APBIO 11-21-08 Lab 1 Better

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Eddy Egan

November 21, 2008

Contreras Pd. ¾ EvenAP Biology

The diffusion and osmosis of water through dialysis tubing (selectively permeablemembrane) and different concentrations of sucrose over 30 minutes, the diffusion of water

and solutes between potatoes and different concentrations of sucrose over ~24 hours and the

plasmolysis and re-hydration of an onion cell

Introduction:

When there is a plasma membrane present, separating solutions of different ion or

molecule gradients, diffusion will occur in order to balance the gradients. Diffusion is the

tendency of any molecule to evenly spread out into the available space. (Campbell and Reece:

p.130) Once the molecules are balanced, they are said to be in equilibrium. Molecules will

randomly move across the membrane until there is an equal amount on either side, moving at

equal rates across the membrane. When in this state it is said to be in dynamic equilibrium. This

form of movement across a membrane is known as passive transport. This means the cell does not

need to expend energy in order to move the molecules across. (Campbell and Reece: p.131) Cells

must be able to regulate the amount of water entering and leaving the cell, similar to how it

regulates movement of molecules. This diffusion of water across a membrane is called osmosis.

The ability of a cell to regulate the amount of water within it is vital to its survival (Campbell and

Reece: p.132). This is because a cell must be able to take in water to assist in metabolic reactions

as in and it must be able to get rid of it when there is a large excess that can cause the cell to lyse.

The same goes for the diffusion of molecules and ions. They must be able to go in or leave a cell

to assist with nerve impulses, muscle contractions and other metabolic reactions.

Solute size and difference in concentration gradients place very important roles in

the diffusion of molecules and ions across a selectively permeable membrane. Solute size is

important in diffusion because it must be small enough to be able to pass through the membrane

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pores, otherwise the solute cannot diffuse. (Campbell and Reece: p.132) An example would be

two sugar solutions separated by a selectively permeable membrane and have different

concentration gradients. Only the water would be able to diffuse because the sugar molecules are

too large for the pores. Concentration gradients also play a large role in solute diffusion. In order

for diffusion to occur, the solutions on either side of the membrane must not be in equilibrium.

When this concentration gradient is present, the ions or molecules will create a net diffusion down

the gradient so that dynamic equilibrium is reached. (Campbell and Reece: p.131) No work is

needed to be done in order for diffusion to occur, resulting in a spontaneous movement. It should

be recognized each substance diffuses down its own concentration gradient, not that of another.

(Campbell and Reece: p.131) An example of the usage of concentration gradients could be the

passing of a filtrate through a kidneys nephron. This is because the sodium chloride in the filtrate

at the loop of the Henle is removed, taking some water with it because of the drop in

concentration gradient as NaCl diffuses. A concept using concentration gradients is tonicity,

which is the ability of a solution to cause a cell to lose or gain water. (Campbell and Reece:

p.132). The solutions, in comparison to water concentration, are named hypotonic, isotonic, and

hypertonic.

Since the membrane is selectively permeable, only certain substances are allowed

to pass through and others are allowed to cross, but have difficulty. When a membrane separates

two solutions of different concentration gradients, it will allow random diffusion with a net

movement down the gradient until both sides have reached equilibrium. It allows such substances

as amino acids, sugars and numerous other nutrients to enter a cell and metabolic waste to leave.

(Campbell and Reece: p.130) It also regulates ion concentration and oxygen concentration. Since

the membrane is a nonpolar lipid bilayer, hydrophobic molecules are able to easily diffuse across.

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Ions and polar molecules, however, pass through much slower since they are charged. (Campbell

and Reece: p.130) In order to help substances that are charged or very large, transport proteins are

embedded in the membrane. There are channel and carrier proteins, which are still part of passive

transport. Sodium-Potassium pumps, however, can pump ions against their concentration gradient

known as active transport, because it uses ATP. In order to help increase the rate of osmosis in

some cells, special channel proteins known as aquaporins greatly assist in the diffusion of water.

(Campbell and Reece: p.130) This would allow two separated solutions to reach dynamic

equilibrium.

