Plant Cells and Water Potential. Data Collecting and Processing

Transcription

1 Carlos Rodriguez October 8 th, 2011 Mrs. Hays Biology Lab Plant Cells and Water Potential Data Collecting and Processing In the following lab our group investigated water potential by immersing potato cores in sucrose solutions and recording the change in mass of the cores. In animal cells, osmosis in and out of the cell is influenced by the concentration of solute on either side of the membrane. In plants, the prediction of the movement of water is complicated because of the presence of the cell wall. As water enters the plant cell the cell membrane is pushed against the cell wall, creating a pressure that opposes the entrance of additional water. Even if the plant cell has a high concentration of solute the pressure will stop the water from entering the cell. This is why knowing the solute concentrations in either side of the cell membrane won t be enough to predict the movement of water; we must also take into account the hydrostatic pressure on either side of the plant cell membrane. Water potential is the physical property predicting the direction in which water will flow, governed by solute concentration and applied pressure. (Glossary) In the picture above the potato cores are in their sucrose solution. The sucrose concentration increases from left to right. An observation our group noticed which is also depicted above shows that as sucrose concentration increased so did the ability for the potato cores to float. In the last three cups the potato cores are floating, compared to the first three. This shows how the sucrose solution became more dense as the concentration increased, therefore allowing the potato cores to float.

2 Raw Data: Initial and Final Mass of Potato Cores in Sucrose Solutions Sucrose Solution in Cup Temperature ºC Initial Mass(g) Final Mass (g) Change in Mass (g) (M) (± 0.5 ºC) (± 0.01g) (±0.01g) (±0.01g) (%) 0.0M 23ºC M 23ºC M 23ºC M 23ºC M 23ºC M 23ºC These were the results of our experiment after having the potato cores immersed in the sucrose solution for 1 day. As seen in the table, 0.0M and 0.2M sucrose solution increased in mass by 27.50% and 5.10%. After 0.2M, the potato cores decreased in mass, the most by the 1.0M, decreasing by 45.50%. We also noticed that the potato cores that were floating suck to the bottom of the cup. Also, when we took out the potato cores out of the concentrations we noticed that as the sucrose concentration increased, the potato cores started to shrink and look more water deprived. Data Processing: Percent Change of Mass For the lab my group used four potato cores instead of one. In order to see if our results were in range of everyone else s results, I ve decided to create a standard deviation between my group s data and the class average. In order to get the class average I had to receive the other group s data and calculate their percent change in mass. This can be done by: Final Mass Inital Mass Initial Mass x 100 = % Cange of Mass After having calculated all the groups percent change, I will average their percent change. I have to do this for each molarity. After calculating the average for each, I will then calculate the standard deviation for each concentration. Having the standard deviation will help me compare my group s data with the class average. Then I added the error bars to make it easier to see if my group s data points were within range of the class averages. Having done all steps need to insure that my data was within range, now I can calculate the water potential of the solution at equilibrium and the water potential of the potato cells. To calculate water potential, the following formula must be used: Ψ = ψ p + ψ S

3 Water Potential = Pressure Potential + Solute Potential In the experiment the ψ p is 0 since the system is open to the atmosphere. The last part of the equation left to find is the ψ S. To find this we use the formula: Where: ψ S= icrt i = ionization constant (for sucrose, this value is 1) C = molar concentration of sucrose per liter at equilibrium (must be determined experimentally) R = pressure constant ( liter bar/mole K) T = temperature of solution in Kelvin s (K = Celsius + 273) (Carolina Biological Supply Company) To complete the calculation we need T and C. The temperature in the experiment was 23 ºC. To convert that to Kelvin s, just add 273, which equals 296. To find C one must look at the graph and find where the line passed 0 at the x-axis. This is the equilibrium point, which happened to be at around.25 sucrose molarity. This is the molar concentration of sucrose that produces equilibrium. So in the experiment ψ S= (296), so ψ S = 6.15 Now we plug in ψ S into the equation to find the water potential. Ψ= , so Ψ= The water potential of the solution at equilibrium was The water potential of the potato cells at equilibrium is also Processed data presentation: (All graphs and table are shown on the next page, due to size and formatting.)

