Arvind Sujeeth

Cradick - +A.P. Bio

2^nd Block

11/21/02

A

Title: The Diffusion of Glucose, Starch and IKI Through Dialysis Tubing

Problem:

How does the size of the pore in dialysis tubing (an artificial permeable membrane) influence the diffusion of sugars, starch and IKI?

Background Information:

All molecules have kinetic energy in the form of thermal motion, which results in diffusion – the tendency for molecules of any substance to spread out into the available space. In diffusion, a substance diffuses down a concentration gradient; in other words, a substance will diffuse from where it is more concentrated to where it is less concentrated. Diffusion across membranes is regarded as passive transport because no energy is required from the cell. However, because cell membranes are selectively permeable, not all molecules can diffuse freely across membranes. Diffusion of specific solutes is called dialysis and can also be accomplished through facilitated diffusion (with the help of transport proteins) and by active transport, the pumping of solutes against their gradients. To do this, the cell must expend energy – kind of like walking up a hill rather than going with the flow. The dialysis tubing used in this experiment has pores of a certain size; that size will determine which of the solutes can diffuse through it and at what rate (smaller molecules will diffuse faster and more easily). Hypertonic is a relative term that refers to a solution that has a greater solute concentration than its environment; in this case, the solute will diffuse out of the solution and water will diffuse in to the solution. Hypotonic refers to the opposite, in which case the solute will diffuse in to the solution and water will diffuse out of the solution. Isotonic refers to a solution having equal concentration to its environment, thus achieving a balance. IKI is an indicator that tests for the presence of starch, but that can also possibly diffuse through the dialysis tubing.

Hypothesis:

After 30 minutes, both solutions (glucose and starch) will diffuse across the dialysis tubing and therefore will be present in both the bag and the beaker. IKI will not diffuse across the tubing to be present in the bag.

Procedure:

The general procedure involves a beaker and a dialysis bag. Tthe dialysis bag contains a solution of the solutes (IV) glucose and sucrose while the beaker contains distilled water and IKI (to test for starch). The bag is placed inside the beaker, and after 30 minutes, the beaker is tested for the presence (DV) of glucose (with Testape) and starch, which will be present if glucose and starch diffused across the tubing. The control group is the starting point of the experiment before any time has elapsed, because the substances have not been allowed to diffuse and thus contain exactly their original contents. The experimental groups are glucose, starch and IKI.

Data Table:

Initial Solution Color Testape Results (Glucose)

Contents Initial Final Initial Final

Bag	15% glucose 1% starch	dark blue (starch present)	blue-black	positive	positive
Beaker	H20 + IKI	amber (no starch)	amber	negative	positive

Analysis:

1. It appears that perhaps a little bit of IKI diffused through the tubing, accounting for the color change in the bag from dark blue to blue-black. I account the color change to this, and not to decreased starch, because there was still no starch present in the beaker after 30 minutes – therefore no starch was able to diffuse out of the bag. And even if starch had diffused out of the bag, the color change would have been to a lighter shade, not a darker one – therefore IKI had to have diffused in. Glucose, the final solute, did appear to diffuse out of the bag into the beaker, based on the Testape results which showed that the final beaker solution contained glucose, while the initial did not.

2. At various intervals, the beaker could have been tested for density, thereby measuring quantitatively the water concentration inside the beaker. If the water concentration went down, water must have been diffusing in to the dialysis bag.

3. I would expect the color outside of the bag to go from clear to dark blue or blue-black, because the IKI would diffuse out of the bag and turn dark in the beaker due to the presence of starch. The bag would remain an amber color throughout, as no starch would be able to diffuse in. Both the beaker and bag would test positive for glucose at the final results, because some of the glucose would diffuse out in to the bag.

Conclusion:

The size of the pore in the dialysis tubing played a critical role in determining what solutes could pass in to or out of the bag. This artificial selectively permeable membrane is a model for the selectively permeable membranes used in real life cells, and shows how membranes can ward off certain molecules while allowing others to diffuse through. My hypothesis was partly incorrect, since IKI did diffuse and starch didn’t. However, because there was only 30 minutes allowed for diffusion, it is possible that the substance that did not diffuse in this experiment (starch) would be able to given more time (i.e., it was diffusing, but because it was a larger molecule, at a much, much slower pace than the other molecules). Through binding and molecular reactions it is also possible to “trap” a substance across a membrane, as we saw with IKI in the bag/beaker that was allowed to diffuse for a full day. If a solute is allowed to diffuse through a membrane and then bind to solutes on the other side and therefore not be able to diffuse back out, it provides an interesting possibility for natural filtration.

