
safe spot on your lab bench, and the 37oC plates in an incubator set
for 37oC. Record the time you start the experiment.

5. After about 1.5 hours, collect each pair of plates and measure the
size of the colored ring around each crystal in millimeters. Record
the radius of each ring in the appropriate box in table 1 below.
Record the time you make the measurements and calculate the time, in
minutes, the plates were sitting.

Start Time

End Time

Length of time (min)

Table 1.

Radius of ring (mm) 4C

Radius of ring (mm)Room

Radius of ring (mm)37C

Potassium permanganate

Methyl Orange

6. Calculate the rate of diffusion for each molecule at each
temperature using the procedure outlined below.

a. Convert the radius from millimeters the micrometers by multiplying
the radius in millimeters by 1000.

b. Divide the resulting number by the number of minutes the plates
were sitting. Record this result in the appropriate box of the
following table.

Table 2. Rate of diffusion (µm/min.)

4C

Room

37C

Potassium permanganate

Methyl Orange

Use this data to answer the following questions.

1. The molecules of which substance diffused more rapidly?

2. The molecular weight of KMnO4 is 158 and the molecular weight of
methyl orange is 327. What relationship is there between the molecular
weight of the substance and the rate of diffusion?

3. What relationship is there between the rate of diffusion and
temperature? What reason can you give to explain this relationship?

Diffusion in a Liquid

In this experiment we will be determining the rate of diffusion of
KMnO4 in water at room temperature. We will then compare this rate
with the rate of diffusion of KMnO4 in agar at the same temperature
(from the table above).

1. Place some room temperature water in a glass Petri dish and place
the Petri dish over a thin, flat metric ruler.

2. Using tweezers place a crystal of KMnO4 directly over one of the
millimeter lines of the ruler and record the time.

3. After 10 minutes, measure the distance the color has moved. Record
the final time, length of time and distance moved in the table below.

4. Calculate the rate of diffusion of the KMnO4 using the procedure
described above and record it in table 3 below.

Table 3.

Start Time

End Time

Length of Time (min)

Distance Moved
(mm)

Diffusion Rate (microns/second)

Water

Agar

Transfer the appropriate data from table 2 above and use the data in
table 3 to answer the following questions.

1. In which experiment is diffusion the fastest?

2. How can you explain this difference in speed?

Diffusion in a Gas

This experiment is an optional demonstration and will be done at the
discretion of your instructor. It should be noted that the two
chemicals involved have a real potential to be harmful and should be
treated with extreme caution. The diagram below illustrates the
experimental setup.

1. Remove the rubber stoppers from the end of the glass tube and
simultaneously dip one of the cotton tipped applicator sticks into
concentrated HCl and the other into concentrated NH4OH.

2. Simultaneously reinsert the stoppers into the glass tube.

3. Look for the formation of a white ring inside the tube. This is
NH4Cl, a white salt formed when HCl and NH3 meet.

4. Measure the distance each gas traveled and record the results in
table 4 below and use the data to answer the following questions.

Table 4.

HCl

NH3

Distance Traveled (mm)


1. The molecular weight of HCl is 36 and NH3 is 17. Which gas did
(should) diffuse the fastest?

2. Calculate the following values: the ratio of the distances, the
ratio of the molecular weights, and the ratio of the square roots of
the molecular weights. Is the rate of diffusion directly or inversely
proportional to the molecular weight or the square root of the
molecular weight?



Diffusion and Cell Membranes – II

Objectives

1. Define the following terms: hypotonic, isotonic and hypertonic.

2. To determine if osmosis and diffusion both occur through a
selectively permeable membrane.

3. To observe the effects of hypotonic, isotonic and hypertonic
solutions on plant cells and animal cells.

4. Given any two solutions of differing osmotic potentials and
separated by a selectively permeable membrane, state which solution is
hypertonic and in which direction the net flow of water will occur.