4 Class Initial and Final Mass of Potato Cores in Sucrose Solution (*Note that Group 1 (my group) used four potato cores instead of one.) Group Initial Mass (g) (±0.01g) Final Mass (g) (± 0.01g) 0.0M 0.2M 0.4M 0.6M 0.8M 1.0M 0.0M 0.2M 0.4M 0.6M 0.8M 1.0M * Group Change in Mass (g) (± 0.01g) Percent Change of Mass (%) 0.0M 0.2M 0.4M 0.6M 0.8M 1.0M 0.0M 0.2M 0.4M 0.6M 0.8M 1.0M * Average (%) STDEV Class Average compared to Group 1 Results Group Percent Change (%) 0.0M 0.2M 0.4M 0.6M 0.8M 1.0M Class

5 Percent Change of Mass (%) Class and Group 1 Percent Change of Mass Average (Error bars are show standard deviation) M 0.2M 0.4M 0.6M 0.8M 1.0M Class Average Group Sucrose Concentration

6 Conclusion and Evaluation In conclusion in the experiment with 0.0M and 0.2M the solute solution was hypotonic to the cell, which caused the water to diffuse into the cell. This showed as an increase in the final mass of the potato core. With 0.0M the potato core mass went from 4.94g to 6.30g and in 0.2M it went from 4.81g to 5.07g. After these two sucrose concentrations, the solute concentration outside the cell was hypertonic, which caused water to exit the cell and resulted in a decrease in the final mass of the potato core. This could have resulted in plasmolysis. Apart from our quantitative data showing this, our qualitative data showed that when we took the potato cores out of the solutions, that as the sucrose concentrations increased the potatoes started to shrink and look water deprived, supporting the fact that water left the potato core because the sucrose solution was getting more hypertonic as the molarity increased. In 0.6M the potato core dropped from 4.99 to 3.66, a loss in weight. In 1.0M the percent change in mass was almost at half, it was 45.5%. After graphing all the data, we found the equilibrium point, which is the sucrose concentration in which the solution is at equilibrium and there is not net movement in or out of the cell, was at.25m. Using the water potential formula I found that the water potential was at bars. Our results were very accurate because our line crossed almost at the same time as the class average line at 0. There was only one molarity in which our data did not go with the class average, it was in 0.8M. Our data point is out of the error bars, showing that our data for 0.8M was not within the standard deviation. There were many weaknesses that could have caused this. The biggest weakness in this lab was that my group took too much time to cut the potato cores, which resulted in the loss of water, which could have affected the initial weight, final weight, and the percent change of mass. This weakness is caused by another weakness, which was having four potato cores. Even though having 4 potato cores showed that our results were accurate and within the standard deviation of the class data, having to cut 4 potato cores took too much time, and may have caused water loss in the potato cores that were cut first, since they all must have been weighted at the same time. Another weakness found in the experiment was being limited to 1 potato. Since we had to cut 24 potato cores that each measured 3cm, it was a very difficult task to try and do this with only one potato. To improve on these weaknesses I would change some of the materials and procedures in the experiment. First off, I would have changed it to one potato core. This would allow for a fast cutting and fast weighing of the potato with a loss of little to no water, since there was no waiting for other potato cores to be cut. This would allow for more valid and realistic results. This would also relieve the group from the pressure of having to cut 24 cores from 1 potato. The last thing I would change would be to have a better device to cut the potato cores. This would be more efficient in the experiment. With these improvements I think the procedure would have flowed more smoothly, also adding more validity and reliability in the results.

7 Works Cited 1. Carolina Biological Supply Company. "Activity C: Plant Cells and Diffusion." Biology Alive. Carolina Biological Supply Company, Web. 9 Oct < P%20Lab%201%20Diffusion%20-%20Osmosis%20Manual.pdf>. 2. "Glossary." Prentice Hall Bridge Page. Web. 09 Oct <