d. This lab clearly demonstrated that cells maximize surface area to volume ratios. With a greater surface area, diffusion takes place over a larger surface and is much more effective. Cells rely on movement of particles through diffusion to carry out activities. If cells were too big, diffusion would take too long because although there is still a large surface in which diffusion can occur, the large volume catches up to the surface area and requires more diffusion. The .5x.5x.5 cm agar cube demonstrated using the ratio of 12:1 as an advantage; the cube completely changed color, meaning diffusion successfully occurred, in only an average of 3.17 minutes. Having a greater surface area than volume in this ratio allowed diffusion to occur over
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Water Potential of 1.0 M NaCl: 0+ -(2)(1.0)(.0831)(295)= -49.03 bars
b. Before being exposed to the 1.0M salt solution, the red onion cells were full of water, and this was clear when looking through the microscope. However, when exposed to the 1.0M salt solution, each of the cells lost water. The cells did not shrink though even though the solution was hypertonic to the cell because of the plant’s firm cell wall. The cells lost water because the water potential of the salt solution was lower than that of inside the cell; the water moved out in attempt to reach …show more content…
This data was found by observing the x-axis intersection of the lines in each graph to approximate the concentration value inside each cell when there is no change in mass, or when the solution is isotonic to the cell. The water potentials are different because the two cells do not consist of the same amount of solute. The concentration of the white potato cell was approximately .23 M of sucrose, while the sweet potato was .59 M of sucrose. While a white potato consists of some sucrose, the sweet potato consists of .36 M more. When both were placed in a 1.0 M sucrose condition, the sweet potato lost only 2.19 grams while the white lost 5.82 because since the white had less concentration of sucrose, it required more water to leave the cell to reach equilibrium with the solution. This exemplifies the effect of molar concentration on the movement of water. A greater solute concentration results in a lower water potential. Since water flows from high to low water potential, more water moved into the sweet potato cell (5.50 grams compared to the white potato’s 2.3 grams) in a 0 M solution because it contains more solute inside; this is why the water potential of the sweet potato is lower than that of the white
Water Potential of 1.0 M NaCl: 0+ -(2)(1.0)(.0831)(295)= -49.03 bars
b. Before being exposed to the 1.0M salt solution, the red onion cells were full of water, and this was clear when looking through the microscope. However, when exposed to the 1.0M salt solution, each of the cells lost water. The cells did not shrink though even though the solution was hypertonic to the cell because of the plant’s firm cell wall. The cells lost water because the water potential of the salt solution was lower than that of inside the cell; the water moved out in attempt to reach …show more content…
This data was found by observing the x-axis intersection of the lines in each graph to approximate the concentration value inside each cell when there is no change in mass, or when the solution is isotonic to the cell. The water potentials are different because the two cells do not consist of the same amount of solute. The concentration of the white potato cell was approximately .23 M of sucrose, while the sweet potato was .59 M of sucrose. While a white potato consists of some sucrose, the sweet potato consists of .36 M more. When both were placed in a 1.0 M sucrose condition, the sweet potato lost only 2.19 grams while the white lost 5.82 because since the white had less concentration of sucrose, it required more water to leave the cell to reach equilibrium with the solution. This exemplifies the effect of molar concentration on the movement of water. A greater solute concentration results in a lower water potential. Since water flows from high to low water potential, more water moved into the sweet potato cell (5.50 grams compared to the white potato’s 2.3 grams) in a 0 M solution because it contains more solute inside; this is why the water potential of the sweet potato is lower than that of the white