With selectively permeable membranes there is Osmotic pressure which is a measure of

the tendency of a solution to take up water when separated by a selectively permeable membrane

from pure water. If the molarity is different on either side of the membrane, then more water

would diffuse across it due to the greater concentration of solutes in the solution. Simply stated,

solutes and solvents will pass through a selectively permeable membrane due to differences in

concentration between the inside and outside of the membrane.

I believe that the higher the concentration of the solution, the larger the amount of water

diffusing in. This is because there are more particles that would need to diffuse into the potato and

dialysis tubing to reach equilibrium. I also believe that the longer the cube and dialysis tubing are

in the solutions, the larger the amount of solution diffusing in because more time allows more

particles to diffuse in order to reach equilibrium. If the potato cylinders are placed in distilled

water, I believe that their mass will decrease. This is because of the starch content in a potato

meaning that more solutes are present in the potato than in the distilled water. This means that

water will rush out of the potato making the water potential negative. If the onion cell is placed in

a salt solution then plasmolysis will occur, conversely, if the onion cell is place in pure water or

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distilled water (hypotonic solutions) then the cell will become engorged with water and a high

amount of turgor pressure will be present, the same pressure that holds a plant up.

Procedure:

Exercise B)

1) Obtain 6 strips of presoaked dialysis tubing.

2) Tie a knot in one end of each piece of tubing to form 6 bags. Fill them each up with

different molar solutions.

3) Rinse, dry and record weight of each bag.

4) Place each bag into a beaker to find the molarity of the solution in the dialysis bags.

5) Now fill each beaker with 2/3 of water or enough to completely submerge the bag.

6) After 30 minutes, remove bags from water and determine their mass.Exercise C)

1) Slice a potato transversely into 4 discs.2) Use potato borer to bore four (4) potato cylinders 5mm in diameter and 3cm long for

each molar solution.

3) Measure and record the mass of the 4 potato cylinders.

4) Pour assigned solution into a beaker.5) Put all 4 potato cylinders into the designated solutions and let sit overnight.

6) Remove cylinders. Place on pre-zeroed weigh boats, and measure and record their

total mass.7) Calculate percentage change from initial to final and graph data.

Exercise E)

1) Prepare a wet mount slide of an epidermis of an onion. Observe and record what yousee.

2) Add a few drops of a salt solution across the slide. Sketch and describe the onion cell.

3) Remove the cover slip and flood the onion cell with water. Observe and describe whathappened to the cell.

Data:

Exercise B)

Table 1.2

Percent Change in Mass of Dialysis Bags –Individual Data

Contents in

Dialysis Bag

Initial Mass Final Mass Mass Difference Percent Change

in Mass

0.0 M Distilled

Water

19.1 grams 19.1 grams 0.0 grams 0.0%

0.2 M Sucrose 22.2 grams 23.8 grams 1.6 grams 7.2%

0.4 M Sucrose 21.8 grams 24.5 grams 2.7 grams 12%




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Exercise C)

Table 1.4

Contents in

Beaker

Initial

Mass

Final

Mass

Mass

Difference

Percent

Change in

Mass0.0 MDistilled Water

4.1grams

3.8grams

.3 grams -7.32%

0.2 M Sucrose 5.2

grams

5.5

grams

.3 grams 5.80%

0.4 M Sucrose 5.3

grams

4.9

grams

.4 grams -7.54%

0.6 M Sucrose 5.5

grams

4.9

grams

.6 grams -10.90%

0.8 M Sucrose 5.5

grams

3.5

grams

2.0 grams -36.36%

1.0 M Sucrose 5.6grams

3.4grams

2.2 grams -39.29%

Exercise E)

1) The cells look like normal plant cells.

2) The vacuoles inside the cells shrunk (shriveled up) with no other changes.

3) The vacuoles increased back to their normal size and the cells appeared to look thesame as our original observation.

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Discussion:

My hypothesis towards the dialysis tubing proved to be correct; when the concentration

of the solution is higher, then the larger the amount of water diffusing into the dialysis tubing. You

can clearly see this in graph 1.1 as the correlation is very positive with a large fluctuation at .8M.

