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How light intensity affects the rate of photosynthesis
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How light intensity affects the rate of photosynthesis The aim of my experiment was to determine whether intensity of light would affect the rate of photosynthesis in a plant. To do this, I placed a piece of Canadian pondweed in varying light intensities, and observed the amount of oxygen being given off. I used Canadian pondweed because of its unusual quality of giving off bubbles of gas from a cut end, when placed in water. Introduction Photosynthesis occurs only in the presence of light, and takes place in the chloroplasts...
repeat readings and find

an average. To extend my enquiries into the rate of photosynthesis, I

could perhaps try to link in some of the other limiting factors to the

same experiment, as well as investigating them in their own right. It

could also be interesting to explore the effects of coloured lights on the

rate of photosynthesis, which could lead to the question of whether or not

other types of light, such as fluorescent lights or halogen lights, would

have a different effect on the rate of photosynthesis.

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INTRODUCTION This piece of... INTRODUCTION This piece of biology coursework is about how cells use osmosis, how, why and what affect it has on the cells in question. The example I am going to explore this in and to illustrate this is the experiment of potato chips of the same length in different concentrations of a sugar, which in this case the sugar is Sucrose. Aim: - I will investigate the effect of varying concentration of a certain sugar solution on the amount of osmotic activity between the solution and a potato chips of given sizes. I will use background theory to help me make predictions and when doing the experiment I will make sure I plan to undertake a fair test and abide by any safety precautions and making sure I control all key variables. I will conduct the experiment and collect the results in a suitable table, and then I will evaluate and conclude what I have found and refer back to the theory. P.6A& P.8A BACKGROUND THEORY Diffusion is the passive movement of particles from an area of high concentration to an area of low concentration until the concentrations are equal. The rate that diffusion happens is dependant on the difference between the two concentrations to start off. The Concentration gradient can decide how quickly diffusion happens. The bigger the difference in the concentration then there is a bigger gradient. The molecules will continue to diffuse until the area in which the molecules are found reaches a state of equilibrium, meaning that the molecules are randomly distributed throughout an object, with no area having a higher or lower concentration than any other. The diffusion and concentration gradient can be shown in an example from the small intestine to the blood capillaries: - This concentration gradient diagrams shows how the food can be diffused from area of high concentration in the small intestine to an area of low concentration which is the blood capillaries. The second diagram shows that the diffusion equals the concentration gradient so in the blood capillaries it increases its concentration of food. Importance of diffusion in biology can be shown in that we need diffusion to stay alive. To stay alive the cells take up oxygen because the oxygen is continually being used up in respiration, it's concentration inside the cell is lower than in the blood or tissue fluids. The concentration gradient results in oxygen molecules diffusing into cells from outside. The same applies to Carbon dioxide but in the other direction. Osmosis is the movement of water molecules across a Semi-Permeable Membrane form a region of high water concentration to a region of low water concentration. Semi permeable membrane, another name for it is a partially permeable membrane, it has small holes in it that allows only water molecules pass through other bigger molecules like glucose can not pass through. The water molecules do actually pass both ways through the membrane this is the water movement. But because there are more water molecules on one side than the other there is a steady flow into the region of fewer water molecules into the stronger solution of glucose. This causes the glucose rich region to fill up with water, the water acts like it is diluting it so the concentrations on either side are evened up or have reached equilibrium this can be explained as the water movement. Water potential: - The tendency for water to move through a partially permeable membrane is called its water potential it is at a maximum in pure water. The addition of solutes to water lowers its water potential. The amount the water potential is lowered depends on how much solute is added. This means that the more concentrated solutions have lower water potentials than more dilute solutions. Water will therefore have a greater tendency to move from pure water to any aqueous solution like and form a more dilute solution to a more concentrated one. The reason in thermodynamics terms is that the potential energy of the water molecules on the left is greater than on the right, the potential energy of the water is called the water potential, the steeper the water potential gradient then the greater the tendency for the water particles to diffuse in this direction. Therefore water potential has the capacity to be a system able to lose water. Plant cells Plant cells always have a strong cell wall surrounding them. When they take up water by osmosis they start to swell, but the cell wall prevents them from bursting. Plant cells become turgid when they are put in dilute solutions. Turgid means swollen and hard. The pressure inside the cell rises and eventually the internal pressure of the cell is so high that no more water can enter the cell. This liquid or hydrostatic pressure works against osmosis. Turgidity is very important to plants because this is what make the green parts of the plant stand up into the sunlight and for opening guard cells. When plant cells are placed in high concentrated solutions they lose water by osmosis and become flaccid. Flaccid means the plant cells have shrunk or have become soft and floppy. This is the exact opposite of turgid. The content of the potato cells shrink and pulls away from the cell wall. These cells are said to be plasmolysed. When the plant cells are put in a solution that has the same osmotic strength as the cells they are between turgidity and flaccidity, this can be called incipient plasmolysis. Example: - stomata cells rely on osmosis to become turgid to open pore and to become flaccid to close pore. Animal cell Animal cells do not have a cell