Introduction

The term tonicity describes the relative concentration of solvent to
solute in two solutions. A solution with the lower solute
concentration is said to be hypotonic relative to the other solution.
Conversely, the more concentrated solution is hypertonic relative to
the first. If the solute concentrations of each solution are equal the
solutions are isotonic with respect to each other. It is important to
remember that these terms are relative terms, that is, the description
of a solution as being hypertonic, hypotonic or isotonic depends on
the solution it is being compared to. Traditionally, in biology, the
cell is the frame of reference. An isotonic solution has the same
solute concentration (and water concentration) as the cell; a
hypertonic solution has a higher solute (and lower water)
concentration than the cell; a hypotonic solution has a lower solute
(and higher water) concentration than the cell.

If a cell in a hypotonic solution (low solute concentration) is
enclosed in a rigid box, for example a plant cell surrounded by the
rigid cell wall, the increasing water pressure inside the cell would cause
water to flow back out of the cell towards the area of lower pressure.
Eventually, equilibrium would be reached when the flow of water into
the cell, due to the concentration differences, equals the flow of
water out of the cell, caused by pressure differences. The pressure at
equilibrium is called the osmotic pressure.

Since all cells contain molecules that cannot cross the plasma
membrane, osmosis always occurs when cells are placed in dilute
aqueous solutions. It is important, then, for cells to be able to
regulate the flow of water into, and out of the cell, a process known
as osmoregulation. In plant cells and bacterial cells, the cell wall
prevents the cell from bursting by providing a rigid casing that helps
regulate the osmotic pressure in the cell. In animals and many
microorganisms, osmoregulatory organs or organelles are found. In
animals the kidney adjusts the concentration of substances in the body
fluids that bathe the cells. In microorganisms, like Paramecium, which
live in freshwater, special organelles, called contractile vacuoles,
accumulate and actively pump out water that flows into the cell by
osmosis.

In this week’s lab you will model a semi-permeable membrane and
investigate tonicity by looking at the behavior of different types of
cells in hypotonic, hypertonic and isotonic solutions.

Diffusion Through a Selectively Permeable Membrane

The plasma membrane of a cell is selectively permeable because it
allows the diffusion of some substances and not others. Small,
uncharged molecules diffuse freely across the plasma membrane, but
charged molecules and large molecules cannot cross the membrane. The
dialysis membrane used in this experiment simulates the activity of
the plasma membrane.

Procedure

1. Obtain a piece of dialysis tubing and make a tight knot in one end
with thread.

2. Fill the bag with solution A, a simulated “liquid” meal containing
10% glucose, 1% starch, 0.5% egg albumin, and 1% sodium chloride.

3. Tie the top of the tube with thread while expelling as much air as
possible. The bag should be limp (flaccid).

4. Rinse the outside of the dialysis tube with distilled water.

5. Place the dialysis tube in a culture dish and add enough solution B
to cover it. Solution B contains 0.5% sodium sulfate dissolved in
water. Let the dish stand undisturbed for about 1½ hours.

6. Based on the recipes for solutions A and B, fill in the “Before”
columns of table 1. Use a + to represent the presence and a – to
represent the absence of a substance.

Table 1.

Substance

Inside Dialysis Tubing

Outside Dialysis Tubing

Before

After

Before

After

1. Starch

2. Chloride Ion

3. Sulfate ion

4. Glucose

5. Albumin

7. After about 1½ hours, remove the dialysis tubing from the culture
dish. Gently agitate the contents of the tubing and note any change in
the tubing (Hint: is it more or less flaccid than when you started?)

8. Rinse the dialysis tubing with distilled water and carefully open
the tubing. Empty the contents into a 100ml beaker. You can now test
which ions and molecules crossed the membrane.

9. Obtain eight test tubes and prepare them as follows:

a. Into each of the first four test tubes, place 10 drops of the
solution from inside the dialysis tubing. Label these I-1 to I-4.

b. Into the second set of four tubes, place 10 drops of the solution
from outside the dialysis tubing. Label these O-1 to O-4.

10. Test for the presence of starch, albumin, glucose, sulfate ions,
and chloride ions in the two sets of test tubes using the following
test:

a. To the first test tube of each set, add 3 drops of IKI to test for
starch. A blue-black color indicates a positive result.

b. To the second test tube of each set, add 1 drop of silver nitrate
(AgNO3) to test for chloride ions. A white precipitate indicates a
positive result.

c. To the third test tube of each set, add 3 drops of 1% barium
chloride (BaCl2) to test for sulfate ions. A white precipitate
indicates a positive result.

d. To test for the presence of glucose, dip a Clinistix into the
fourth test tube of each set. Compare the results to the color chart
on the side of the container.

e. To test for the presence of albumin, dip an Albustix into the
fourth test tube of each set. Compare the results with the color chart
on the side of the container.