The results we had expected resembled what our graph had shown, except that one point at .8M.

Although the point at .8M does seem to be out of place, it still trends with the data and a positive

correlation there was just a larger influx of water into the tubing. When forming my hypothesis

about the potatoes, I had not had in mind that the pressure potential inside the potato would be

greater than the water outside. This is because of the starch content in a potato. More solutes are

present in the potato than in the distilled water, meaning that water will rush out of the potato

making the water potential negative. Graph 1.2 shows the negative correlation we had seen. The

peak in percent change of mass could have happened if there were different concentrations of

starch in the potato. A low concentration would make the water potential inside the potato positive

and allow water to be absorbed. The results that I had expected did not follow our graph. I thought

that the graph was going to start of relatively positive and then have negative values thereafter.

When the data line crossed zero we could determine the molarity of our solution. My hypothesis

had also stated that the longer the time in the solutions, the larger the amount of solution diffusing

in. Our experiment did not prove nor disprove this statement and was found to be irrelevant to

what we were experimenting.

The results of the onion experiment proved my hypothesis to be correct. As soon as the

salt solution was added to the wet slide, plasmolysis was very prevalent and could be witnessed

under the microscope. When water was flushed back onto the wet slide, the cell contents were

pressed against the cell wall once again and went back to their original state. All of these results

can be located in exercise E above.

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Water is the primary molecule needed in order to sustain life. Different amounts of

water affect cells of plants and animals differently. When an animal cell is in a hypotonic

solution, water will fill up the cell, commonly causing it to burst, or lyse. (Campbell and Reece:

p.133) This is because the concentration of solutes outside the cell is small and water diffuses

into the cell to make it even. When an animal cell is in an isotonic solution, the cell remains

normal. This is because there is a balanced amount of solutes in and outside of the cell. In a

hypertonic solution, the water will leave the cell, cause it to shrivel up. This is because water

inside the cell must leave in order to balance the solutes inside the cell. In plants cells, however,

the more water the better. When a plant cell is in a hypotonic solution, water enters the cell

making it turgid, which is the healthy state for plant cells. (Campbell and Reece: p.132) This is

because the uptake of water is eventually balanced when the plant’s cell walls push back on

themselves. In an isotonic solution, water will enter and leave the cell by diffusion. This causes

the plant cell to become flaccid and limp because there is no net tendency for water to enter. In

a hypertonic solution, water will leave the cell, causing it to plasmolyze. (Campbell and Reece:

p.133) This in turn will cause the plant to wilt and can be fatal for the plant.

Osmotic potential is directly related to solute concentration. This is because osmotic

potential is the potential for water to diffuse across a selectively permeable membrane. For

example, when an animal cell is in a hypotonic solution, there is a greater solute concentration

inside the cell compared to outside of it. Water would have the potential to travel down its

gradient into the cell in order to balance the solute concentration inside and outside. In a

hypertonic solution, there is a greater concentration of solutes outside than inside. For this case,

water would travel down its own gradient outside of the cell to decrease the concentration

outside and raise it inside until there is dynamic equilibrium.

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The molarity of the sucrose in the bag determines the amount of water that either moves

into or out of the bag, which in turn, changes the mass. For example, when the bag contained a

0.2M solution, water entered the bag to make the concentrations inside and outside of the bag

more equal. As this happened, the mass rose by 1.6 grams, a 7.2 percent increase. If each of the

bags were placed into a 0.4M solution instead of distilled water, the masses of the bags would

have changed in different ways. The mass of the bags filled with distilled water and 0.2M

sucrose would have gone down because water would have left the bag instead of entering it. The

mass of the 0.4M bag would have stayed the same because the concentrations are now equal. The

masses of the 0.6M, 0.8M, and 1.0M bags would have increased because water would have

moved into the bag to equalize the concentrations. In the experiment the percent change in mass

was calculated to show how much the mass increased due to the addition of water, which was

trying to equal the concentrations in both the bags and the cups.