wall and can easily burst if they are put into pure water because they take in so much water through the process of osmosis. The cell membrane in animal cells is partially permeable. An example of an animal cell bursting due to osmosis can be shown in an example of red blood cells: - The red blood cells have been placed in distilled water. Their cytoplasm is a strong solution. Water passes through into the cells by osmosis, but animals do not have a cell wall to stop them swelling up too much so they burst, this is called haemolysis. In active transport particles e.g. ions can be moved against the concentration gradient. This movement of ions or any particles against the normal concentration gradient uses a lot of energy. This type of transport is called active transport; it involves the use of energy from respiration, adenosine triphosphate ATP. Cells which that can do more active transport usually have more mitochondria than in other cells. Mitochondria are involved in aerobic respiration and the production of ATP so this is why cells with more mitochondria do more active transport. Energy is needed for active transport: - Isotonic point is the point when the potato chips is either not increasing ion mass or increasing in length, this is where no osmosis is taking place, both the potato cells and the solution have the same molar concentration. Hypertonic is where there is one thing having a higher osmotic pressure than a surrounding medium or fluid under comparison. Hypotonic is where one thing e.g. a potato cell has a lower osmotic pressure than a surrounding medium or fluid which in this case is the sugar solution. P.6A Predictions and extended theory P.8A For this particular investigation I think that when there is a lower concentration of the sugar solution sucrose in the test tube the bigger the length of the potato chip will be. The reason for this is that the water molecules pass from a high concentration, in the water itself, to a low concentration, in the potato chip. Therefore, the chips in higher water concentrations will have a bigger length than in higher sugar concentrations. I predict that the higher the concentration of sucrose sugar then there will be more length lost in the potato chips. I predict at the highest concentrations of sucrose there will be no further decreasing in size because the potato cells have plasmolysed and no more water can leave the cell. If the concentration in the solution is more concentrated than in the vacuole of the potato cell, then water passes out of the cell due to osmosis. When a solution is outside the cell is more concentrated than a vacuole of the cell water passes out due to osmosis, as cytoplasm is pulled away from the cell the entire cell becomes plasmolysed, the vacuole also shrinks. This piece of theory proves that my prediction to be true because it shows the water will pass out of the cell when there is higher concentration out of the cell than in the vacuole of the cell. I predict if a cell is surrounded by pure water or by a sugar solution whose solute concentration is lower than and water concentration is higher than that of the cells contents water will flow into the cell and the cell swells up. The external solutions is said to be HYPOTONIC to the solution in the cell. In contrast I predict that if the cells in the potato are surrounded by a sugar solution that solute concentration is higher and the water potential lower than that of the potato cells then water will flow out of the cell, the cell shrinks. In this case the external solution is said to be HYPERTONIC to the solution in the cells. If the cell has the same solute concentration and water potential as its surrounding solution then there will be no net flow of water. The external solution is said to ISOTONIC. P.4A Key variables & FAIR TEST The Independent variables the things that I am able to change: - Concentration of sugar solution: - The thing that I will be changing in this investigation into the factors affecting osmosis is the concentrations of the solution that helps me to investigate osmosis with potato tissues The different sugar sucrose concentrations will be changed to see the result of the potatoes when placed in sugar solutions placed in different molarities strengths. Dependent Variable things that I have to measure: - The things that I will be measuring in this experiment on osmosis are the initial mass and overall length change of the potatoes placed in different molarities of sugar solutions. I will record the masses in grams by using a balance and the length changes by using a ruler. Weight: - this is one of the dependant variables and so is the final length it is dependant on the rate of activity of osmosis on the tissue of the potato chips. Fixed Variables The things that I will keep the same The things that I have chosen to keep the same in each individual experiment to keep it the experiment a fair test is: - Initial length: - The initial length for all the potato chips will all be the same, this is so that this can be called a fair test and this is to keep the same sizes of each potato tissue. Temperature: - The temperature throughout the experiment will be at room temperature. The room temperature will remain constant because the experiment will be in only one room so the temperature will not fluctuate and affect the results, this will be part of a fair test. Type of potato: - I will also be keeping the same potato through out the whole experiment to as using a different one might give me anomalous results as some might have more water in it then others. Time: - The time or duration that all the chips will be in the different solutions will be the same, it will be 2 days. The time has to be the same because each chip will have the same amount of osmosis, and it will constitute in a fair test. Amount of sugar solution: -The amount will all reach up to 40ml so it is a fair test and there are all equal parts, e.g. one part water and three part sucrose, but the solution should still reach 40 ml. For the experiment to be fair and the results to be considered reliable there has to be a fair test. A fair test can make the results be considered to be reliable and within the experiment reduce the number of errors. All the non variables will be needed to be kept as non variables so the results are not affected and so the results show how only the concentration of the sucrose affects osmosis and not any thing else. I will make sure each solution will have the correct amount of distilled water and sucrose so that each test tube has a solution that comes exactly to 40ml. The measurements for the solutions have to be perfect as to not change the out come of the experiment. I have to ensure that every time I handle the potatoes my hands are clean and dry. This is to stop any kind of contamination