Category: 1st Semester

 

 

Introduction:

 

Atoms and molecules are the building blocks of cells. Both have kinetic energy and are constantly in motion. They continually bump into one another and bounce off into new directions. This action results in two important processes, diffusion and osmosis.

Diffusion is the random movement of molecules from an area of higher concentration of those molecules to an area of lower concentration. Cells have selectively permeable membranes that only allow the movement of certain solutes. Diffusion is vital for many of life’s functions in a cell. It allows oxygen and carbon dioxide exchange in the lungs and between the bodies of intracellular fluid and cells. Diffusion also aids in the transport of nutrients and water in the xylem and phloem of plants. In those plants, it permits for the absorption of water into roots. An example of this process is the diffusion of a smell in a room. Eventually dynamic equilibrium will be reached. This means that the concentration of the molecules carrying the smell will be approximately equal through out the surrounding enclosed area and no net movement of the molecules will occur from one area to another.

Osmosis is special kind of diffusion. It is the diffusion or movement of water through semi-permeable membranes from a region of higher water potential (hypotonic solute) to a region of lower water potential (hypertonic solute). Water potential is the measure of free energy of water in a solution. There are three types of solutions. Isotonic solutions have an equal concentration of solute on both sides of the membrane, and dynamic equilibrium has been reached in the solution. Hypertonic solutions have a higher concentration of solute on one side of the membrane than the other. Hypotonic solutions are the opposite of hypertonic solutions. A solute is what is being dissolved by the solvent (water is the most common solvent) in a solution.

Water will always move from an area of higher water potential to an area of lower water potential. An important factor effecting of diffusion and osmosis is water potential. Water potential measures the tendency of water to leave one place in favor of another place. Water potential is affected by two physical factors. One factor is the addition of solute, which lowers the water potential. The other factor is pressure potential. An increase in pressure raised the water potential. The water potential of pure water at atmospheric pressure is defined as being zero. The Greek letter psi is used to represent water potential. The following formula can be used for calculations:

ψ (Water potential) = ψp (Pressure potential) + ψs (Solute potential)

Movement of water into and out of a cell is influenced by the solute potential on one side of the cell membrane relative to the other side. Plasmolysis is a phenomenon in walled plant cells in which the cytoplasm shrivels and the plasma membrane pulls away from the cell wall when the cell loses water to a hypertonic environment. This leads to a loss of turgor pressure (the force directed against a cell wall after the influx of water and the swelling of a walled cell due to osmosis) and eventual death of the plant. If water moves into the cell, the cell may lyse, or burst (in animal cells, plant cells are equipped to handle large intakes of water). Water movement is directly proportional to the pressure on a system. Pressure potential is usually positive in living cells and negative in dead ones.

Diffusion and osmosis are not the only processes responsible for the movement of ions or molecules in an out of cells. Active transport is process that uses energy from ATP to move substances through the cell membrane. Normally, active transport moves a substance against its concentration gradient, that is to say from a region of low concentration to an area of higher concentration.

 

Hypothesis:

 

Osmosis and diffusion will continue until dynamic equilibrium is reached and net movement will no longer occur. Diffusion is effected by the solute size and concen-tration gradient across a selectively permeable membrane. Water potential greatly determines the results in sections of the experiment.

 

Materials:

 

Exercise 1A

For this exercise, the following materials are required: a 30 cm of 2.5 cm dialysis tubing, 250 ml beaker, distilled water, funnel, 2 dialysis tubing clamps, 15 ml of 15% glucose/1% starch solution, 4 pieces of glucose tape, 4 ml of Lugol’s solution (Iodine Potassium-Iodide or IKI), a timer, paper and pencil.

Exercise 1B

This exercise of the experiment requires six strips of 30 cm dialysis tubing, 250 ml beaker, 12 dialysis tubing clamps, funnel, six cups, distilled water, an electronic balance, timer, paper towels, and about 25 ml of each of these solutions: distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 M glucose, and 1.0 M glucose. For recording results, paper and pencil are necessary.

