Osmosis. An experiment to find the concentrations of salt solution which is the same as the concentration of cell sap in a potato.
An experiment to find the concentrations of salt solution which is the same as the concentration of cell sap in a potato. Concentration of salt in solution/ % Mass of potato core at start/ g Mass of potato core at end/ g Change in mass/ % % change in mass/ % 0 2.03 .99 - 0.04 - 1.97 8 2.13 2.05 - 0.08 - 3.75 6 2.14 2.12 - 0.02 - 0.93 4 .95 .93 - 0.02 - 1.02 2 2.16 2.11 - 0.05 - 2.31 0 2.17 2.15 - 0.02 - 0.92 Investigation > To find out about osmosis in potato cells. > To find the concentration of salt solution which is the same as the concentration of cell sap in the potato. Prediction > The higher the concentrations of salt solution, will make the potato go flaccid. > However, the lower concentrations of salt solution will make the potato go turgid, increasing the length and mass. Biological knowledge which supports prediction > Osmosis- diffusion of water molecules from a dilute solution to a concentrated solution across a semi-permeable membrane. > Potatoes don't have a very salty taste so there is a probably a more dilute solution of salt solution in the cell sap. > When there is a high concentration of salt solution, the water will diffuse from the potato across the semi-permeable membrane, into the salt solution. Independent variable: (thing that changes) > Concentration of salt solution Dependent variable: (what you are
Osmosis Coursework
Osmosis Coursework Planning Before I started this coursework I learnt about osmosis. Osmosis is the diffusion of water molecules from a region of high concentration to a region of low concentration through a semi-permeable membrane. In a high concentration there is a lot of solvent (water) and less solute (salt). This is called a dilute or weak solution. In a low concentration there is less solvent (water) and more solute (salt). This is called a strong or concentrated solution. First of all I will cut the potato with the knife. Then I will use the scalpel to cut the potato into twenty equal strips and I will use a ruler to cut them into the same length. Then I will put two strips of potato into each of the ten petri dishes. I will then take salt solutions and pour different concentrations from 0.5%-5% into the petri dishes. I will then leave them in the dishes for the same amount of time and then take them out and measure their lengths. I will then record the results in a table and then draw a graph of the results. Equipment Salt solution of different concentrations 0 petri dishes used to put salt solution and potato in Potato cut up into pieces to be put in salt solution Scalpel used to cut up potato into equal strips Ruler used to measure length of potato before and after experiment Cutting tile used to cut potato on
Osmosis, My aim for this experiment is to see the results of potato tissue's mass difference, when placed in different concentrations of sugar solutions. Variables involved:
Osmosis Background Information Osmosis is defined as the movement of water molecules from an area of high water concentration to an area of low water concentration, across a semi-permeable membrane. In a high concentration of water the amount of solute (e.g. sugar) is low. This could be called a weak or dilute solution. In a low concentration of water the amount of solute (e.g. sugar) is high. This could be called a strong or concentrated solution. When two such solutions are divided by a semi-permeable membrane the water will move from the area of high concentration to the area of low concentration, until both sides are equal (have reached equilibrium). This can be seen in living cells. The cell membrane in cells is semi-permeable and the vacuole contains a sugar/salt solution. So when a cell is placed in distilled water (high water concentration) water will move across the semi-permeable membrane into the cell (lower water concentration) by osmosis, making the cell swell. This cell is now referred to as turgid. If done with potato cells the cells would increase in length volume and mass because of the extra water. If these potato cells were placed in a solution with a low water concentration, then the opposite would happen. Water would move out of the cell into the solution. In extreme cases the cell membrane breaks away from the cell wall and the cell is referred to
Enzymes at work
Enzymes at work The aim of my investigation is to find out what pH the enzyme named peroxidase works best at. We can find the enzyme peroxidase if we grind up celery, or most other animal and plants. A chemical reaction in plants and animals produce hydrogen peroxide, this can poison them if it is allowed to accumulate. The enzyme peroxidase acts as a catalyst for decomposing hydrogen peroxide, it is broken down into water and oxygen. The reaction is: 2H20 = 2H20 + O2 (Aq) (L) (G) I plan to measure the amount of oxygen given off by the celery at different pHs and whichever pH helps the enzyme produce the most oxygen. Before I do this I must do a preliminary plan to see what my other variables should be so that I can get the best results. I predict that the optimum pH will be around 7 because if it goes any lower or higher than this it could change the enzymes shape and/or charge properties of the substrate so that either the substrate cannot bind to the active site or it cannot undergo catalysis. Different enzymes have different optimum pHs and it is hard to say what it will be because some enzymes work best at low temperatures and others at high temperature. As you can see from the graph the two different enzymes, represented by red and green lines, have a different optimum pH. When the pH is altered is will become denatured or it may be optimum it depends