and make sure that I do not pass on any extra water onto the potato. If I do not keep my hands clean and dry them then this will affect the results and make my results inconclusive. I will also use the same potato in the experiment so there will less a chance of anomalous results, because of the use of different potatoes. P.2ASAFETY Safety in any experiment is always needed, because if there is no safety then the person doing the experiment and others who are not taking precautions may become injured during the experiment. In the experiment I will use some sharp pieces of equipment like scalpel, peeler, and knife; if they are misused then this could result in injuries. Whilst peeling the potato I should make sure my fingers are not in the way of the peeler or the kitchen knife as they are sharp and can cut the skin on the fingers causing deep wounds. Whilst cutting the potato, extreme care and precision had to be taken with the scalpel as it is very sharp and could easily cause a serious wound. Also I should not be holding the potato in my hand and cutting it, the potato should be put on a white tile so no damage is caused to any surfaces and the person's hands and fingers. When carrying the knife or any sharp equipment I should not run and I should face the sharp end down this could stop any injuries if I were to slip or fall. The other precautions are health ones where after the experiment the people should wash there hands as they have been in contact with any solutions or the potatoes, and if the person conducting the experiment does not wash there hands then they may have a lot of germs on their hands. P.4B APPARATUS & appropriateness of measuring equipment → Sugar solution: - the sugar that will be used in this experiment is Sucrose. → Distilled water: - used to change the molar or concentration of the sucrose. → Measuring cylinders: - measure exact amount of sucrose and distilled water. → Test tubes: - this is where the sugar solution and the potatoes will go in. → Test tube racks: - used to hold all the test tubes. → Potato → Peeler: - used to peel the skin of the potato. → Scalpel or knife: - used to cut the potato into the chips. → Clipper → cling film paper: - to cover the test tubes so no air can come into the test tubes. → Ruler: - to measure the lengths of the cut potato chips to make sure they are 3cm. → Pan balance: - an accurate electronic weighing machine to weigh the potatoes. → Masking tape and pen: - to make the labels for the different solutions. → Tile: - to cut the potato on. → Sieve: - to drain out the solution over a sink. → Paper towels: - to dry potato chips after when I take them out of the solutions. The least count for the ruler that I will use will be 0.1cm because I want to measure the cut potato chips the rulers lowest measurement is 0.1cm or 1mm so this will be the least count that I will use. METHOD P.4B To do the experiment start off by working in a group of three so each person can do a separate thing at the same time. The first person should start off by making the labels for the test tubes: - → cut out some masking tape and label each test tube from the letters A-F → then should note down what the letters solution is made up of: - A range of sucrose sugar solutions will be prepared with concentrations 0 molar, 0.15 molar, 0.25 molar, 0.5 molar, 0.75 molar and 1 molar. This will be done by adding varying amounts of distilled water to varying amounts of sucrose solution: - A B C D E F SUGAR SOLUTION ml 40 ml 30 ml 20 ml 10 ml 5 ml 0 ml WATER ml 0 ml 10 ml 20 ml 30 ml 35 ml 40 ml MOLARITY molar 1.00 m 0.75 m 0.50 m 0.25 m 0.15m 0.00 m "¢ The second person will make the solutions using the Sucrose sugar and the distilled water. You should use the above table to make the solution "¢ I will do this by using the measuring cylinders and measuring the correct amount of water and sugar solution. "¢ Then I will put the correct molarities I have made into the test tubes, I will do this for one at a time so there is no confusion. The third person should be dealing with the potatoes: - "¢ Get an average sized potato. "¢ Using the knife or the peeler, peel of the skin of the potato. "¢ Then cut the potato into a block which has as all an equal width and length. "¢ Then on the tile using a sharp scalpel and ruler cut ten chips all the same length of 3cm on the white tile. "¢ This part of the preparation must be done very accurately as a change in the surface area may allow more or less osmosis to occur. "¢ Then I would weigh the chips individually on the pan balance and record the weight of the chips and make sure the difference is no more than +0.5 "¢ After this I would then place two potatoes in each of the test tubes. "¢ The top of the test tubes will be covered with cling film paper. "¢ Then I would put the test tubes in the test tube rack. I would leave them for 2 days as this will allow sufficient time for osmosis to happen and the experiment to take place. After these 2 days I will have to take out the chips to measure them: - "¢ First I will do test tube A "¢ I will drain out the solution from the test tube over the sink using a sieve. "¢ Then I will dry out the chips on the paper towels and then measure the chips and then make a note of the results when the chips are dry. "¢ I will put the result as test tube-A results, chip 1, and chip 2, recording the results of the lengths. "¢ I then will do this same procedure for the rest of the test tubes and taking down the results. "¢ I will do it one by one because there will be less chance of any mistakes or errors within the experiment and the recording of the results. P.6B NUMBER OF READINGS & APPROPIATE RANGE The table format that I will show my results in: - Sucrose Solution A-E m=molar Volume of sugar ml Volume of water ml Initial length of chip cm To 1.d. Final Length of chip cm to 1.d.p Change in length of chip 1dp + - cm % change in length of chip to 2.d.p no unit Average % change in length of chip no unit to 1.d.p 1 2 1 2 1 2 1 2 A 1.00 m B 0.75 m C 0.50 m D 0.25 m E 0.15 m F 0.00 m This is the table format I have chosen. The table shows that I have appropriate range because I have used 2 readings for each concentration so I can have reliable results. The 2 readings per concentration can give me an average so I can get a group of reliable results. Each column has the correct heading and the correct measurement unit; it also has the decimal point the numbers will consistently and appropriately be rounded to. I have got a sufficient amount of readings as I have chose 6 different molars concentrations this will make it have a wide and suitable range. P.8b PRELIMINARY WORK secondary sources of info Previous studies and work have shown that the predictions I have made are true. These secondary results show what the percentage of mass change in potato