Exercise 1C

This part of the experiment requires a large potato, potato corer (about 3 cm long), 250 ml beaker, paper towels, scale, six cups, knife, paper, pencil and about 100 ml of each of these solutions: distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 glucose, and 1.0 M glucose.

Exercise 1D

This section requires a calculator, paper, pencil, and graphing paper.

Exercise 1E

This section of the experiment requires paper, pencil, paper towels, onionskin, dye, microscope, slide, cover slip, salt water (15%), and tap water.

 

Methods:

 

Exercise 1A

First, soak the dialysis tubing in distilled water for 24 hours. Before handling the tubing, wash dirty hands thoroughly to prevent getting oils on the dialysis tubing and changing the results. Remove the tubing and tie off one end using the clamp. To use the clamp, twist the end of the bag several times and then fold it onto itself. Next, open the other end of the tubing by rubbing the end between two fingers. Fill it with the glucose/starch solution using a funnel. Use the glucose tape by dipping it into the solution. Record the color change of the tape and the color of the bag. Tie of the end with the tubing clamp. It is necessary to leave space for expansion but no air. Fill the beaker with distilled water and add the 4-ml of Lugol’s solution. Record the color change. Use glucose tap to test for any glucose in the water. Record these results. Set the dialysis tubing in the beaker and let it sit for about 30 minutes. Remove the bag and record the change in water and bag color. Use the last two pieces of glucose tape to measure the glucose in the water and bag. Record results.

Exercise 1B

First, soak the dialysis tubing for about 24 hours. Again be sure to cleanse hands. Tie off one end of each tube with the clamps. Next, fill each tube with a different solution (distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 glucose, and 1.0 M glucose) with the funnel and tie off the end again leaving empty space, but no air. Weigh each bag separately on the electronic balance and record the masses. Soak the bags in separate cups filled with distilled water for about 30 minutes. Remove the bags and gently blot dry with paper towel. Reweigh, and record the mass.

Exercise 1C

First, slice the potato into to 3-cm discs. Use the potato corer to core out 24 cores. Weigh 4 cores together and record their mass. Fill each cup with one of the following solutions: distilled water, 0.2 M glucose, 0.4 M glucose, 0.6 M glucose, 0.8 glucose, and 1.0 M glucose. In each cup put 4 potato cores, and allow them to sit over night. Take out the cores and blot them dry. Again weigh them on the electronic scale. Record the change in mass. Calculate the information for the table. Compare the results with another group.

Exercise 1D

First, determine the solute potential of the glucose solution, the pressure potential, and the water potential. Graph the information given about the zucchini cores.

Exercise 1E

Prepare a wet mount slide of dyed onion skin. Observe under a light microscope and sketch how the cells appear. Add a few drops of the salt solution using a paper towel to wick the solution under the slip. Observe how the cells are effected and make another sketch.

 

Results:

Exercise 1A

 

 

Table 1: Change of Color of Dialysis Tubing and Beaker

 

Initial Glucose Tests Final Glucose Tests

 

 

Which substance(s) are entering the bag and which are leaving the bag? What experimental evidence supports your answer? Iodine Potassium Iodide and water enter the bag. This is proven by the color change (starch test) and the increase in the size of the bag. Glucose left the bag and this is proven by a positive test on the surrounding water.

 

Explain the results you obtained. Include the concentration differences and membrane pore size in your discussion. The results show that the water, glucose, and IKI molecules were small enough to pass through the selectively permeable membrane. The starch didn’t leave the beaker because its molecules were too large to pass through the selectively permeable membrane’s pores.

 

Quantitative data uses numbers to measure observed changes. How could this experiment be modified so that quantitative data could be collected to show that water diffused into the dialysis bag? The bags could be massed before and following their immersion in the solution. The volume of the solution in the beaker could be found before and after the immersion of the bag by using a graduated cylinder.

 

Based on your observations, rank the following by relative size, beginning with the smallest: glucose molecules, water, IKI, membrane pores, and starch molecules. The smallest substance was water, then the IKI molecules, glucose, the membrane pores, and the largest substance was the starch molecules.

 

What results would you expect if the experiment started with a glucose and IKI solution inside the bag and only starch and water outside? Why? Based on the size of the molecules, the glucose and IKI would move out of the bag and the water would go in. The large starch molecules would be left in the beaker.