B

Title: The Effect of Varying Levels of Sucrose Concentrations On Osmosis Through

Dialysis Tubing

Problem:

What effect does solute concentration have on osmosis?

Background Information:

Osmosis is specifically the passive transport (diffusion) of water. Osmosis is a critical function of life because living cells rely on the balance of water between the cell and its environment. Water potential is the measure of free energy of water in a solution. When a cell’s water balance is maintained, the cell wall expands until it is turgid, or stiff; if the plant loses water and is unable to maintain a healthy water balance, its cells become flaccid. One result of becoming flaccid is plasmolysis, which is the separation of a cell’s plasma membrane from its cell wall, which is usually lethal to the plant. Thus, if a plant is unable to maintain a sufficient system of osmoregulation to control its water balance, the effect on the plant can be devastating. Molarity is a measure of the density of the solute inside the solution, to show increasing and decreasing levels of the solute. Similar to hypertonic and hypotonic, hyperosmotic and hypoosmotic are relative terms referring to the amount of water in a solution relative to the amount of water in its environment. This particular experiment works because the dialysis tubing is large enough to allow water to diffuse through, but small enough to disallow sucrose (the solute whose concentration is being tested) from diffusing through.

Hypothesis:

The solution of 1.0M sucrose will have the highest % change in mass, because that solution has a higher solute-to-water ratio than any of the solutions, and therefore has the lowest relative water potential and will have the most water diffusing in to it.

Procedure:

Dialysis bags contains varying concentrations (from .2 to 1-M, in intervals of .2) of sucrose (IV) are immersed in solutions of distilled water and let sit for 30 minutes. The mass (DV) of the bags are measured after the 30 minutes, which serves as a measure for the amount of water that diffused in to the bag. The greater the % change in the mass, the greater amount of water that diffused in to the bag. The control group is distilled water, and the experimental group is the varying solute concentrations (from .2 to 1-M).

Data Table:

On separate sheet.

Analysis:

1. There is a direct relationship between the increase of molarity of sucrose within the dialysis bags and the increase in mass of the dialysis bags. As the bags’ molarity increased in the experiment, more water diffused through the tubing into the bag through osmosis, resulting in a greater increase in mass.

2. If all the bags were placed in a solution of .4-M sucrose instead of distilled water, the bags would not always gain mass, because osmosis and diffusion is relative to the environment of the solution. Since the bags with less than .4-M sucrose would be hypotonic to the beaker, they would actually lose water to the beaker, and thereby lose mass. The bag with .4-M would be isotonic to the beaker, and there would be no net movement of water due to their equal water potentials. The bags with greater than .4-M of sucrose would by hyperosmotic and achieve the same results of a net gain in mass (although not as much) as water moves out of the beaker into the bag.

Conclusion:

My hypothesis proved correct: as the solute concentration increased within the bag, the flow of water through osmosis also increased. This holds true to the basic principles of osmosis and diffusion which state that water will move from an area of lower concentration to an area of higher concentration; the bags all had a lower concentration of water than their distilled water environment (the beaker), therefore all gained water through osmosis. Those with the least concentrations of water (1.0-M) gained the most through osmosis, and those with the greatest (.2-M) gained the least.

C

Title: Observing and Calculating the Changing Water Potential Through Osmosis of

Potato Cells

Problem:

What is the water potential of a potato cell?

