Ph And Surface Area Of The Potato example essay topic

Introduction In this experiment, I will be studying the activity of the enzyme catalase. Catalase breaks down of Hydrogen peroxide H 2 O 2 into water H 2 O and oxygen O 2. catalase 2 H 2 O 2 2 H 2 O + O 2 catalase Hydrogen peroxide water + oxygen Enzymes are biological catalysts. They are made up of protein. They will only act upon one substance, is substrate. Enzymes only work at their optimum in a specific environment.

There are many different factors in their environment, which affect an enzyme's ability to decompose their substrate. Enzymes work at an optimum if you find the right balance between these factors. Some examples of these factors are temperature, pH and surface area of the potato. They are three main groups of enzymes involved in breaking down food molecules Proteases, Carbohydrases and Lipases. Proteases breakdown proteins, pepsin and trypsin are proteases. Carbohydrases breakdown carbohydrates, amylase is a Carbohydrases.

Lipases breakdown fats. Enzymes aren t just used inside the body they re also used outside the body in many different ways. Examples of these are + Biological detergents + Fruit industry + Photographic industry + Dairy industry + Brewing industry + Baby foods + Paper industry + Starch industry + Rubber industry + Baking industry At the moment the hypothesis about how enzymes work is the Lock and Key Theory. This theory states that the active site of the enzyme and the substrate are extract ly the same size. In the right environment (right pH and temperature) it is thought that the active site of the enzyme will catalysis the reaction.

If one of the variables is changed like the substrate or the pH the enzyme won t catalyse the reaction. This is a diagram of the lock and key theory. It shows the enzyme's active site and the substrate are the exactly the same size and shape By binding together the reaction takes place. The products of the reaction then leave the active site and the enzyme can look for another substrate molecule. In the experiment that I m doing the products of the reaction are water (H 2 O) and oxygen (O 2). The enzyme is catalase, from potatoes, and the substrate is Hydrogen peroxide.

Problem What factors influence the catalase's ability to decompose Hydrogen peroxide into water and oxygen Factors There are many different factors that affect this reaction. They are as follows: The factors that we can test are pH, surface area of potato containing catalase and temperature of the solution. I have chosen to keep the temperature constant because it's know that catalase will work best at body temperature. That is its natural environment. I have also chosen to keep the surface area constant. This is because the larger the surface area of potato quicker the better the reaction.

This is why the body physically breaks down food so those enzymes contained in bile have a larger surface area to work on. I have chosen to vary the pH because I don t know exactly how it affects the enzymes' ability to breakdown Hydrogen Peroxide. Question What will happen to the rate of the reaction catalyse by catalase if I vary the pH Prediction I predict that the catalase will work at optimum at a neutral or slightly acidic environment. I say this because catalase is present in liver cells.

It is in the liver because for this reason. The liver job in the digestion system is to neutralise any toxins in the bloodstream. Hydrogen peroxide is a toxic by-product of digestion so it is there are neutralise Hydrogen peroxide. The products of this reaction (water H 2 O and oxygen O 2) are essential to the function of the body. Catalase is present in all our liver cells.

This means it's natural environment is a neutral or slightly acidic pH. This is backed up by my preliminary work (e.g. the effect of catalase on liver cells) in which I ve studied the rate of the reaction. In this case the rate of reaction is fastest at a neutral pH. The theory behind this prediction is the Lock and Key theory. This theory states that the active site of the enzyme and the substrate are exactly the same size. If one of the variables is changed or at an extreme like the substrate or the pH the enzyme won t catalyse the reaction, because if difference in the environment alters the hydrogen bonds, ionic charge or whatever is making it hold its shape.

This has a particular effect on the active site of the enzyme changing its shape so it and the substrate won t fit. This means that the enzyme has been denatured. Therefore, this is why I think at pH 2, 4 and 10 will have a slow rate of reaction or no reaction will take place. I predict that my results will look like this graph: The catalase with work at optimum at pH 7 (7.07) and the rate of the reaction will drop down steadily either side of that pH. Plan I will collect the pieces of equipment I need and set them up as shown in the diagram. I will collect 20 milliliters of hydrogen peroxide and of each of the five pH buffers.

I will measure the pH's using a pH meter making sure that they are the same pH. I will weigh two five-gram pieces of potato (I will not use a 10-gram piece of potato because the shape would mean that not all of the potato would be in contact with the solution). I will make sure the surface area is the same with all the potato pieces by using the same cork pourer. I will collect these quantities because than I have a 1: 2 ratio between grams of potato and Hydrogen peroxide so the differences in the reaction are apparent. Any ratio that is below this won t show the affects of the differing environments. Also in other experiments were the ratio was less the results weren t conclusive enough to prove anything.