Exercise 1B

 

Table 2: Dialysis Tubing Mass Change Results: Individual Data

 

 

Table 3: Dialysis Tubing Mass Change Results: Group Data

 

Graph 1: Percent Change in Mass of Dialysis Tubing in Sucrose Solutions of Different Molarities

Explain the relationship between the change in mass and the molarity of sucrose within the dialysis bags. The molarity is directly proportional to the percent change in mass. As the mass percentage increases, so does the molarity.

 

Predict what would happen to the mass of each bag in this experiment if all the bags were placed in a 0.4-M sucrose solution instead of distilled water. Explain your response. They are inversely proportional because whenever the sucrose molarity inside the bag is more concentrated, it will become more dilute and vice versa. The solutions will reach equilibrium somewhere between the two concentrations.

 

Why did you calculate the percent change in mass rather than simply using the change in mass? Each group began with different amounts of solution for their initial mass. Therefore, results cannot be based on those numbers. The differences in mass don’t deal with the proportional aspect of the solutions, making the real results less accurate. The percent was calculated to give the exact difference, along with considering the quantities of solution.

 

A dialysis bag is filled with distilled water and then placed in a sucrose solution. The bag’s initial mass is 20g, and its final mass is 18g. Calculate the percent change of mass, showing your calculations in the space below. 18g (final mass) – 20g (initial mass) / 20g (initial mass) = 2/20g x 100 = 10% change of mass

Exercise 1C

 

Table 4: Potato Core: Individual Results

 

Table 5: Potato Core Results: Class Data

 

 

 

Graph 2: Percent Change in Mass of Potato Cores at Different Molarities of Sucrose

 

 

Exercise 1D

Graph 3: Percent Change in Mass of Zucchini Cores at Different Molarities of Sucrose

 

If a potato is allowed to dehydrate by sitting in the open air, would the water potential of the potato cells decrease or increase? Why? The water potential of the potato would decrease because water moves from a high water potential region to a low potential region, and a dehydrated potato cell is hypertonic in comparison with the environment. The concentration of solute would increase and osmotic potential would decrease.

 

If a plant cell has a lower water potential than its surrounding environment, and if pressure is equal to zero, is the cell hypertonic or hypotonic to its environment? Will the cell gain water or lose water? Explain your response. If the plant cell has lower water potential, that means the water will come into the cell; the cell is hypertonic to its environment. This cell will gain water because water follows its concentration gradient.

 

In figure 1.5, the beaker is open to the atmosphere. What is the pressure potential of the system? The pressure potential is zero.

 

In figure 1.5, where is the greatest water potential? The greatest water potential is within the dialysis bag.

 

Water will diffuse_________the bag. Why? Water will diffuse out of the bag because the inside of the bag has the highest water potential.

 

Calculate solute potential of the sucrose solution in which the mass of the zucchini cores does not change. Show work. ψ s = -iCRT ψ s = (-1)(0.36 mole/liter)(0.0831 liter bar/mole K)(300 K) ψ s = -8.975 bars

 

Calculate the water potential of the solutes within the zucchini cores. Show work. ψ = ψ s+ ψ p ψ =0 + -8,975 , ψ = -8.975 bars

 

What effect does adding solute have on the solute potential component of that solution? Why? Adding solute to a solution would increase the solute potential and decrease the water potential.

 

Consider what would happen to a red blood cell placed in distilled water:
a. Which would have the higher concentration of water molecules? The

 

distilled water would have the higher concentration of water molecules.

 

Which would have the higher water potential? The distilled water would also have the higher water potential.

 

What would happen to the red blood cell? Why? The red blood cell would take in a lot of water and might lyse due to pressure inside. This is because animal cells lack tolerance under hypotonic situations.

 

Exercise 1E

 

Describe the appearance of the onion cells. The onion cells appear to have great turgor pressure, spread out, thick and bright in the inside. The cell walls were very defined and it was clear where one cell ended and another began.

