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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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Aim I am going to...Aim I am going to investigate the effect that different concentrations of sugar solution have on the amount on osmotic activity between the solution and potato chips of a given size. Scientific theory and diagrams For this particular investigation I think that the lower the concentration of the sugar solution in the test tube the larger the mass of the potato will be. This is because the water molecules pass from a high concentration i.e. in the water itself, to a low concentration i.e. the potato chip. Therefore, the chips in higher water concentrations will have a larger mass than in the higher sugar concentration. As a result of this the potato will decrease in mass because there are less water particles in the potato cell than there is outside. In this instance the potato will go flaccid. I think that the lower the concentration of the sugar solution in the test tube the larger the mass of the potato will be. This is because the water molecules pass from a high concentration i.e. the water itself, to a low concentration i.e. in the potato chip. Therefore, the chips in the higher water concentrations will have a larger mass than in the higher sugar concentrations "“ this is referred to as being turgid. Water molecule Movement. Semi permeable Membrane. This is an example of Osmosis. The water particles move through the partly permeable membrane from the region of high concentration to the region of low concentration. The diagram on the right is the potato chip. Equipment I will need: Core borers Ruler Knife Sugar solution diluted to 5 different strengths Balance ½ a large potato. Access to boiling equipment e.g. boiling tubes. Method Firstly I got ½ a small potato and used a core borer to remove the pieces of the potato. Once I had got 15 pieces of potato I cut them and obtained their mass in gram. Through careful and precise cutting and boring techniques I attempted to obtain potato samples that were nearly identical in mass in the region of 0.65g. I then got 5 test tubes and filled them with the correct solution here are the 5 solutions:- · 20ml water 0M · 15ml water 5ml sucrose 0.25M · 10ml water 10ml sucrose 0.5M · 5ml water 15ml sucrose 0.75M · 20ml sucrose 1M Then I put 3 pieces of potato in each test tube and labelled the tubes so I knew which one was which. I left them for 24 hours and then weighed them again and plotted a table of results. Diagram of the experiment Was it a fair test? I tried to make it a fair test by trying to make all the pieces of potato the same size and put the same number in each test tube. I used electric scales so they were more accurate and made sure I left them all for exactly the same amount of time. When I weighed them at the end I made sure they were all dry so it would be a fair test. To create a fair test certain aspects of the experiment will have to be kept the same with only one key variable to be changed. They variable that will be changed in my experiment will be the concentration of the sugar solution. By doing this I will obtain a varied set of results from which I hope to draw my conclusion. If any of the following non "“ variables are not kept constant my experiment would not be a fair test. · Length of potato samples to be constant thereby preventing variation in surface area. · I am going to carry out the experiment at a constant room temperature. · I shall treat the potato in the same way e.g. have all been cored without being washed or peeled. · I will the same set of electronic scales and measure in grams to 2 decimal places. · I will measure the potato samples before it is put in the solution and after. · The volume of the solution will be the same in each test tube. Number and range of results I bored 15 pieces of potato using the corer and measured them so they were nearly all the same weight. I put 3 in each test tube so I could get a better set of results. This gave me a wider range of results to work with so I could give a better explanation. My results will be accurate because I have got a lot of results and I have used electric scales so my results are more accurate and reliable. I repeated my results 3 times so they were accurate. I have tabulated and graphed the results on the following pages. Preliminary work Before I did this coursework I did nearly the same experiment but with sultanas, I did not measure the results when I used sultanas I just did it as a trial coursework so I had some experience in what to do. Final prediction and plan I knew roughly what would happen because I knew that one would swell up and that one would go down so I had a plan really because I did some work in class. Safety procedures v Be careful with core borers to prevent injury. v Wear safety goggles to protect the eye from food products and the solutions. v Be careful carrying glass around the room. v Make sure there is no electricity near water. v I washed my hands every time I handled the potatoes to make sure they were clean and dry to prevent contamination of any kind. I thoroughly to make sure I did not pass on extra water onto the next potato sample. What I found out I found out that the potato cells increase in mass in solutions with a high water concentration and decrease in mass in solutions with a low water concentration. This is demonstrated clearly in my results with an average percentage change of -27.3% at 1 molar concentration and a 56.5% increase at 0 molar concentration. My results also match with my initial predictions. Evaluation The experiment was very successful in my opinion. I obtained a large quantity of very accurate results from which I was able to create an informative graph. I think I took easily enough results for the amount of concentrations that I was using, and the time that I used for the experiment to last was enough to allow sufficient osmosis to occur. However if I was to repeat the experiment I might well increase the time of the result to allow more osmosis to happen and possibly find out the saturation point of the chips. The range of concentrations was adequate but I would possibly create more concentrations if I repeated the experiment