chips in different concentrations molars. Bibliography: - Britannica encyclopaedia 2001 CD-ROM edition. Percentage change in accordance to the varying solutions Concentration Average % change in mass 0.00m +4.9 0.25m -2.1 0.50m -5.2 0.75m -7.9 1.00m -9.7 These results show that the varying concentration of the solution gives a gain or loss of % of mass. The results suggest to me that as the molar or concentration is high then the more mass has been lost. It also shows that the lower the concentration of sucrose the less % of mass lost and when the solution has no sucrose 0 molar then the potato chips actually gain mass. A graph to show the preliminary results: - The result from the graph can explain part of my theory. For example on 0 molar concentration of sucrose which is all water the mass has increased to as my theory states as the higher water concentration the bigger the mass will be and the potato cells have become turgid. The graph shows as my theory states that the higher the concentration of the sugar solution then the potato chip will have a lower mass and flaccid. The graph shows that at 1.00 molar of sucrose the potato cells have plasmolysed and no more water can leave the cell, this is shown because this is were the graph shows that there is no more decreasing in mass. This agrees with my theory and predictions but it proves it about the mass and to some extent a link with the length, but I want to know if this applies to the length so I will conduct my experiment, but this has made me understand and given me an overall impression. OBSERVATIONS O2A, O4A, O4B, O6A, O6B, O8A General observations from the experiment: - After I had taken out the potato chips out of the test tubes I had realized some of the potato chips had swelled up while others had shrunk, this is all due to osmosis. The test tube-E potato chips had volume of water of 40 ml and I could see that the chips in this test tube swelled up and increased its length. In test tube-A there was 40 ml of sugar and I noticed that the potato chips in this test tube had shrunk in length. RESULTS TABLE: - Sucrose Solution A-E m=molar Volume of sugar ml Volume of water ml Initial length of chip cm to 1.d.p Final Length of chip cm to 1.d.p Change in length of chip 1d.p + - cm % change in length of chip to 2.d.p no unit Average % change in length of chip to 1.d.p no unit 1 2 1 2 1 2 1 2 A 1.00 m 40 0 3.0 3.0 2.8 2.7 -0.2 -0.3 6.66 % 10.00 % -8.3 % B 0.75 m 30 10 3.0 3.0 2.7 2.9 -0.3 -0.1 10.00 % 3.33 % -6.6 % C 0.50 m 20 20 3.0 3.0 3.0 2.8 0.0 -0.2 0.00 % 6.66 % -3.3 % D 0.25 m 10 30 3.0 3.0 3.2 3.1 +0.2 +0.1 6.66 % 3.33% +5.0 % E 0.15 m 5 35 3.0 3.0 3.3 3.3 +0.3 +0.3 10.00 % 10.00 % +10.0 % F 0.00 m 0 40 3.0 3.0 3.5 * 3.5 +0.5 +0.5 16.66 % 16.66 % +16.6 % The results table shows that there are hardly any anomalous results which make the results reliable. In each column I have used the same decimal place to make it consistent and to keep the data in an accurate record. There was one unreliable reading and it was changed to a suitable one the reading marked with * shows that it has been changed to a suitable reading. The reading to begin with could have been unreliable because of slight human error which is within experimental error. All the lengths are written in accordance to the least count of the measuring devise. For the lengths of the chips all are to 1.dp because the ruler I used had a least count of 1mm or 0.1cm. The final column I put to 1 decimal place 1.d.p because this will be the most suitable to plot on a graph paper. ANALYSIS A.2A: - From the results of this experiment I have found out that the higher the concentration of the sucrose sugar solution then the bigger the % increase in length of the potato chips. I have also found out that the lower the concentration of the sugar sucrose the more % length of chips is lost in relation to initial length. A.4A A.6A: - graph and line of best fit A.4B: - Pattern in readings: - From the readings results we can get a certain idea of what pattern there is. In the readings of 0.00 molar which is 100 % water this is the point where the maximum amount of water was taken in through osmosis and therefore as the reading shows us the potato chips in this solution had the biggest % increase in length it was + 16.6 %. From the table of readings we can also see that at 1.00 molar of the sucrose solution this is the concentration where the potato chips had lost the most % amount of length it was at "“ 8.3 %. From the readings we can see that results do not double so we know this relationship is not directly proportional. Pattern on graph: - The shape of the graph is a one of a curve, going from + y to - y and in the + x. On the y axis the value go from + 16.6 to "“ 8.0. The graph shows the y-axis is inversely proportional to the x-axis so this indicates to us that the % change in length of the potato chip is inversely proportional to the concentration of the sucrose sugar solution. The patterns on the graph show that as the higher the concentration of the sucrose solution the more % of length is lost, but only to a point. On the graph this point is 0.90 molar, after this point on the graph till 1.00 molar the % change in length stays at "“ 8 %, this point is where the potato cells in the chips have fully plasmolysed so no more mass is lost and the cell stops shrinking in length. At low concentrations of sugar or high concentrations of water the graph gets steeper, for example at the points of 0.1 molar and 0.2 molar on the graph, they show a very steep gradient suggesting the lower the concentration of sugar the higher the % gain in length. Also at the point where there is all water 0.00 molar of sucrose this are where most % gain of length is and the maximum point of the cells in the amount of water they can take in through osmosis. The gradient does change in my graph, it is not consistent this show it is inversely proportional. It gets less steep as x-axis gets bigger or the molar/concentration gets stronger. This is because the potato chip is becoming as flaccid as it possibly can, and so the change in length of each molar concentration is becoming closer and closer together. A.6B: - EXPLANATION OF RESULTS WITH THEORY & A.8A: -EXPLANATION OF RESULTS WITH DETAILED THEORY Explanation of readings results and patterns: - At the concentration point of 0.00 molar the % change in length was + 16.6 % this result shows that the most water taken up by osmosis is in a solution which is 100 % water and the potato cells became turgid so the length has increased. It did this because in osmosis the water tries to create equilibrium to go from one high water concentration which was the 0.00 molar solution of sucrose to a low concentration which is the cells of the potato. At 0.00 molar there is high water potential because in pure water