 

Describe the appearance of the onion cells after the NaCl was added. The plasma membrane shriveled from the cell wall, or in other words, plasmolysis occurred.

 

Remove the cover slip and flood the onion with fresh water. Observe and describe what happened. The onion cells were again hypertonic to their environment and were restored to their original state of appearance.

 

What is plasmolysis? Plasmolysis is the separation of the plasma membrane from the cell wall in a plant cell.

 

Why did the onion cell plasmolyze? The environment around the cell was hypertonic to the cell so water left the cell to reach dynamic equilibrium with the NaCl solution. With all the water leaving the cell, the cell membrane separated from its cell wall.

 

In the winter, grass often dies near roads that have been salted to remove ice. What causes this to happen? The salt causes the grass’s environment to become hypertonic, and the water leaves the plant cells, causes withering and eventually death of the plant. The high concentration of salt in the soil also speeds the death of the plant.

 

 

Sketch of Onion Cells Onion Cells + NaCl

 

Error Analysis:

 

Several could have possibly been made throughout the lab.

Exercise 1A

The data collected in this lab experiment did not appear to contain any errors, however, an error in the results may have unknowingly occurred. If there was a leak where the tubing was twisted shut or a tear in the dialysis tubing, all of the data would be inaccurate.

Exercise 1B

This section of the lab had to be repeated because of incorrect data (that is to say it did not “harmonize” with the other groups’ data). If the person handling the dialysis tubing did not wash their hands thoroughly and accidentally touched the portion of the tubing to serve as the permeable membrane, the oils from their hands could have blocked pores on the tubing, effecting the data.

Exercise 1C
Some mistakes that could have taken place are mathematical miscalculations while finding the initial and final masses. A piece of potato skin could have been left in the beakers along with the potato. This causes problems in the data tables. Another possible source of error could be that the students did not pat dry the potato sample well enough and increased the masses of the cores. Numerous may have occurred while using the electronic balance.

Exercise 1D

In this part of the lab, only calculations were made. Simple mathematical errors are bound to occur in this section of the lab.

Exercise 1E

In part 1E, after adding the NaCl solution to the onion cells, the cells should have reduced in size, but no reaction appeared to take place. This may have occurred in part because the onion itself was already dried out and dehydrated, or while the onion was being looked at through the microscope, the heat from it may have caused the cells to loose water. Another possibility is that the reaction took place so quickly that those witnessing could not see it.

Discussion and Conclusion:

 

Exercise 1A

The data shows what molecules can and cannot diffuse across a selectively permeable membrane. The color change showed that the Iodine Potassium Iodide was small enough to pass through the pores of the membrane. It is shown that the water and glucose solution moved out of the dialysis bag because water is small enough to pass through the membrane and the Testape tested positive for glucose inside the beaker. The glucose started out inside the bag and tested negative with the Testape inside the beaker before the immersion.

Exercise 1B

It can be concluded from the results gathered during the experiment that sucrose cannot pass through the selectively permeable membrane, but instead water molecules must move across the membrane to the area of lower water potential to reach dynamic equilibrium.

 

Exercise 1C

The results provided information that leads us to conclude that potatoes do contain sucrose molecules. This is known because the cores took in water while they were emerged in the distilled water. This means they had a lower water potential and higher solute potential than the distilled water.

Exercise 1D

The calculations made it evident that all of the results could be determined and proven correct with the simple equations and formulas. Performing these mathematical computations helped give a better understanding of water and solute potential.

Exercise 1E

This particular part of the lab illustrated the shrinking of the plasma membrane from the cell wall in a plant cell, or, in other words, plasmolysis. It shows how plant cells react in a hypertonic environment, the NaCl solution. The turgor pressure decreases as water leaves the cell. This shows how the onion cells had high water potential so water moved to the area outside the cell with lower water potential. Then, after adding water back to the cells, water moved back into the cells, restoring turgor pressure.

Overall

Water potential and concentration gradients are the two phenomenons that effected the results of the experiments. There are many important facts pertaining to water potential. Water potential is used by botanists to determine the movement in and out of a cell. It is effected by two components, pressure and solute potential. Water moves from areas of higher water potential (higher free energy and more water molecules) to areas of lower water potential (lower free energy and less water molecules). Water diffuses down a water potential gradient. Pure water has an atmospheric pressure of zero which is important when using the formula ψs = -iCRT. Water potential is inversely proportional to solute potential. These facts led to or effected the results gained in each section of the lab.

