Glucose levels were measured at 0.52, 0.57, and 0.67 showing that an increase in surface area corresponded with a non-significant (p=0.096) increase in the glucose diffusion rate. Introduction Diffusion is the random movement of substances from an area of higher concentration to an area of lower concentration (down a concentration gradient) and this occurs to stabilize the cell (Kaisa, 2011). Diffusion occurs until the particles are completely evenly distributed. Several factors can alter the diffusion rate of substances, but specifically several factors can influence the diffusion rate of glucose from potatoes. Changes in temperature, surface area, soaking time of the samples, and many other factors may influence the diffusion rate of glucose.
We let them soak for 20 minutes than we removed them from the test tubes. Next we measured their new weights and recorded them. They were; 10% NaCl decreased by 4%, 15% NaCl decreased by 8%, and 20% NaCl decreased by 12.5%.The results showed us the conclusion that the potatoes were hypertonic to the deionized water and were hypotonic to the sodium chloride solutions. Introduction: The objective of this experiment is to detect diffusion and osmosis in potato cells in 3 solutions. Diffusion is the spontaneous spread of molecules from an area of high concentration to an area of low concentration.
If we place the potato in a sucrose solution with a similar solute concentration as a potato, then the least amount of water will diffuse into or out of the potato cells. This occurs because this scenario will be initially closest to an isotonic solution, allowing us to determine the solute concentration of a potato. Materials: * Knife/ Cork Borer * Seven 30 cm pieces of Dialysis tubing * Thirteen 250 mL Beakers * 15% Glucose Solution * Glucose Test Strips * 1% Starch Solution * Distilled water * Lugol’s solution * 25 mL of: * .2 M sucrose * .4 M sucrose * .6 M sucrose * .8 M sucrose * 1 M sucrose * Paper Towels * Clock * Potatoes * Plastic
To grow the E. coli, 190ml of 2% glucose nutrient media was added to a 250nl Erlenmeyer flask. 10ml of nutrient media was used to blank the spectrophotometer set at OD600. 10ml of E. coli stock culture were added to the 190ml nutrient media, creating a 1:20 dilution. The inoculated nutrient media was then placed into a 37C orbital shatter to incubate. All solutions measured with the spectrophotometer were placed in spectrophotometer tubes.
The liquid of homogenate was filtered into a beaker through Miracloth (2 layers cloth) to remove large plant components and 1 ml of the filtrate was transferred to a conical tube. 8.4 g of ammonium sulfate was slowly added to the 40 ml of the filtrate as it was stirred on a stir plate for 15 min to achieve 37% saturation (210g/L of solution). The solution was then centrifuged at a speed of 9000 x g at 4oC for 15 min to sediment the proteins. The resultant supernatant 1 was transferred to a beaker with 1 ml transferred to a conical tube and the obtained pellet 1 was resuspended in 4 ml of distilled water and transferred into a dialysis bag to remove the salt. Then, 3.4 g of ammonium sulfate was slowly added to the supernatant 1 as it was stirred for 15 min to achieve 50% saturation (85g/L of solution).
The digestive system helps break down this food into small pieces, the enzymes within the digestive system breaks the protein into amino acids, fatty acids and carbohydrates into glucose. The sugar, amino acids and fatty acids are able to be used as energy sources in the human by body cells. Energy requirements are ordinarily expressed in terms of calories. The calorie used in nutritional discussions is actually the "large calorie “Calorie. This is really a kilocalorie the amount of heat energy required to raise the temperature of one kilogram (about 1 quart) of water one degree Celsius.
The fungal culture was grown in 100 ml PDB for 5 days. The culture was harvested and the mycelial mat was separated by filtration using Whatman No. 1 filter paper. Then mycelium was ground separately in pestle and mortar using liquid nitrogen. About 50 mg of the powdered mycelium was transferred into a microtube contained 500 µl of TES (100 mMTris, pH 8.0, 10 mM EDTA, 2% SDS).
. Design a) Defining the problem and selecting Variable I) Aim/Research Question: To investigate the effect of solute concentration on the length of potato strips when submerged in a range of concentrations of sugar solution. II) Hypothesis: If the concentration of the sugar solution increases, then the osmotic gain of the potato strip will decrease. III) Scientific Explanation: Osmosis is a type of passive transport where passive movement of water molecules occurs along a concentration gradient, across a partially permeable membrane. (Damon, )Since the movements of substances in passive transport take place from an area of high concentration to an area of lower concentration until both areas reach equilibrium, the water molecules will move from the potato strips to the sugar solution when there is less water concentration in the solution.
Research Question: How will the different potato wedges submerged in various salt solutions affect on the weight and length of the potato wedges? What is the DV and how exactly will you measure it and how many trials will you do: %∆ Change weight and length of the potato chips after they have been placed in different salt concentrations after 24 hours What factors could affect the DV: What? | How? | Why? | Temperature | Keeping the temperature the same by placing the petri dishes in the refrigerator.
If the area around a cell has a higher water concentration it will gain water by osmosis. This is because the water molecules surrounding the potato cell are able to pass freely across the cell membrane in both directions; therefore more water will enter the potato than will leave it so it will swell up. Furthermore, if the area around the potato’s cells has a lower concentration of water than the cell will loose water by osmosis due to more water diffusing out through the partially permeable membrane than entering the cell, this will lead to a reduction in the size of the cell. When the concentration of water molecules in the potato is near equal to that of the area around it, the rate of osmosis will decline and the cell will stay the same size, as there is an equal amount of water entering the cell than there is leaving it. Below is a drawn diagram showing the process of osmosis.