this is the maximum. As the results in the table show when the addition to solutes which in this case is sugar sucrose it lowers the water potential which means there is less water moving across the solution to the potato. This is why in the results as it shows that as the concentration of the sugar increases the less water is taken into the cells of the potato from the solution. From the results of the table of reading you can see that at strong concentrations of sucrose sugar at 1.00 molar this is the point concentration were most % of length has been lost. The reason for this is as my theory and predictions state that when plant cells are placed in high concentrations of sugar solution they lose water through osmosis because the concentration gradient makes the water from the cells to go to dilute the sugar solution, this makes the potato cells become flaccid means that they have shrunk in size and they are called plasmolysed cells. The diagrams below show the gradient at which the water from the cell goes to dilute the strong concentration of sucrose sugar by the process of osmosis: - First stage: - the original state. Second stage: - the osmosis makes the loss of water in the potato so the potato shrinks in size as it becomes flaccid. The sugar solution has know been diluted. Explanation of graph results and patterns: - The results from the graph are similar to the ones from the readings table. The reason why the graph gets steeper as the molar increases is because as the concentration is increased in the solution the water in the cell is more than in the solution so the water net movement is from the cell to the solution to dilute the solution. The potato chip becomes flaccid but as much as it can because there is a limit so this I why the points become closer together. There is a limit to how flaccid the potato chip cells become because this I the isotonic point were there is hardly any more increase or decrease the cells in the potato and in length as no osmosis is taking place because the solution and cells in the potato have the same molar concentration. The graph results show that as the higher the concentration of the sucrose solution the more % of length is lost but to a point, this point on the graph is 0.90 molar, after this point on the graph till 1.00 molar the % change in length stays at "“ 8 %, this point is where the potato cells in the chips have fully plasmolysed so no more mass is lost and the cell stops shrinking in length, this is why the graph towards the end of the x-axis straightens out horizontally. The reason why the cells in the chips have fully plasmolysed is because the concentrations of the water or sucrose have both reached an equilibrium point or the isotonic point. Therefore from the results and explanation the isotonic point on the graph is 0.9 molars of sucrose solution. At the point on the graph were there is high concentration of sugar solution the results on the graph show there is a lot of % loss of length of the potato chips. The detailed reason behind this is that if there is a solution that is more concentrated than a vacuole of the cells of potato, water passes out due to osmosis. As the cytoplasm is pulled away from the cell the entire cell becomes plasmolysed and the vacuole shrinks. As the vacuole shrinks, so does the cell and therefore the potato chip decreases in length. In this case the solution is said to be hypertonic. On the graph at low concentrations of sugar the graph gets steeper for example from the concentrations of 0.15 to 0.37, it suggests that lower the concentration of sugar the higher the % gain in length. The reason behind this is that because the potato is surrounded by a sugar solution whose solute concentration is lower than and the water potential is higher than that of the cells contents, then water will flow into the cell and the cells swell up so the potato chips increase in length. The external solution is said to be hypotonic to the solution of the cells. This can be shown on the graph of the results of 0.00 molar, where there is the most % gain in length this is also the maximum point of water the cells of the potato can take up. The points on the graph which show a % gain in length are 0.00 molar to 0.37 molar. A.8A: - My predictions were correct as the theory within my prediction was verified by the results that were obtained from this experiment. The results were backed up by my theory and stated the same ideas and principles that my predictions did. Throughout the analysis of my results I have found results from both the readings and the graph match and this has led me to an overall conclusion of the analysis of results which backs up my original prediction and theory. Therefore the statements I made in the analysis have already been proven by theory and the results from both the readings and the graph; this proves my predictions are correct. The evidence that shows my predictions are correct: - In my prediction I said "I think that when there is a lower concentration of the sugar solution sucrose in the test tube the bigger the length of the potato chip will be." This prediction is supported by my result that is "At the concentration point of 0.00 molar the % change in length was + 16.6 % this result shows that the most water taken up by osmosis in the lowest concentration of sucrose sugar." Another prediction that I made was "the higher the concentration of sucrose sugar then there will be more length lost in the potato chips." This is supported by my results which can be seen from this passage, "From the results of the table of reading you can see that at strong concentrations of sucrose sugar at 1.00 molar this is the point concentration were most % of length has been lost." Another prediction I made was "I predict at the highest concentrations of sucrose there will be no further decreasing in size because the potato cells have plasmolysed and no more water can leave the cell." The results that proved this correct were "The graph results show that as the higher the concentration of the sucrose solution the more % of length is lost but to a point, this point on the graph is 0.90 molar, after this point on the graph till 1.00 molar the % change in length stays at "“ 8 %, this point is where the potato cells in the chips have fully plasmolysed so no more mass is lost and the cell stops shrinking in length, this is why the graph towards the end of the x-axis straightens out horizontally." In my prediction I also said, "If the cell has the same solute concentration and water potential as its surrounding solution then there will be no net flow of water. The external solution is said to ISOTONIC." The result that proved this is "The reason why the cells in the chips have fully plasmolysed is because the concentrations of the water or sucrose have both