In plant and animal cells, loss or gain of water can have different effects. In an animal cell, it is ideal to have an isotonic solution. If the solution is hypertonic, the cell will shrivel from lack of water intake. Inversely, if the solution is hypotonic the cell could take in too much water and the cell will lyse and break open. For a plant cell, the ideal solution is a hypotonic solution because the cell takes in water increasing turgor pressure. Turgor pressure is important for plant support and maintaining shape. If the solution is hypertonic, the cell will plasmolyze and died from lack of water. In an isotonic solution, the plant cell does not have enough turgor pressure to prevent wilting and possible death. The information gained through this lab is important in understanding the effects of different solutions on organisms in our environments, including ourselves.

 

 

Introduction:

Atoms and molecules are constantly in motion. This kinetic energy causes the molecules to bump into each other and move in different directions. This motion is the fuel for diffusion. Diffusion is the random movement of molecules from an area of higher concentration to an area of lower concentration. This will occur until the two areas reach a dynamic equilibrium. When this dynamic equilibrium is reached the concentration of molecules will be approximately equal and there will be no net movement of molecules after this point. The molecules will still be in motion but the concentrations will remain the same.

Osmosis is a special kind of diffusion in which water moves through a selectively permeable membrane. A selectively permeable membrane allows diffusion for only certain solutes (the substance being dissolved) and water, the most common solvent (a dissolving substance). The most common selectively permeable membrane is the cell membrane. Water moves from an area of high water potential to an area of low water potential. Water potential is the measure of free energy of water in a solution and is represented by the symbol ψ (psi). Water potential is affected by two physical factors: the addition of a solute (ψs) and pressure potential (ψp). The addition of solutes to a concentration will lower the water potential of that solute, causing water to move into the area. Water movement is directly proportional to the pressure potential. Water potential can be determined by the equation:

ψ = ψp + ψs

Pure water has a water potential of zero. The addition of solutes will cause the water potential value to be negative, while an increase in pressure potential will cause a more positive water potential value.

There are three relationships that can occur between two solutions. When two solutions have equal solute concentrations, they are isotonic and no net movement of solute occurs. There is also no net movement of water. If the two solutions differ in solute concentrations, they will either be hypertonic or hypotonic. The hypertonic solution has a lower concentration of solute. Water will move out of a hypertonic solution, while solute will move in, (moving up the concentration gradient-similar to water potential). This depends on the selective qualities of the membrane. Pertaining to cells, this will cause the cell to shrivel or become flaccid. The hypotonic solution has a higher concentration of solute, and therefore has less water. This solution will gain water, while losing solute. This movement between the hypotonic and hypertonic solutions will continue until the point of dynamic equilibrium is reached. A hypertonic cell may also undergo plasmolysis. Plasmolysis is the shrinking of the cytoplasm in a plant cell in response to the diffusion of water out of the cell. When a cell is hypotonic it may lyse. In plant cells, it creates turgor pressure against the cell walls keeping the plant from becoming wilted.

Besides osmosis and diffusion, molecules and ions can be moved by active transport. This process includes the use of ATP to drive molecules in or out of a cell. Active transport is generally used to move molecules against a concentration gradient, from an area of low concentration to an area of higher concentration of molecules.

 

Hypothesis:

In this experiment, diffusion and osmosis will occur until dynamic equilibrium is reached. This experiment is done in a theoretical condition with no other variables affecting the movement of the solute except water potential.

 

Materials:

 

Exercise 1A

This exercise requires a 30 cm of 2.5 cm dialysis tubing, 250 ml beaker, distilled water, 2 dialysis tubing clamps, 15 ml of 15% glucose/1% starch solution, 4 pieces of glucose tape, 4 ml of Lugol’s solution (Iodine Potassium-Iodide or IKI), and clock or timer.