The effect of sucrose solution concentration On osmosis in potato chips
The effect of sucrose solution concentration On osmosis in potato chips Plan I am going to cut six pieces of potato using a borer. They will all the same weight. I will make them all 0.7g. I will be able to make sure they are all the same weight-using scales. If a piece of potato is too big I will use a scalpel and a tile to trim it so they are all the same weight and then place these into a test tube each. I will hold the test tubes in a test tube rack. Using the measuring cylinder I will measure 10ml of each strength of sugar solution rinsing the measuring cylinder after each measure. This will then be added into separate test tubes with the pieces of potato. E.g. a piece of potato will have 0.0M strength; a piece of potato will have 0.2M strength and a piece of potato will have 0.4M strength etc. I will put a sticky label on each test tube stating what strength of sugar solution it contains. I will then leave these for 2 hours and see what happens then record my results. APPARATUS A potato Sugar solutions (0.0, 0.2, 0.4, 0.6, 0.8, and 1.0) Borer Scalpel Tweezers Scales 6 test tubes Test tube rack Measuring cylinder I will measure the amount of sugar solution in the test tubes. I will change the strength of sugar solution in each test tube. To make the experiment a fair test I will keep the weight of the pieces the same weight I will also only add the same
Investigation to work out the concentration in potato cell sap
Sarah Carr Biology Coursework 2001 Investigation to work out the concentration in potato cell sap Preliminary work As an introduction to osmosis, an experiment was carried out with a model cell. The model cell experiment shown how osmosis occurs, demonstrated with two cells - Cell A Cell B Water diffuses into 'cell' Water diffuses out of 'cell' Visking Visking Tubing Tubing (Partially (Partially Permeable Permeable Membrane) Membrane) Cell A was filled with sugar solution, cell B was filled with water, and then the two cells were weighed. Cell A was placed into a beaker of water and cell B was placed into a beaker of sugar solution. They were both left for 30 minutes, and then the cells were weighed again. It was discovered that cell A (containing sugar solution), increased in weight. This was because water was allowed to pass into the cell and into the strong solution by osmosis. Therefore cell B, decreased in weight because water had passed out of the cell and into the strong solution by osmosis again. As it has been identified that cells can increase in weight with osmosis, it is fair that we can say cells of a potato core will behave in the same way as the model cell. Aim My aim is to work out the concentration in potato cell sap. I am to find the increase/decrease in weight of
How nitrogen-containing substances are made available to and are used by living organisms.
How nitrogen-containing substances are made available to and are used by living organisms. (25 marks) Nitrogen is taken up by living organisms in the soil for the purpose of manufacturing proteins, nucleic acids and other nitrogen-containing compound. However, a vast minority of living organisms can use up the nitrogen gas directly, plants are only capable of absorbing the nitrogen-containing substances as nitrate ions, which are taken up from the soil via active transport. And so nitrogen enters the ecosystem through the absorption by plants and so theses plants are then eaten and digested by animals. For nitrogen gas from the surrounding atmosphere to be available for living organisms, especially plants it needs to be nitrified down to nitrate ions. There are specific organisms in the soil that feed on ammonium-containing compounds which release ammonia which then forms ammonium ions in the soil. The next stage is nitrification which is where ammonium ions in the soil are converted into nitrite ions with the aid of nitrifying bacteria, and then the nitrite ions are then converted into nitrate ions. It is an oxidation reaction and so energy is released in this stage. Another pathway by which nitrogen-containing substances are made available to living organisms is when the nitrogen gas in the surrounding environment is converted into nitrogen-containing compounds in a
An Experiment to investigate change in the' rate of reaction' of potato Catalase enzymes due to varied concentrations of hydrogen peroxide.