reached an equilibrium point or the isotonic point. Therefore from the results and explanation the isotonic point on the graph is 0.9 molars of sucrose solution." In my prediction I said "I predict that if the cells in the potato are surrounded by a sugar solution that solute concentration is higher and the water potential lower than that of the potato cells then water will flow out of the cell, the cell shrinks. In this case the external solution is said to be HYPERTONIC to the solution in the cells." This is proven by the results: - "On the graph at low concentrations of sugar the graph gets steeper for example from the concentrations of 0.15 to 0.37, it suggests that lower the concentration of sugar the higher the % gain in length." The predictions that I have made have been supported and been proven by my results, therefore with all this conclusive evidence it can be concluded that my predictions were correct. EVALUATION E-2A The experiment and procedure that I undertook was quite interesting and enjoyable to carry out. It was interesting and enjoyable to do the experiment because the experiment is important and think it has a real relevance as I know what importance osmosis has in biology. The procedure was suitable in this investigation because it was fair and it had all the requirements like the correct equipment and method to find the particular results that were needed. E-4A EVALUATION OF ACCURACY AND ANOMALOUS RESULTS From the graphs line and the plotted points from the results table there is some difference as the graphs line may not go exactly straight through the point. The points are mainly on the graph line but a few are not, these points are not anomalous because they are not far away from the graph line of best fit. This shows that my results are accurate and reliable. Some of the results that were off the line of best fit could have been down to human error in the procedure or little problems in the method. The results just of the line of best fit show some experimental error and the results that I obtained are not anomalous. If you look back on page 13 on the graph you can see I have indicated the points which have shown an error but an error within experimental error and not an anomalous result. On the graph the result points that are off the line of best fit are not very far away from the line of the graph they are in fact quite close, this shows that there are no anomalous results as the results from the readings and results from the graph are the same or similar. E-4B EVALUATION OF METHOD If I were to do this experiment again there would be some definite changes that I would make to make sure that there is as little experimental error and human error as possible so that this will not come up in the results. In the method I would have changed the way I cut the potatoes, next time I should use a machine or device that can cut the potato into an equal square shape where all the dimensions and weight are equal. I would also use another device that cuts the chips into the same length and width, as having chips with the same dimensions will not affect the surface area and affect the rate of osmosis, when I did it I used a knife and scalpel and the dimensions of the chips were not 100 % the same this is due to my human error, this could had affected the results. If I were to do the experiment again I would use a wider range of concentrations, for example I would do 0.0 molar, 0.1 molar, 0.2 molar, 0.3 molar, 0.4 molar, 0.5 molar, 0.6 molar, 0.7 molar, 0.8 molar, 0.9 molar and 1.0 molar. This is a complete range of concentrations of the sucrose solution. This will give me even more reliable results and will back up my predictions more so as the results will be to every 0.1 molar which is very accurate and will allow me to find the isotonic point far more accurately. The other things I will change in the method is I will reconsider the time I allow the potato chips to be in the solution, I will decide the time when I know what the saturation point of the potato chips are so I can decide the appropriate duration the chips should be in the concentration. I could also improved the method by changing the way I measured the different liquids to make the solution and also by changing the way I dried the chips. Some of the small indifferences in the results could have been down to me not drying the chips for the same time and the same way, so if I were to do it again I would find another way to do this. The way I could next time measure the solutions could be with a burette or a pipette, this is so I could measure the amount of fluids accurately, so no mistakes are made on my behalf. E-6A RELIABILTY OF RESULTS Detailed comment on accuracy Concentration Of Sucrose. % From Experiment 1.d.p % From Graph 1.d.p % ERROR 1.d.p 0.00 molar + 16.6 % + 16.6 % 0.0 ÷ 16.6 × 100 = 0.0 % error 0.15 molar + 10.0 % + 9.6 % 0.4 ÷ 9.6 × 100 = 4.1 % error 0.25 molar + 5.0 % + 5.0 % 0.0 ÷ 5.0 × 100 = 0.0 % error 0.50 molar - 3.3 % - 3.2 % 0.1 ÷ 3.2 ×100 = 3.1 % error 0.75 molar - 6.6 % - 6.6 % 0.0 ÷ 6.6 × 100 = 0.0 % error 1.00 molar - 8.3 % - 8.0 % 0.3 ÷ 8.0 × 100 3.7 % error Average % error = 1.8 % The table show that my results are within the acceptable 5 % of experimental error, therefore this proves that my results are reliable and are accurate. Many of my results had no errors and my average % error is only 1.8 % this is small and well inside the 5% accepted experimental error, so there are no anomalous results in context. This tells us that my results in general were accurate and reliable. E-6B FUTHER WORK In this investigation and experiment that I undertook I looked at how the varying concentrations of the sugar sucrose affects osmotic activity in the solution and cell, but more importantly how it affects the length of the potato chips. For further work I would go along this line but investigate some different things. The things that I would investigate are looking into how the varying concentrations of sugar solution affect the % change in mass. I could use the same sugar or I could use a different one like glucose. The other things that I would also change the potato and use a soft fruit like a strawberries or a peach. Other further work which I could under take is an experiment with visking tubing; it will involve a beaker full of pure water and visking tubing containing a sugar solution, there is also a tube there and as the water enters the visking tubing the water rises up the tube. These are just some of the things that I can do to extend my investigation in this area.   

INTRODUCTION This piece of biology coursework is about how cells use osmosis, how, why and what affect it has on the cells in question. The example I am going to explore this in and to illustrate this is the experiment of potato chips of the same length in...