Exercise 1B

This experiment requires six strips of 30 cm dialysis tubing, 250 ml beaker, 12 dialysis tubing clamps, distilled water six cups, scale, timer or clock, paper towels, and about 25 ml of each of these solutions: distilled water, .2 M glucose, .4 M glucose, .6 M glucose, .8 glucose, and 1.0 M glucose.

Exercise 1C

This experiment requires a large potato, potato corer (about 3 cm long), 250 ml beaker, paper towel, scale, six cups, knife, and about 100 ml of each of these solutions: distilled water, .2 M glucose, .4 M glucose, .6 M glucose, .8 glucose, and 1.0 M glucose.

Exercise 1D

This experiment requires a calculator, paper, pencil, and graphing paper.

Exercise 1E

This experiment requires onion skin, dye, microscope, slide, cover slip, salt water (15%), and tap water.

 

Methods:

 

Exercise 1A

First, soak the dialysis tubing in distilled water for 24 hours. Remove the tubing and tie off one end using the clamp (twist tubing end about 7 times and fold end on self, slide into the clamp). Next, open the other end of the tubing (rubbing end between fingers) and fill it with the glucose/starch solution. Use the glucose tape and record the color change of the tape and the color of the bag. Tie of the end with the tubing clamp (leave empty space, but no air). Fill the beaker with distilled water and add the 4-ml’s of Lugol’s solution, record the color change. Use glucose tap to test for any glucose in the water (record). Set the dialysis tubing in the beaker and let it sit for about 30 minutes. Remove the bag and record the change in water and bag color. Use the last two pieces of glucose tape to measure the glucose in the water and bag and record results.

Exercise 1B

First, soak the dialysis tubing for about 24 hours. Tie off one end of each tube with the clamps. Next, fill each tube with a different solution (distilled water, .2 M glucose, .4 M glucose, .6 M glucose, .8 glucose, and 1.0 M glucose) and tie off the end (leave empty space, but no air). Weigh each tube separately and record the masses. Soak the tubes in separate cups filled with distilled water for about 30 minutes. Remove the tubing, blot dry, reweigh, and record the mass.

Exercise 1C

First, slice the potato into to 3-cm discs. Use the potato corer and core out 24 cores (don’t get any). Weigh 4 cores together and record the mass. Fill each cup with a different solution (distilled water, .2 M glucose, .4 M glucose, .6 M glucose, .8 glucose, and 1.0 M glucose). In each cup put 4 potato cores and let it sit over night. Take out the cores and blot them dry. Record the change in mass. Calculate the information and compare.

Exercise 1D

First, determine the solute potential of the glucose solution, the pressure potential, and the water potential. Then, graph the information given about the zucchini cores.

Exercise 1E

First, prepare a wet mount slide of dyed onion skin. Observe under a light microscope and sketch what the cells. Add a few drops of the salt solution, observe, and sketch the change.

 

Results:

 

Exercise 1A

 

Table 1    Change of Color of Dialysis Tubing and Beaker

 

 

Which substance(s) are entering the bag and which are leaving the bag? What experimental evidence supports your answer?

 

Glucose is leaving the bag slowly; this is shown buy using the glucose tape in the beaker. Iodine is entering the dialysis bag; this is shown by the change of color in the bag. Water enters the bag; the bag becoming fatter, which shows this.

 

Explain the results you obtained. Include the concentration differences and membrane pore size in your discussion.

 

The substances moved and out of the bag, according to the gradient. Some were small substances and moved in and out of the bag quickly and easily. The larger substances, like starch and glucose, were slow or didn’t enter or leave the bag at all.

 

Quantitative data uses numbers to measure observed changes. How could this experiment be modified so that quantitative data could be collected to show that water diffused into the dialysis bag?

 

The mass of the bag could be recorded before and after soaking.

 

Based on your observations, rank the following by relative size, beginning with the smallest: glucose molecules, water, IKI, membrane pores, and starch molecules.

 

Water→ IKI→ Glucose Molecules→ Membrane Pores→ Starch Molecules

 

What results would you expect if the experiment started with a glucose and IKI solution inside the bag and only starch and water outside? Why?