InitialMass

FinalMass

Mass Differences Percent Change inMass

20.0

grams

18.0

grams

2.0 grams 18 – 20 = 2 ; 2 / 20 =

0.10 ; 0.10*100=10%change in mass

The water potential of the potato core after dehydrating it will decrease because the

water within the potato would evaporate and therefore lower the water potential. The solute

concentration of the plant cell is hypertonic because the solute concentration is higher than the

water concentration. Because of this, water will diffuse into the cell to reach dynamic

equilibrium.

As discussed before, plasmolysis is the loss of water and turgor pressure within a plant

cell and can be easily seen when the cells contents are pulled away from the cell wall, exactly

what we saw in our observations of the onion cell. When distilled water was then flushed over

the onion cell, removing all of the NaCl, the contents of the cell were then replenished with

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the water and were once again forced against the cell wall increasing the turgor pressure to

normal rates. The reason that the onion cells plasmolyzed was because the area surrounding

them had lower water potential thus water should have moved out of the cells. The same thing

can be correlated to grass that often dies near roads that have been salted to remove ice. The

water is drained from their cells as water moves out of the grass cells and into the hypertonic

NaCl area around it, killing the grass.

Four sources of error could be found in our lab report. This first error has to due with

surface area to volume ratios. We were asked to approximate two thirds of dialysis tubing to fill

and approximation lead to variation in diffusion of water across the membrane due to

differentiation of surface area of the dialysis tubing. The knots that were tied on each end also

led to this differentiation since uniform knots are not practical and relatively impossible to

make. Along with this was the diffusion of molecules when the dialysis tubing was in the

solutions. The tubing was not laid out flat or suspended in the solution therefore, overlapping

had occurred and other parts of the tubing were pressed against the side of the beaker. The sides

of the tubing that had overlapped or been pressed against the glass did not diffuse water and

surface area was lost leading to inconsistencies in diffusion rates. In order to fix this problem,

slight changes can be made. Instead of using dialysis tubing as our selectively permeable

membrane, a more rigid capsule-like membrane could be used. The capsule would ensure that

the surface area in all trials and experiments would be the same. Furthermore, no overlapping

would occur and it could be easily suspended in a solution rather than pressed against the side

of the beaker. The microscopes used in our experiments were common light microscopes. The

problem with light microscopes and wet slide observations is that the light produces heat and in

turn evaporates the water in the wet slide. This had happened in our experiment. Before NaCl

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was even added to our sample, plasmolysis could be faintly seen since our lamp was so hot.

Also, since the lamp was so hot when we added the NaCl to the solution, water evaporated to

such a degree that salt crystals were forming around the cover slip. It effectively broke the

bonds between NaCl’s polar covalent bonds with H20. It could then be stated that there was no

water potential to draw the water solution out of the cell. A light microscope with a light source

that stays cool could be used to simply fix these problems.

There are many further studies that could be branched off of this simple one. An

exciting one could be different surface area to volume ratios in relation to the diffusion of water.

This thought came from my knowledge of the different shapes of cells within the human body,

such as all of the different kinds and shapes of epithelial cells. Our experiment would be

relatively the same but only one solution would be used. Instead of using a core borer to attain

our potato samples, careful precise measurements would be made to ensure that the volume of

each same is the same but the surface areas are different. This way we can see which cell is

capable of diffusing the water the best. Another experiment includes the affects of the solutions

we had used on animal cells instead of plant cells. We would simply replace our onion cells

with an animal cell such as a red blood cell or a muscle cell. More powerful microscopes would

need to be used but the results could in turn be very rewarding and we could see the lysing of a

cell clearly, no cell wall would be present to prevent the cell from bursting.

References:

Campbell, Neil A., and Jane B. Reece. Biology: AP* Student Edition. 7th ed. San Francisco, CA:

Pearson Education, Inc., 2005. p. 130, 131, 132, 133

Contreras, Gretchen. "Diffusion and Osmosis." AP Biology Lab 1 Handout. Hightstown High

School. 2007.

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APBIO 11-21-08 Lab 1 Better