Background Information:

As mentioned in section B, Water potential is the measure of free energy of water in a solution. To go further in depth, however, water potential is more concisely the combination of the pressure potential and the osmotic potential of a solution. The pressure potential of a solution is a measure of the exertion of pressure (tension) – either positive or negative – on a solution. Osmotic potential is a measure of the solute concentration of a solution. Water potential is abbreviated by the Greek letter “psi”, and is measured in units called bars. In the case of pure water open to the atmosphere, both osmotic potential and pressure potential are 0; 0 + 0 = 0, and thus the water potential of the open pure water is also 0. A solution can have a negative water potential if its pressure potential is 0 and its osmotic potential is negative due to the presence of a solute, for example. In this experiment, the water potential of potato cells will be determined by comparing it with various molar concentrations of sucrose; the concentration at which no net movement of water by osmosis can be observed has a water potential equivalent to the potato cells. Once water potential is shown to be equivalent, its quantitative value is calculated through the formula psi = - iCRT (i = ionization constant, in this case 1, C = osmotic molar concentration, R = pressure constant, .0831, T = temperature in Kelvins). This actually only calculates osmotic potential, but because the pressure potential in this experiment is 0, the water potential is equivalent to the osmotic potential. Again, water always diffuses by osmosis from a region with a greater water potential to a region with a lower water potential, and follows the rules and patterns of osmosis described in section B.

Hypothesis:

The water potential of .4-M sucrose will be equivalent to the water potential of the potato cells.

Procedure:

Begin by pouring 100 mL of the solution (IV –> each of the varying concentrations) into a 250-mL beaker. Use a cork borer to cut four potato cylinders, each skinned and about 3 cm in length. Take the mass of the four cylinders together (control group). Place them in the beaker and let stand overnight. The following day, record the temperature of the liquid in the beaker. Remove the cores from the beakers and take their mass again (DV).

The experimental group, is, of course, all of the solutions used to saturate the cores overnight.

Data Table:

Contents in Beaker	Temperature in Celsius	Initial Mass in grams	Final Mass in grams	% Change	% Change (class avg.)
Distilled water	20 degrees	2.6	3.63	39.6	31.3
.2-M sucrose	20 degrees	2.5	2.79	11.6	12.3
.4-M sucrose	20 degrees	2.5	2.1	-16	-7.6
.6-M sucrose	20 degrees	2.5	1.08	-56.8	-24.4
.8-M sucrose	20 degrees	2.2	1.5	-31.8	-23.8
1.0-M sucrose	20 degrees	2.4	No data	No data	-18.3

Analysis:

1. Personal Data:

Molar concentration of sucrose = .3-M, based on an estimated trend line equation of y = -102.71X + 30.79, yielding an x-intercept of .29976.

Class Average Data:

Molar concentration of sucrose = .335-M, based on an estimated trend line equation of y = -53.3X + 17.866, yielding an x-intercept of .335208.

2. Osmotic potential of this sucrose solution = -iCRT =

Personal: (1)(.3)(.0831)(293) = 7.30449

Class: (1)(.335)(.0831)(293) = 8.1567

3. Using the class average data as the more valid set of data, the water potential of the potato cells is equal to about 8.1567 bars.

4. The water potential would become lower because the cells would lose osmotic potential to its environment; i.e., the amount of water per molecule of solute would decrease drastically, thereby decreasing the amount of free energy per mole of water, which is the definition of water potential. In simpler terms, if water evaporated faster than the solute, there would be less water and the same amount of solute, which causes an even more negative measure of osmotic potential.

5. If a plant cell has a lower water potential than its surrounding environment, then it is hypertonic in terms of solute concentration to its surrounding environment. Because pressure potential is given to be 0, the only way for the environment to have a higher water potential would be to have a higher osmotic potential, which would require less solutes than the plant cell itself. Therefore, to have a higher water potential, the plant cell is required to have more solutes than its environment – it must be hypertonic to its environment.

Conclusion:

My hypothesis, based on the assumption that the water potential of a potato cell would probably reach equilibrium with the water potential of the sucrose solution near a moderate concentration of that solution proved nearly accurate – as accurate as I could have hoped for, given the available background. It actually reached equilibrium somewhere between .2-M and .4-M sucrose – at .335-M sucrose if the class average data is accurate. Either way, it is in the ballpark; even my group’s data, which consisted of some fairly “out there” numbers, yielded an equilibrium of molar concentration at .3-M, which is fairly close. I attribute the differences in data collection to a variety of atmospheric and human error factors that are present in almost all experiments and can never be completely prevented against. The final results, determining water potentials of the plant cells of 8.15 and 7.30 for the class and personal data respectively, do not appear to be very far off. Not knowing enough about water potential and the pressure measurement of bars, though, I cannot accurately determine the % error or the extremity of the difference between the two values – an avenue for further study, perhaps.