The IKI would have left the bag and changed the color of the solution in the beaker. Eventually dynamic equilibrium would be reached and there will be no net movement.

Exercise 1B

 

Table 2    Dialysis Bag Results: Individual Data

 

* To Calculate: Percent Change in Mass = Final Mass – Initial Mass Initial Mass * 100

 

Table 3

Dialysis Bag results: Class Data

 

 

 

Graph 1

Percent Change in Mass of Dialysis Tubing in Glucose Solutions of Different Molarity

Explain the relationship between the change in mass and the molarity of sucrose within the dialysis bags.

 

They are directly proportional. The percent change of mass goes up as the molarity of the sugar goes up.

 

Predict what would happen to the mass of each bag in this experiment if all the bags were placed in a 0.4 M sucrose solution instead of distilled water. Explain your response.

 

There will be no net movement when 0.4 M is in the dialysis bag. When the concentration is above 0.4, the bag will lose water. When the concentration is below 0.4, the bag will gain water.

 

Why did you calculate the percent change in mass rather than simply using the change in mass?

 

The volumes of the solutions were not exactly the same.

 

A dialysis bag is filled with distilled water and then placed in a sucrose solution. The bag’s initial mass is 20g, and its final mass is 18g. Calculate the percent change of mass, showing your calculations in the space below.

 

Percent Change in Mass = Final Mass – Initial Mass    X  100 = 18-20 x 100 = 10%

Initial Mass                               20

 

 

The sucrose solution in the beaker would have been hypertonic to the distilled water in the bag.

 

Exercise 1C

 

Table 4     Potato Core: Individual Data

 

 

Table 5     Potato Core Results: Class Data

 

 

 

Graph 2

Percent Change in Mass of Potato Cores at Different Molarities of Glucose

 

Exercise 1D

 

If a potato is allowed to dehydrate by sitting in the open air, would the water potential of the potato cells decrease or increase? Why?

 

The water potential of the cells will decrease, the osmotic potential will decrease, and the solute will increase. This occurs because the cells have become dehydrated.

 

If a plant cell has a lower water potential than its surrounding environment, and if pressure is equal to zero, is the cell hypertonic or hypotonic to its environment? Will the cell gain water or lose water? Explain your response.

 

The environment is hypotonic, so it will gain water. This is because it has less water than the surrounding environment.

 

In figure 1.5, the beaker is open to the atmosphere. What is the pressure potential of the system?

 

The pressure potential is zero.

 

In figure 1.5, where is the greatest water potential?

 

The dialysis bag.

 

Water will diffuse out of the bag. Why?

 

The water will diffuse out because there is higher water potential inside the bag.

 

Zucchini cores placed in sucrose solutions at 27° C resulted in the following percent changes after 23 hours:

% Change in Mass Sucrose Molarity

20% Distilled Water

10% 0.2 M

-3% 0.4 M

-17% 0.6 M

-25% 0.8 M

-30% 1.0 M

Graph 3

Percent Change in Mass of Zucchini Cores of Sucrose Solutions of Different Molarity

b) What is the molar concentration of solutes with in the zucchini in cells?

About .36 M

 

Refer to the procedure for calculating water potential from experimental data.

Calculate solute potential (ψs) of the sucrose solution in which the mass of the zucchini cores doesn’t change. Show work.

 

ψs =-iCRT
ψs =-(1)(0.35)(0.0831)(295)
ψs =-8.580075

ψ =0+ ψs
ψ =0+(-8.580075)
ψ =-8.580075

 

Calculate the water potential (ψ) of the solutes within the zucchini cores. Show work here.

 

ψ = ψs + ψp

-8.580075 = ψs + 0

-8.580075 = ψs

 

What effect does adding solute have on the solute potential component of that solution? Why?

 

Adding more solute will increase the solute potential and decrease water potential by making it more negative.

 

Consider what would happen to a red blood cell placed in distilled water:
a) Which would have the higher concentration of water molecules?

 

Distilled Water

b) Which would have the higher water potential?

 

Distilled Water

c) What would happen to the red blood cell? Why?

The red blood cells would pull in water and lyse.