11. Record your results in the “After” columns of table 1. Use a + to
represent a positive test and a – to represent a negative test. Use
these results to answer the following questions.

1. At the start of the exercise, which solution (A or B) was
hypertonic compared to the other (that is, which had the higher
concentration of solutes)?

2. Which solution gained water in the course of the exercise (A or B)?

3. Which of the substances (starch, chloride ions, sulfate ions,
glucose, albumin, and water) were able to pass through the membrane
(in either direction)?

4. Which substance(s) moved out through the membrane?

5. Which substance(s) moved in through the membrane?

6. Why did each substance move in the direction it did?

7. By what process did the substances move across the membrane?

8. Why did some substances fail to pass through the membrane?

9. Would you expect all of the molecules of a diffusible substance to
move across the membrane? Why?

10. Which of the following statements best describes the situation at
equilibrium if you let the system stand for a long time?

a. No molecules move across the membrane.

b. All molecules cross the membrane equally often in either direction.

c. Molecules to which the membrane is permeable cross equally often in
either direction.

d. Only water molecules cross the membrane equally often in either
direction.

e. Molecules to which the membrane is permeable move across the
membrane from a region of high concentration to a region of low
concentration.

11. Did water move across the membrane? What is your evidence?

12. What is misleading about trying to equate the results of this
exercise with how the cell membrane regulates passage of material?

13. Dialysis membrane is permeable to iodine (IKI). What result would
you expect to see if you put IKI in solution B at the start of the
exercise?

Osmosis and Tonicity—If set up…

As you discovered last week, osmosis (the diffusion of water) occurs
whenever two solutions of different solute concentration are separated
by a selectively permeable membrane. The difference in solute
concentration between the two solutions determines both the direction
and rate of water flow. Water always diffuses from a hypotonic
solution to a hypertonic solution; consequently, a cell placed in a
hypotonic solution will gain water and a cell placed in a hypertonic
solution will lose water.

The next three experiments explore tonicity (the solute concentration
of a solution) using potato strips, red blood cells and Elodea cells.

Osmosis and Tonicity—If set up…

As you discovered last week, osmosis (the diffusion of water) occurs
whenever two solutions of different solute concentration are separated
by a selectively permeable membrane. The difference in solute
concentration between the two solutions determines both the direction
and rate of water flow. Water always diffuses from a hypotonic
solution to a hypertonic solution; consequently, a cell placed in a
hypotonic solution will gain water and a cell placed in a hypertonic
solution will lose water.

The next three experiments explore tonicity (the solute concentration
of a solution) using potato strips, red blood cells and Elodea cells.


Procedure: Potato strips

1. Using the provided cork borer cut 6 tubes of potato, each
approximately 3 cm in length. Use a razor blade to cut the tubes to
length; remove any skin from the ends of the tubes.

2. Label five test tubes 0, 0.1, 0.2, 0.3, 0.4, and 0.5. Place one
potato tube to each test tube.

3. Fill the test tube labeled “0” with distilled water to cover the
potato tube, and fill the remaining test tubes with sodium chloride
solutions of the appropriate concentration to cover the potato tubes.

4. After at least 1 hour, observe the potato tubes for limpness (water
loss) or stiffness (water gain) and answer the following questions.

1. In which tube(s) has the potato become limp? Why did the water
diffuse out of the potato? How would you describe the relationship
between the solution and the potato (use the correct scientific term)?

2. In which tube(s) has the potato become stiff? Why did the water
diffuse into the potato? How would you describe the relationship
between the solution and the potato (use the correct scientific term)?

3. Is there any tube(s) in which the potato appears to have neither
gained nor lost water? How would you describe the relationship between
the solution and the potato (use the correct scientific term)?

4. Based on these results, what is the approximate concentration of
solutes in potato cells?

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