"Party Gel" Transport Pre-Lab:
a.) The in class demonstration indicated that when the phenolthalin (base indicator), was added to the agarose gel, it formed a hard, hot pink solution. Although , once white vinegar (acid), was added, the solution became clear. The pink color,initially, was due to the presence of the phenolthalin. However, when the acidic vinegar was added to the solution of the agarose gel, the solution became either neutral or slightly acidic, and caused the phenolthalin to be deactivated.
b.) Once the vinegar reacted with the agarose gel and phenolthalin, there was a slight clear border around the edges of the shapes. The border eventually widened as the time passed. The clear border formed from the edges toward the middle. In the morning, about 24 hours later, the shapes were completely clear. Surface area did to play a role in the reaction of both the shapes. The rate at which the shape reacted was based upon their surface area. The more exposed area on the shape, the quicker it would react. The star had a higher surface area-to-volume ratio than the heart, and that led it do be affected by the vinegar quicker.
b.) Once the vinegar reacted with the agarose gel and phenolthalin, there was a slight clear border around the edges of the shapes. The border eventually widened as the time passed. The clear border formed from the edges toward the middle. In the morning, about 24 hours later, the shapes were completely clear. Surface area did to play a role in the reaction of both the shapes. The rate at which the shape reacted was based upon their surface area. The more exposed area on the shape, the quicker it would react. The star had a higher surface area-to-volume ratio than the heart, and that led it do be affected by the vinegar quicker.
Lab Procedure:
1. Fill each 50 mL beaker with 30 mL of sucrose-water solution
2. Core the sweet potato, and then cut each of the cored sweet potatoes into 1 in strips
3. Weigh each piece and calculate the mass
4. At the same time, place a 1 in sweet potato strip into each of the 5 beakers, labeled A-E.
5. Leave the sweet potatoes in the solution for 15 minutes
6. Weigh each piece again and calculate the new mass.
7. Put the sweet potatoes back into the different solutions for another 5 minutes
8. Take out each sweet potato from its solution and weigh them again
9. Record data
Variables:
Independent Variable: Concentration of sucrose in the sugar water
Dependent Variable: weight of the sweet potato cores
2. Core the sweet potato, and then cut each of the cored sweet potatoes into 1 in strips
3. Weigh each piece and calculate the mass
4. At the same time, place a 1 in sweet potato strip into each of the 5 beakers, labeled A-E.
5. Leave the sweet potatoes in the solution for 15 minutes
6. Weigh each piece again and calculate the new mass.
7. Put the sweet potatoes back into the different solutions for another 5 minutes
8. Take out each sweet potato from its solution and weigh them again
9. Record data
Variables:
Independent Variable: Concentration of sucrose in the sugar water
Dependent Variable: weight of the sweet potato cores
Hypothesis:
The potato with the highest percent increase in weight during the total time interval was placed in the solution with the lowest sucrose concentration. The potato with the lowest percent increase in weight during the total time interval was placed in the solution with the highest sucrose concentration.
Data: Sweet Potato Weight Change
Conclusion Questions:
a.) After conducting the lab, it was found that the least concentrated solutions were C and E. Whereas the most concentrated solutions were found to be A, B, and D. This grouping of the solutions were made possible by considering osmosis. If the sucrose solution in the beaker was lower than the sucrose
concentration of the cell then water would enter the cell in an attempt to reach an isotonic state. This explains that the core's weight would increase. The sucrose solution in the beaker if higher than the sucrose concentration of the cell would result in water leaving the cell to reach an isotonic state. This indicated that the core's weight would decrease in size. According to the data, the no change in the weight of solutions A,B, and D made them match with the water potential of the potato.
b.)