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Introduction: In this piece of...Introduction: In this piece of coursework I will be attempting to discover what affects the digestion of starch. There are four main factors that affect the digestion of starch. 1. Temperature of the solution 2. PH of the solution 3. Surface area of the starch 4. Concentration of starch As the temperature of the starch gets cooler the rate of digestion will decrease. This is due to the number of collisions occurring during the reaction. If the pH value of the solution is quite high then the active site is affected and can be damaged and the hydrogen bonds are also changed. This reaction or occurrence is known as denaturing. This will result in a poor digestion of starch because the active site forms the gap for the starch particle to go into. The surface area of the starch affects the digestion, the reason for this very simple. If the surface area is greater then the particles are able to react on a larger area so more gets digested. The concentration of the solution will play a great part in the digestion of starch. If the concentration of starch to water is quite low then the reaction will not be as quick because there are not as many particles of the enzyme to react with the starch particles active site. Digestion: Amylase and carbohydrase found in your saliva, works best in slightly alkaline conditions. At a PH of about 7·5. The main types of carbohydrate which you eat are: cellulose, starch, and sucrose. You cannot digest cellulose at all because its molecules cannot be broken down in your digestive system so your body is not able to absorb them. Starch on the other hand is easily digested. The process of digesting starch begins in the mouth. Saliva contains an enzyme called amylase which begins to break down the starch molecules into maltase. You do not usually keep food in your mouth long enough for your amylase to finish digesting any starch in it. However when the food enters the stomach the amylase stops working. This is due to amylase not being unable to work in acidic conditions. Active site: The high degree of specificity shown by enzymes, suggest that the combination of substrate and enzyme is very exact. It is thought that each enzyme molecule has a precise place on the surface, the active site, to which the substrate molecules become attached. In our experiment this is going to be one of the main processes occurring during the practical. The area on the enzyme where the substrate or substrates attach to is called the active site. Enzymes are usually very large proteins and the active site is just a small region of the enzyme molecule. This is what is meant by the "lock and Key Theory" that is the active site. This the active site in its three stages of alteration when it is being heated. The three stages are travelling from left to right. Stage 1= the yellow diagram, Stage 2=the Orange diagram, Stage 3= the whit diagram. This image was gathered from: www.scripps.edu/~ulrich/ three_structures_large.jpg For my experiment I will only be altering one of these factors; temperature. The reason I chose to use temperature as my variable is that I feel that it will give us the most interesting results. However keeping the temperature of the solution the same throughout the time span will not be easy and will probably slightly affect the accuracy of the results. To try and keep my results as accurate as I can and to keep the test fair I will inserting a digital temperature probe into the solution that I am heating and once it has reached the designated degree then I will keep the solution at that constant temperature by withdrawing and advance the Bunsen burner to the heated solution. There will be three other pupils taking part in the experiment with me so I was able to use them to help assist me with making sure that the test was fair. One person was the time keeper so was able to tell me when to take a pipette drop from the solution. Another person had the responsibility of ensuring that the temperature of the solution was at a constant. I will be participating in three experiments, which means I will be doing each temperature three times. This should give me more accurate results and if there are any anomalies then the final results would not be affected because an average would be used to work out the final result and draw the graph. The concentration of the solution will stay the same throughout the experiment so that would be kept fair. Prediction: Amylase is an enzyme found in your saliva and is produced by your salivary glands in the mouth. The purpose of amylase in the body is to break down starch and turn it into maltose molecules, however there is not enough time to completely digest the starch into glucose because the food starch is swallowed before the digestion is finished. By the time the starch has reached the stomach the reaction has stopped. However in our experiment we will have enough time, and so the experiment and digestion of starch will finish. Starch is a polymer, which means that it is a compound that is made up of a number of simple molecules. As soon as the starch ids digested the starch is turned into a monomer single molecule. This substance is known as glucose and is small enough so that it can be absorbed into the blood. Amylase is a 'Biological Catalyst', this means that it will speed up the reaction within the body. Amylase is also 'Temperature Specific' this means that the reaction rate varies in conjunction to the temperature of the solution it is working in. This would mean that there would be a variation in the rate of reaction. As the solution cools the starch and amylase molecules lose energy and so the rate of the digestion of starch decreases. The processes of denaturing is shown below: In this investigation, I predict that as the temperature increases the amount of time taken for the experiment to finish will decrease. So on a graph the results would be placed roughly along a straight line. But I also predict that the reaction will get to a certain stage where the rate of the reaction will reach a stage where it cannot increase any more. This is due to the enzyme denaturing. Another reason for the rate of reaction increasing at first is due to the particles in the solution gain more energy from the heat source Bunsen burner and so more of the bonds are destroyed and broken. This rise in temperature also affects the bonds of the active site and damages them, as a result the active site is damaged to a certain extent that it cannot digest the starch any more. This is known as denaturing. Denaturing is when the active site is damaged to a certain extent due to such a high temperature environment that the rate of the reaction is so long that it will take a long time for the reaction to finish. So, on a graph it would look something like this: This graph was extracted from "BIOLOGY A functional Approach Second Edition M.B.V. Roberts" My reason for this is that usually, as the temperature rises the rate of reaction increases. This is due to the particles gaining greater energy as they are heated and so are causing far more collisions. As a result the amount of time for the reaction to finish will decrease. But I also predict that the reaction will get to a certain stage where the rate of the reaction will reach a stage where it cannot speed up any more. This is due to the enzyme denaturing. Denaturing is when the active site is damaged to a certain extent due to such a high temperature environment that the rate of the reaction is so long that it will take a long time for the reaction to finish. Some of the key scientific reasons are from the following resources: Mary Jones & Geoff Jones and the Robert Nelsons Biology a functional approach. Method: For this investigation I will be testing how the concentration of the solution affects the digestion of starch. The experiment will be carried out by using the following equipment and substances: · Starch solution · Amylase · 1x Boiling tube · 1x