so that I would have more varied results, i.e. 0.10m, 1.15m, 1.20m, and so on. This way would have allowed me to also find out the isotonic point far more accurately. Obtaining the same size sample of the potatoes was the most difficult part of the experiment because even though I was recording my results by mass, it could well have affected the surface area and so the overall rate of osmosis. If I were to repeat the experiment I would have possibly found a machine to obtain precise sample sizes to ensure that all potatoes would be the same weight and dimensions. As well as the potato I could have found a more accurate way to measure out the solutions and to determine the molar concentrations. I could have used a burette or dropper pipette. This would ensure that I have an accurate amount of fluid in each test tube. I could also weigh each chip on a more accurate scale, e.g. not to 0.00g but to 0.0000g. There were not any results that were wildly out of the ordinary. When the potato chips were removed from the test tubes and dried I may well have dried some potatoes more thoroughly than others and so some would have more excess water, which would add to the mass. If the experiment was repeated I could find another way to dry the potatoes that would ensure that all were dried in the same way for the same time. Despite this I think that the experiment was successful and I was very pleased with the complete comparison of my results with my initial prediction. Table to show the results of the experiment and the difference in mass of the potatoes. This table shows the weight before the potato was put in the solution, the weight after the potato was in the solution for 24hours, and the difference between the weight before and the weight after and the concentration in molars. There are 3 results in each table because I put 3 pieces of potato in each test tube. There are not really any odd results they are all fairly accurate. I have recorded my results to 2 decimal places for accuracy and consistency. Mass before sample 1 Mass before sample 2 Mass before sample 3 Mass after sample 1 Mass after sample 2 Mass after sample 3 Difference sample 1 Difference sample 2 Difference sample 3 % change sample 1 % change sample 2 % change sample 3 Average % change 20ml sucrose 0.68 0.65 0.64 0.49 0.47 0.47 -0.19 -0.18 -0.17 -27.94 -27.69 -26.56 -27.40 15ml suc 5ml water 0.62 0.63 0.63 0.58 0.51 0.57 -0.04 -0.12 -0.06 -6.45 -19.05 -9.52 -11.67 10ml suc 10ml water 0.63 0.65 0.67 0.63 0.61 0.67 0 -0.04 0 0.00 -6.15 0.00 -2.05 5ml suc 15ml water 0.62 0.63 0.63 0.66 0.69 0.69 0.04 0.06 0.06 6.45 9.52 9.52 8.50 20ml water 0.68 0.64 0.67 0.77 0.72 0.83 0.09 0.08 0.16 13.24 12.50 23.88 16.54   

Aim I am going to investigate the effect that different concentrations of sugar solution have on the amount on osmotic activity between the solution and potato chips of a given size. Scientific theory and diagrams For this particular investigation I think that the lower the concentration...

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Aim: I am going... Aim: I am going to investigate how different surface areas, and therefore enzyme concentration, of the same volume of potato affects rate of reaction, when placed in 10ml hydrogen peroxide. Prediction: As the surface area increases, so do the amount of enzymes, which are biological catalysts that are on show, free for the hydrogen peroxide to collide with. Because the catalase and hydrogen peroxide are a complimentary pair, and the enzymes are the ones that break down the substrate, then the more enzymes there are, the more hydrogen peroxide will be reacting at once, and once that has been broken down, then another hydrogen peroxide molecule will 'lock' into the catalase's active site, and the process will start again. Therefore the more enzymes or catalysts available, the faster the rate of reaction will be. As the reaction is taking place, oxygen and water are made from the hydrogen peroxide. The oxygen is given off as a gas, this is what I will be measuring, but the water will be left in the test tube along with the hydrogen peroxide. As the reaction goes on, there will be less hydrogen peroxide, because it has reacted, so there is less chance of collision between substrate and enzyme because the concentration of hydrogen peroxide has decreased. Making the chance of collision even lower is the water that has been broken down from the hydrogen peroxide, which is now in the way of the substrate still left trying to collide and react, which is now colliding into the water molecules, and taking longer to collide with an enzyme, making the reaction slow down. Results Volume Pieces Size of sides cm Surface Potato no. of potato cm of potato h:w:d area cm 1 2 3 Average Average 3sf 1 1 1 1 1 6 Oxygen 1.4 1.1 1.3 1.266666667 1.27 1 2 1 1 0.5 8 produced 1.8 2.1 1.8 1.9 1.90 1 4 1 0.5 0.5 10 ml 2.7 2.7 2.4 2.6 2.60 1 8 0.5 0.5 0.5 12 3.3 3.1 3.2 3.2 3.20 1 16 0.5 0.5 0.25 16 4.4 3.9 4.2 4.166666667 4.17 Skill A: Analysis My results show that, as surface area increased, the rate of reaction also increased, with 6cm² producing the least amount of oxygen in 4 minutes, and 16cm² surface area producing the most oxygen in 4 minutes. The relationship between the two surface area and oxygen produced is that, they are almost directly proportional, but with rate decreasing just a little at each recorded surface area, at around 0.1ml from 6cm² until 12cm². When it reaches 16cm², the oxygen produced is about 0.5ml less than if oxygen produced was directly proportional to surface area in that, from 6cm², oxygen produced goes up 0.3ml every 1cm². If I investigated further, with higher surface areas, I would expect to start seeing the graph level off gradually, as it had started to at 16cm², and eventually, the rate at which it would level off, would increase rapidly, until almost completely level. My results support my prediction up to just below 16cm² surface area. Below this point, the results were both directly proportional and oxygen production doubling, as surface area doubled, with a few points slightly out, which is to be expected. At and therefore I would expect, after 16cm², the results started to become lower than the line of direct proportionality, and I expect that the doubling