A. 24.5145(C) = 24.5145 (0.8 M) = 19.6116
B. 24.5145(C) = 24.5145 (0.2 M) = 4.9029
C. 24.5145(C) = 24.5145 (0.6 M) = 14.7087
D. 24.5145(C) = 24.5145 (0.4 M) = 9.8058
E. 24.5145(C) = 24.5145 (1.0 M) = 24.5145
The list did not match the actual calculations. There was either no change or a very slight change among the potatoes in the solution.
c.) If plant cells were left in a hypotonic solution over the weekend, they would be unlikely to burst. Plant cells have large contractile vacuoles, that expand when water enters the cell in large amounts. Unlike an animal cell, plant cells are less likely to burst because their vacuoles are able to hold the water in.
d.) If this experiment was conducted with animal cells or tissues, it would not be as effective. Animal cells may not have the same high sucrose concentrations as the potato did. This would lead to different results, unlike the plant cell who either stayed at the same weight or gained weight. Animal cells might loose weight. Also the ability of plant cells to take in large amounts of water without bursting is not shared with animal cells. An animal cell would be more likely to burst with the weight gain.
concentration of the cell then water would enter the cell in an attempt to reach an isotonic state. This explains that the core's weight would increase. The sucrose solution in the beaker if higher than the sucrose concentration of the cell would result in water leaving the cell to reach an isotonic state. This indicated that the core's weight would decrease in size. According to the data, the no change in the weight of solutions A,B, and D made them match with the water potential of the potato.
b.)
A. 24.5145(C) = 24.5145 (0.8 M) = 19.6116
B. 24.5145(C) = 24.5145 (0.2 M) = 4.9029
C. 24.5145(C) = 24.5145 (0.6 M) = 14.7087
D. 24.5145(C) = 24.5145 (0.4 M) = 9.8058
E. 24.5145(C) = 24.5145 (1.0 M) = 24.5145
The list did not match the actual calculations. There was either no change or a very slight change among the potatoes in the solution.
c.) If plant cells were left in a hypotonic solution over the weekend, they would be unlikely to burst. Plant cells have large contractile vacuoles, that expand when water enters the cell in large amounts. Unlike an animal cell, plant cells are less likely to burst because their vacuoles are able to hold the water in.
d.) If this experiment was conducted with animal cells or tissues, it would not be as effective. Animal cells may not have the same high sucrose concentrations as the potato did. This would lead to different results, unlike the plant cell who either stayed at the same weight or gained weight. Animal cells might loose weight. Also the ability of plant cells to take in large amounts of water without bursting is not shared with animal cells. An animal cell would be more likely to burst with the weight gain.
Sources of Error:
There were several sources of error in the mitosis lab. The scale used to weigh each potato could have caused much error itself as it was only capable of measuring to the nearest tenth of a gram. The measurements were general, as the small mass of each potato piece were calculated. Had there been a more accurate reading of the mass of each potato, then there would have been a better sense of disparity between it and the solution.
While experimenting, several water droplets fell on to the scale, by mistake. The water droplets on the scale could have led to altered measurements. The water also was factored into the measurements of the mass of the potato as it was on the scale . Because it is unclear of exactly how much the water droplets themselves weighed, they cannot be subtracted from the scales calculations. A small source of error could have been that the person who was calculating the mass was not rounding correctly.
The choice to use sweet potatoes rather than the other choices could have obscured some of the lab results as well. The sweet potato is naturally more sweet, and this caused the results of the reaction to be less noticeable. The higher sucrose content of the sweet potato when mixed with the water and sucrose solution had less of an effect than that of another potato would have, because the sweet potato decreased in disparity .
While experimenting, several water droplets fell on to the scale, by mistake. The water droplets on the scale could have led to altered measurements. The water also was factored into the measurements of the mass of the potato as it was on the scale . Because it is unclear of exactly how much the water droplets themselves weighed, they cannot be subtracted from the scales calculations. A small source of error could have been that the person who was calculating the mass was not rounding correctly.
The choice to use sweet potatoes rather than the other choices could have obscured some of the lab results as well. The sweet potato is naturally more sweet, and this caused the results of the reaction to be less noticeable. The higher sucrose content of the sweet potato when mixed with the water and sucrose solution had less of an effect than that of another potato would have, because the sweet potato decreased in disparity .