I will be investigating the concentration of the cell sap in a potato using osmosis.
The cytoplasm of a plant cell and the cell sap in its vacuole contains salts, sugars and proteins which effectively reduce the concentration of free water molecules inside the cell. The wall of a plant cell allows water and dissolved substances to pass through it. The cell membrane of the cytoplasm is partially permeable. Because of osmosis if a plant cell is surrounded by water or a solution more dilute than its contents, water will pass through into the vacuole. If the cells of a potato are in a solution with a lower concentration than themselves, then water will diffuse from the cell sap, across the membrane and cell wall into the surrounding liquid. Even though water molecules are free to pass across the membrane in both directions in both directions it is likely more water will enter the cell than leave it. As water passes into the cell, the volume of the vacuole increases and begins to press against the cytoplasm and cell wall. When the vacuole presses against the cell wall it is turgid, however the cell wall prevents the cell from bursting.
If the potato is surrounded by a more concentrated solution than the contents of its cells then most water will diffuse out of the cells until the concentration outside is the same as the concentration inside. When water is lost the vacuole shrinks and the cells becomes flaccid and soft as water has diffused out of them by osmosis. However the cell wall cannot decrease in size so the cell membrane shrinks from the cell wall, leaving a gap between the membrane and the wall. These cells have been plasmolysed. The potato cells will decrease in size and mass.
Pilot Study
In my pilot study I used a knife to cut and peel a potato which I then cut 3 potato chips out of and recorded the mass and length of each. I took 3 test tubes and filled one with distilled water, one with weak sugar solution (o.3 M) and the last with a strong sugar solution (1 M). I placed one chip in each test tube and left the apparatus for 30 minutes. After 30 minutes I took the chips out of the test tubes and dried them on paper towels, I then recorded the mass and length again.
My pilot study proved that when the potato is left in a solution of a lower concentration the potato chip increased in mass but if the surrounding solution has a lower concentration it decreases.
This pilot study helped me decide not to measure the lengths of the potato chips but to concentrate on the mass because it is easier to distinguish the change and will give me hopefully more accurate results.
To find the concentration of the cell sap I need to find the concentration of the surrounding solution when there is no change in the mass of potato so equilibrium is achieved. I know that the concentration of the cell sap is fairly dilute as the potato in the 0.3 solution decreased only be 1g so I predict the concentration is approximately 0.35.
Prediction Graph
Variables In Experiment
The independent variable in my experiment is the concentration of the sucrose measure in moleurs, so there is more or less water outside of the potato chip.
My dependant variable is the change in mass of the potato chip, the uptake and decrease of mass in grams.
The factors I will try to keep constant, to make it a fair test are:
- The temperature of all the test tubes
- The amount of liquid in the test tubes
- The time the chips are left in the solution
- The size and shape of the chips
- I will try to use the same potato for all chips
- I will dry to dry off excess water which may affect the mass.
- I will use the same balance, method and equipment.
- I will not use the skin of the potato on my chips.
Method
I will make up 50g of 6 solutions of different concentrations. They will be 0M, 2M, 3M, 4M and 5M. I will be testing mostly more dilute concentrations because in my pilot study I found that the concentration of the cell sap is not very strong.
I will make up my solutions as follows:
Analysis
My experiment shows that when it is in a very low concentration a potato will gain mass, and as the concentration gets higher the potato loses more mass.
When the concentration was 0M and 1M, the cell sap in the potato is more concentrated than the surrounding liquid so water molecules diffuse into the vacuole making the cell turgid and therefore increasing the weight of the potato chip.
When the concentration of my liquid was 2M or higher the liquid surrounding the potato was of a higher water potential than the cell sap, so water diffused from the cell sap, across the membrane and cell wall into the liquid so that the concentration inside and outside the cell were equal. The cells in the potato became flaccid as water had diffused out by osmosis, and lost mass making the chips lighter.
AS the Molar of the concentration got higher the potato cells lost more mass, as my graph shows a negative correlation. As the points are close to a straight line it shows that the results are proportional to each other. According to my graph the concentration of cell sap was 0.135, as this was where it crossed the x-axis. This means that the concentrations were already at equilibrium , so there was no net movement of water molecules. Water molecules were passing in and out of the cells but in equal amounts, so the chip did not increase or decrease in mass.
My predictions were close to being right, but the actual results were probably different because I was more accurate because I tested more solutions than in my pilot study.
Introduction
I will be using osmosis to find the concentration of cell sap in a potato cell. From previous research I know that by the process of osmosis the water will travel into the cells making them swell and become turgid . therefore increasing the mass of the potato chip. I will know the concentration of the cell sap when there is no movement between the cells, which I will be able to tell because there will be no change in the potatoes mass. At this point the solution of sug