would also. This is due to the breakdown of hydrogen peroxide, which leaves less substrate for collision, and also producing water, making the solution more dilute, so collisions become less frequent, after the more hydrogen peroxide that has reacted. If you double the surface area of enzymes then the rate of reaction doubles as well. When the surface area is doubled form 8cm² to 16cm², the rate goes up from 1.9ml "“ 4.17ml of oxygen produced in 4 minutes, which is an increase of 2.19 times. I would expect the rate to keep increasing but a bit less rapid than as previously from 16cm² onwards. The 0.19 more than two times rate can be put down to experimental error, and could be due to anomalous results. All the active sites can never all be being used at the same time, because they all may vary slightly in the time it takes to break down their individual substrate, and some may not be being used, or some finishing as some are starting, plus it takes time for the substrate to fit in and leave the active site, so you could never really reach the theoretical maximum value of oxygen that could be produced for a specific volume to surface area ratio because of this. As the surface area increases, so do the amount of enzymes, which are biological catalysts that are on show, free for the hydrogen peroxide to collide with. Because the catalase and hydrogen peroxide are a complimentary pair, and the enzymes are the ones that break down the substrate, then the more enzymes there are, the more hydrogen peroxide will be reacting at once, and once that has been broken down, then another hydrogen peroxide molecule will 'lock' into the catalase's active site, and the process will start again. Therefore the more enzymes or catalysts available, the faster the rate of reaction will be. As the reaction is taking place, oxygen and water are made from the hydrogen peroxide. The oxygen is given off as a gas, this is what I will be measuring, but the water will be left in the test tube along with the hydrogen peroxide. As the reaction goes on, there will be less hydrogen peroxide, because it has reacted, so there is less chance of collision between substrate and enzyme because the concentration of hydrogen peroxide has decreased. Making the chance of collision even lower is the water that has been broken down from the hydrogen peroxide, which is now in the way of the substrate still left trying to collide and react, which is now colliding into the water molecules, and taking longer to collide with an enzyme, making the reaction slow down. This is why I expect a slow down in the increase of oxygen production as surface area becomes greater than 16cm². I kept the conditions for each experiment the same, with the exception of surface area, which I was measuring. If I had altered any of the other variables, then it would not have been a fair test, as the enzymes work at different rates in different conditions, with an optimum level for each variable, and also a point in each variable where the enzymes start to denature. I kept the temperature throughout the experiment at a constant of 24ºC. If I had increased the temperature then the hydrogen peroxide would have had more kinetic movement energy, and therefore would have moved faster, and as collision theory is basically how the rate of reaction depends on how often and how hard particles collide with each other. So in order to react they need to collide hard enough. So with the increase in temperature, the hydrogen peroxide will have more energy to move faster, so they have more collisions, and also hit harder. Faster collisions only come with increasing temperature. If the temperature is too low, then the particles do not have enough kinetic energy to hit hard enough to react with each other. Enzymes also work better in warmer temperatures; so unless in their optimum temperature environment, temperature is a limiting factor, and the enzymes will not work at their optimum rate. However, if the temperature goes too high, or too low, the enzymes will become denatured, and therefore start to stop working, and may start to die off, which decreases rate of reaction. Concentration is another variable that changes the rate of reaction. If there are many particles of substrate between water molecules, then there is bound to be more collisions of enzymes and substrate, whereas, if there are few particles of substrate between the water molecules, then there is less chance of reaction of the complimentary pairs. If there are more packed in together, then they are closer and are more likely to collide with their reactant. Enzymes also work best in an optimum pH value, and if not at that, like temperature, pH is a limiting factor. But as the enzymes were left in their tissues/optimum environment, then it was not a factor in this experiment. Hydrogen peroxide breaks down on its own without catalase. Catalase is just a catalyst, which just speeds up a reaction, by reducing the initial amount of energy needed for a reaction; it breaks down the substrate faster, in this case the hydrogen peroxide. An enzyme is like a lock, for which it only has one specific substrate, that being the 'key'. As the substrate fits into this 'lock', it goes into what is called the active site, the part of the enzyme that actually does the breaking down of the molecules. But after the reaction, the catalase is not used up, or does not differ after the reaction; it can be used again and again after countless reactions. It also does not cause the reaction, but simply speeds the reaction up. I also kept the same potato, for each set of results I took three, as using lots of different ones could have produced unreliable results, as different potatoes' enzymes' reaction times may differ slightly. I did 5 different experiments each on 3 different potatoes 5 experiments on each, to obtain accurate results incase one potatoes enzymes were very different from most.   

Aim: I am going to investigate how different surface areas, and therefore enzyme concentration, of the same volume of potato affects rate of reaction, when placed in 10ml hydrogen peroxide. Prediction: As the surface area increases, so do the amount of enzymes, which are biological catalysts that...

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