 

Exercise 1E

 

Prepare a wet mount of a small piece of epidermis of an onion. Observe under 100x magnification. Sketch and describe the appearance of the onion cells.

 

 

 

Describe the appearance of the onion cells after the NaCl was added.

 

The plasma membrane shriveled from the cell wall, causing plasmolysis.

 

Remove the cover slip and flood the onion with fresh water. Observe and describe what happened.

 

The onion cells absorbed water and increasing in turgor pressure.

 

What is plasmolysis?

 

Plasmolysis is the shrinking of the cytoplasm of a plant cell in response to diffusion of water out or the cell and into a hypertonic solution surrounding the cell.

 

Why did the onion cell plasmolyze?

 

The environment became hypertonic to the cell and the water left the cell running with its concentration gradient because of the salt. With all the water leaving the cell, it shrank, leaving behind its cell wall.

 

In the winter, grass often dies near roads that have been salted to remove ice. What causes this to happen?

 

The salt causes the plant cells to plasmolyze.

 

Error Analysis:

Exercise 1A

Error that could have occurred in this experiment is that some of the glucose/starch solution could leaked into the beaker before the dialysis bag was inserted, but after the glucose test.
Exercise 1B

Error that could have occurred in this experiment is that the sugar could mot have been mixed completely with the water or some sugar could have been lost during the mixing. Another mistake could be that the bags were not bloated dry well enough.
Exercise 1C

Error that could have occurred in this experiment is that the sugar could mot have been mixed completely with the water or some sugar could have been lost during the mixing. Another mistake could be that the potato cores were not bloated dry well enough. Also the measuring of the liquids’ volumes may not have been accurate.
Exercise 1D

Errors that could have been made in this exercise could be that the numbers were put in the calculator wrong.
Exercise 1E

Error that could have been made in this exercise could have been that the onion skin could have dried out before the salt water was added, thus affecting the results.

 

Discussion and Conclusion:

 

Exercise 1A

In this experiment, the ability of substances to move across a selectively permeable membrane was viewed. A glucose/starch solution was put in the dialysis bag. The glucose molecules leaked out of the bag (learned from before and after test with glucose tape). Using IKI to test for starch, the change in color for the bag only shows that the starch molecules were too large to escape out of the dialysis bag, but the IKI molecule were small enough to enter the bag.

Exercise 1B

In this experiment, the study of hypotonic and hypertonic solution was tested. When the bags were placed in a hypotonic solution, they gained water. This could be viewed by massing the bags before and after the soaking.

Exercise 1C

In this experiment, potato cores were proven to have some sugar in them. When placed in low or no sugar environments, they gained water. When placed in high sugar environments, the cores lost water.

Exercise 1D

In this experiment, all these conclusion made in previous experiments were reinforced with scientific equations.

Exercise 1E

In this experiment, the turgor pressure of a plant cell’s plasma membrane was observed. The plasma membrane gives the cell shape and form. When salt is added, the cytoplasm loses water and causes the plasma membrane to shrink. This causes the plant to wilt.

Concentration gradient and water potential affected all aspects of this experiment. Water potential is used by many scientists to study the effects of different substance on plants, for good or bad. Pressure potential and solute potential are the two main components. Water moves to different areas based on water potential. Water will move from high water potential to low or low water potential to high. This change is called a gradient. Ψs = -iCRT is the formula used for solute potential. Water potential and solute potential are inversely proportional. When water potential goes up, solute potential is low. All these factors affected the results of these experiments.

Plant and animal cells react differently to different environment. A hypertonic solution will cause an animal cell to shrink in size. A hypotonic solution for an animal cell will cause it to lyse. Isotonic solutions are ideal for animal cells. Plant cell’s plasma membrane shrinks if in a hypertonic solution, causing the cell wall to loose shape or plasmolyze. While an isotonic solution is good, it doesn’t provide quite enough support for the cell wall. Hypotonic solutions are the best for plant cells. The plasma membrane presses against the inside of the cell wall giving it a lot of support. Without this pressure, turgor pressure, the plant will wilt and die. These experiments enable us to better understand living things, including our own bodies, and be able to take care of them.