water bath · 1x gauze · 1x Heat proof mat · Bunsen burner · 1x pipette · 2x Spotting tile · Iodine solution · 1x measuring cylinder To increase the level of accuracy of the experiment a digital thermometer will be used to provide me with better and more accurate results. I will also perform the experiment 3 times so that if there are any anomalies then they will be clear and not affect the averages. The pipette will be used to extract some of the solution so that it can be tested for starch. The measuring cylinder will be used to measure accurately the volumes of amylase and starch. This is how the experiment will be set up and what it will look like: Preliminary Study We decided to test out our method by doing a preliminary study. For this test we used a range of temperatures from 25-100°C, and with 6 different temperatures in all. This was a good range as on the last temperature it was clear that the enzyme had denatured. However the temperature that the enzyme denatured at was higher than expected. In the real experiment I will be repeating this experiment three times as this will rule out anomalies and make the results more accurate. The method worked well and there were no apparent mistakes or problem with it. Temperature Time taken for starch to disappear Seconds 25 120 40 90 55 70 70 60 85 60 100 +360 The chosen temperatures seemed to work well or this preliminary test so I will be using them for my final investigation. And this also provides me with the information that the enzyme will denature at a temperature between 85 and 100°C and that it will take longer than 360 seconds. Results Here are my results from my the three experiments that I carried out: Temperature ºC wanted Actual temperature ºC Time taken for the experiment to stop Exp 1 Exp 2 Exp 3 Average Exp 1 Exp 2 Exp 3 Average 20 26 25.5 24.5 25.3 120 120 120 120 40 40 39 40.5 39.83 100 100 100 100 55 53 54 56 54.3 80 70 60 70 70 70 70 71.5 70.5 60 50 50 53.3 85 83 85 83.5 83.83 60 60 60 60 100 95 93.5 94.5 94.3 Still not finished after 360 seconds These results were all gathered using the same apparatus and layout like that of the one explained in the method. These results portray quite an accurate set of results. There are no anomalies so the results are very pleasing. It was interesting to discover that the experiment did not finish when the temperature of the solution was at 100ºC. This is most likely be due to the active site being deformed. This is known as denaturing. Denaturing is the word given for change of an enzyme and the our case the amylase is the enzyme for the digestion of starch. The graph's are on the next two pages! Graph 1. "“ This graph displays how the temperature of the solution affects the amount of time it takes for the reaction to finish. As you can see the results follow along a slight descending curve, and then suddenly at around 85-94ºC there is a sudden drastic change in the length of time it takes for the reaction to finish. This would suggest to me that the form of the active site would have been deformed. This would result in the time taken for the reaction to finish would take far longer. Graph 2. "“ This graph is displaying the rate of the reaction in contrast to the average temperatures of the experiment. This graph clearly displays that the fastest time it too for the reaction to finish was 70ºC. From then onwards, the length of time for the reaction to finish dramatically increased. The protein had denatured. In a difference of about 11ºC the rate of reaction had increased by 0.140 on the graph. The reason for this sudden change in reaction rate is due to part of the amylase enzyme active site not being able to brake down the starch molecules efficiently and properly. This is due to the temperature being too high. Conclusion: Like in my prediction there was a certain point in the digestion of starch that the enzymes will not be able to work in the usual way. This is due to the temperature the enzyme Amylase is being worked in is too high and the protein of the enzyme is not able to digest the starch molecule properly. The place where the starch molecule is digested is called the active site. As the temperature of the solution of amylase enzyme and starch molecule increases the active site becomes deformed and is not able to digest the starch molecule properly. Coming back to the lock and key theory, the area on a precise place of the large protein is known as the active site to which the substrate molecule is attached I have discovered that this place becomes deformed when placed in a solution of a high temperature. So, as a result, the rate of reaction as the temperature reaches +70ºC starts to decrease. In our experiment when we had managed to get the temperature of the solution up to nearly boiling point 94.5ºC the time it took for the reaction to finish took so long that we had to stop it early. And as a result the breakdown of the starch is not being catalysed. The protein had denatured and it would have taken a very long time before the reaction had finished. The amylase that we had been using was in fact bacteria amylase instead of salvia amylase. Bacteria amylase has a higher denaturing temperature compared with the salvia amylase. And so the results were slightly different to that of my prediction, however the results did produce a similar graph to that of my prediction. Compare the graphs of my prediction and that of my rate of reaction graph from my results. Evaluation: Throughout this investigation I have tried to be as accurate and fair in everything that I have done whether it be taking readings off the thermometer or measuring out the volumes of the amylase and Starch solutions. The careful detail for accuracy prevailed, as the results did not contain any anomalies, and the graphical results were as predicted. I also feel that they were as accurate as they could have been using the relatively limited equipment available. However, if I had access to more advanced and more accurate equipment then the results would be more accurate and precise. An enclosed environment that was able to change in temperature would be ideal for keeping the temperature of the solution at a constant. Not allowing the solution to fall or rise, unless there was a change in temperature conducted by the scientist. An electric water bath would be a good piece of equipment for this, as it will keep the water at a constant temperature. This constant precise temperature that would be supplied by the water bath would create a graph that would be more similar to that that I used in my prediction. A timed thermometer reader would also give me the actual reading at that particular time so the results would be more accurate on the graph. I think it would be perfect idea if we conducted the same experiment but using saliva amylase and compare the differences with the bacteria amylase that I used in this experiment. Another way in which I could have improved this experiment is to have increased the sampling rate to every 5 seconds. This would have given more accurate results because a lot can happen in every second. However it would not be practical if I took the samples every second because I would not have had enough time to second the date in a time space of 1 second as I would have extract some of the solution and put a drop of it into a spotting ti le with one drop of iodine in it. Ptyalin is an enzyme that occurs in the salvia and is used to help convert the starch molecules into sugar. It would be interesting to discover the role and difference that this has in comparison to when you only use the bacteria amylase. some of the information was gathered from: "A dictionary of Science" by E.B.UBAROV, D.RCHAPMAN, ALAN ISAACS   

Introduction: In this piece of coursework I will be attempting to discover what affects the digestion of starch. There are four main factors that affect the digestion of starch. 1. Temperature of the solution 2. PH of the solution 3. Surface area of the starch...

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