I will then put these quantities into a conical flask and attach a gas syringe to see how much oxygen gas is envolved. The gas syringe measures from 0 to 100 ml's. I will measure the volume of oxygen gas given off every minute in milliliters. I will stop measuring after ten minutes because in previous experiments that I ve done the reaction tends to slow down after this time.

I will repeat the experiment three times with all five pH buffers to make sure that any anomalous results won t affect the average result. Fair Test In this experiment I will make it a fair test because I will do many things, such as 1. I will use the same concentration of hydrogen peroxide 2. I will use the same mass of potato (10 grams) 3.

I will use the same surface area of potato 4. I will have the solution at the same temperature (27 degrees / room temperature) 5. I will use the same volume of hydrogen peroxide (20 milliliters) 6. I will use the same volume of the pH buffer Safety I will make sure that I will do this experiment safely because I will do the following things: + I will wear safety specs when ever I m doing an experiment + I won t lean over the bench so that I won t spill any of the solutions + I will be careful when carrying any solution or liquid + I won t run around the classroom + I will tuck in my tie so it won t interfere with any experiment + I will wash my hands after handling any substances (potato, Hydrogen peroxide). + I will be careful when pouring any substance + I will be careful when cutting the potato to size Method I will collect the following pieces of apparatus a Glass syringe, Conical flasks, a Stopwatch, a Clamp, a tile, a Clamp stand, two five gram pieces of Potato, 20 ml of Hydrogen peroxide, scales, a knife, 20 ml of the pH buffers (pH 2, 4, 7, 9, 10), measuring cylinder and a cork pourer.

I will set them up like the diagram shows (below), making sure that the gas syringe is horizontal so it doesn t slip down. I will get the potato, tile, scales and the cork pourer. I will cut out cylindrical pieces of potato using the cork pourer on the tile. I will weigh them to make sure that they are five grams. I will then make any necessary adjustments using the knife. I will then take the measuring cylinder and measure 20 ml of Hydrogen Peroxide and the pH buffer (2, 4, 7, 9, 10).

I will put them together in a conical flask. I will reset the stopwatch. Simultaneously, I will start the stopwatch and place the potato pieces in the solution. I will attach the gas syringe by pushing the cork of the gas syringe in the conical flask.

I will then observe the reaction to make sure that it's going to plan e.g. equipment is working. After a minute I will measure the oxygen envolved by the solution. I will measure the oxygen envovled every minute for ten minutes cumulatively. I will do this for every pH buffer three times to get rid of any anomalous results. Diagram Apparatus list 1. Glass syringe 2.

Conical flasks 3. Stopwatch 4. Clamp 5. Clamp stand 6. Potatoes 7. Hydrogen peroxide 8. pH buffers (2, 4, 7, 9, 10) 9. measuring cylinder 10. cork pourer 11.

Knife 12. Tile 13. Scales Conclusion My results show that pH 7 is the optimum, that pH 2 and 4 are the worst. This means that a neutral environment is optimum for catalase and, acidic environment denatures catalase. This does agree with my prediction, that pH 7 would be the best environment for catalase, that pH 2 and 4 are the worst environment for catalase. This means that my prediction is correct, therefore, pH 2 and 4 denatures catalase and pH 7 is catalase's natural environment My results show that pH 10 is the best this is an anomalous result.

I say this because then compared with other primary and secondary sources; pH 10 was one of the worst environments. There are many factors that could have made this result. Examples of these are + We did the experiments at different times, in the morning and afternoon. Therefore, the chemicals might not have been at the same temperature. + We had different types of potato; this resulted in a change in the concentration of catalase in the potato. + Inaccurately of the scales, gas syringe + Temperature fluctuation in room My experiment has as accurate as I could have made it in the time available and the equipment available.

To make the experiment more accurate I would need more time and better equipment. For example we could used a more precisely calculated syringe e.g. plastic syringe. The results that I have collected show that the Lock and Key theory is correct. Therefore, this is why pH 2 and 4 had a slow rate of reaction. Diagram Evaluation I think that my plan has worked well because I have been able prove my prediction. Also I think that the range of pH's that I used was good because it gave me a good idea of what the enzymes activity in different environments is.

My results were good they were as accurate as I could make them. An example of a factor beyond my control, the extra gases pressure when putting the cork into the conical flask. The only result which was not accurate was pH 10, it is an anomalous result. I think this is because of the following reasons: + We did the experiments at different times, in the morning and afternoon. + Inaccurately of the scales, gas syringe + Temperature fluctuation in room Apart from this anomalous result I think that my results are sufficient to support that pH 7 is the optimum. If I did this experiment again I would try to make my results more accurate by repeating my results as many times as I could in the period of time available.

I already know that pH 7 is the optimum I would not have such a large range of pH's I would have a smaller range from pH 5-9. I would expand the experiment by using different enzymes and their substrates to see if there is any relationship with the substrate and the environment in which the enzyme works at an optimum.