An Experiment to investigate change in the' rate of reaction' of potato Catalase enzymes due to varied concentrations of hydrogen peroxide. Definition of an Enzyme: - Enzymes are biological catalysts which control all the chemical reactions inside cells How Enzymes work. Enzymes are biological catalysts; these are in all living creatures, these catalysts speed up the rate of reactions. Most enzymes are globular proteins; their molecules are round in shape. The enzymes speed up the process of conversion of substrates into products. The shape of the enzyme molecule is what is important; they have an area, like a gap in the molecule, which is, called the active site. The substrate molecule(s) "plug in" to the active site on the enzyme molecule. The molecules can then be broken down (or built up) far more quickly than if they were moving around randomly. An enzyme will only catalyze it's "own" particular reaction i.e. it is very specific. This is explained by the fact that each enzyme has it's own particularly shaped active site and only one type of substrate will fit in. A cell contains many hundreds of enzymes, all catalyzing different reactions. Even though the enzyme takes part in the reaction it is freed afterwards and can be used again and again. Because of this enzymes are needed in very small amounts. There are various factors that affect the rate of
I am going to measure the amount of gas given off when certain amounts of potato react with Hydrogen peroxide.
Biology SC1 Getting the information I am going to measure the amount of gas given off when certain amounts of potato react with Hydrogen peroxide. I will time each reaction until it has given off a certain amount of gas. Apparatus > Water bath > Boiling tube (with tube) > Bung > Measuring cylinder > Ruler > Borer > Scalpel > Pestle and mortar > Beaker > Potato > Hydrogen peroxide > Stop watch Method . Set up the equipment like above. 2. Take a cylindrical piece of potato and cut off the skin from both ends of the cylinder. 3. Measure how much you would want to take off and cut with scalpel. 4. Place the potato in the boiling tube. 5. Add the hydrogen peroxide to the potato in the tube, which will also help to push the potato down. 6. Put the bung on the top of the boiling tube ASAP. 7. Take the measurement of how much water is up the measuring tube. 8. Time until the amount of o2 is given off that you want. 9. Record results on a piece of paper. How to make it a fair test I have to make this test fair otherwise there would be no point in doing the investigation. The most important thing, which may sound obvious, is that you clean the equipment with distilled water. This means that the potato does not start giving off gas from the hydrogen peroxide from the last experiment. I will also have to make sure that the amount of potato that I cut up is the
An experiment to show the effect of hydrogen peroxide on the catalase.
An experiment to show the effect of hydrogen peroxide on the catalase Aim An investigation to see the volume of oxygen produced when hydrogen peroxide was added to the catalase Method * The equipment was set up * With a cork borer several tubes from the potato were cut out * Using a ruler, 3mm long pieces of potato were cut up * The pieces of potato were put into the round bottomed flask * With the tap closed, 10cm³ of hydrogen peroxide was poured into the reservoir * The tap was opened until all the hydrogen peroxide had poured into the round-bottomed flask and then it was closed firmly. * The pipette was put into the test tube of water * The timer was started for 3 minutes * The amount of bubbles were counted, using a calculator * After 3 minutes the number of bubbles were recorded * The round bottomed flask was washed out with distilled water * The experiment was repeated 3 more times Results table Bubbles produced in 3mins Number of potato pieces 3mm st 2nd 3rd 4th Average 5 20 20 9 6 8.75 0 39 35 44 47 41.25 20 54 41 49 52 47.50 30 40 32 23 67 31.70 40 54 204 205 94 89.25 50 214 223 213 204 213.75 60 75 05 74 230 71.00 Interpreting results From the graph, we can see that as the number of potato pieces increase so does the amount of oxygen given off until 50 potato pieces. At 10 and 20 potato pieces,
