The enzyme potato catalase Essay Example for Free

The enzyme potato catalase Essay Analysis Instead of filtering out poor data for my results I have decided to select the best result to concentrate on. I removed quite a few results. The reasons for removing results was either because of missing data, anomalous data (not following the trend of others that I believed to be correct), and unusual entries such as amount of oxygen collected decreasing as the experiment went on. This most likely was because of inaccurate readings as other explanations, such as the oxygen dissolving in the water, are unlikely. These are the result I have chosen: The effect of hydrogen peroxide (substrate) concentration upon the rate of oxygen production in the presence of Catalase Concentration of Hydrogen Peroxide (vol) Volume of gas collected in inverted measuring cylinder at end of each successive minute (cm3) over a period of 4 minutes Student InitialsThis is an example of one of the results I did not include: From my selected results I also split them into the 4 different concentrations of hydrogen peroxide tested for ease of use which went as follows:5cm2 10cm2 1 Minute 2 Minutes 3 Minutes 4 Minutes 1 Minutes 2 Minutes 3 Minutes 4 Minutes Minutes 2 Minutes 3 Minutes 4 Minutes 1 Minutes 2 Minutes 3 Minutes 4 Minutes 1I did not include this as some information is missing, possible implying that the way in which the experiment was performed was faulty. This could lead me to a false conclusion. This is another example of the type of results I did not include:   As you can see the amount of oxygen collected has decreased throughout the experiment, this is very unusual, and likely suggests that the readings were incorrectly read, for this reason I do not want inaccurate data to lead me to a poor conclusion. Read more: Essay on  Potato Catalase After that I found the standard deviation of all my results to further check there were no anomalous data, these are my findings: CONCENTRATION 5cm2 Entry/Minute Sum of Standard Deviation   Standard Deviation0CONCENTRATION 10cm2 Entry/Minute Sum of Standard Deviation CONCENTRATION 20cm2 Entry/Minute Sum of Standard Deviation 7. 26 5. 98 7. 45 8. 72 Standard Deviation   Although there are a few entries with high deviation (for example entry number 6 on the 5cm2 concentration table) all his/her results are consistently off the standard deviation, this suggests that there is nothing wrong with their collection of data, so there for I decided to leave them in. After selecting my results, I have taken the mean of the selecting results, it is as follows: Concentration/Time This graph shows that as the concentration of hydrogen peroxide increases so does the amount of oxygen produced. It increases at roughly the same rate throughout the reaction and the amount of oxygen produced is generally a higher amount with a higher concentration. This half matchs with my hypothesis, I predicted that the reaction would start to slow down after the initial reaction had occurred, this does not however seem to be the case. However, the initial rate of reaction is a lot higher The possible reasons for this could be that the reaction did not have enough time to start to level off or slow down as there was still a lot of substrate left over and the reaction could still be performed at maximum rate, if this was the case it would not start to slow down until there was significantly less substrate available, obviously this has not been the case. These results do match my hypothesis in that I said as concentration increased, so would the amount of oxygen produced and the rate of reaction would generally be greater. Here is a graph to show the initial rate of reaction for different concentrations of hydrogen peroxide As you can see, as the concentration increases, so does the initial rate of reaction, this is because a greater amount of hydrogen peroxide is available, which means more substrate molecules come in contact with the enzymes (and thus their active site) and can be separated into their products, this is explained simply by collision theory in the introduction. The reason that later on in the reaction the rate of the reaction may be different is because there is likely to be less substrate left over as the reactions occurring would have separated them into their products, hence the reaction rate would not be going as fast. Appendix Mean Added up all the entries, then divided them by the number of entries there were. E. g. The mean for   Would be (1. 1+1. 3+1. 2+1. 4)/4 which equals 1. 25 Standard Deviation For the single entries: (Entry-(mean of all results in that concentration and minute))to the power of 2 For the total standard deviation: Square root of (Sum of all single standard deviations/number of entries 1). Bibliography Biology 1 (Cambridge Advanced Sciences) Internet URLs: http://www. clunet. edu/BioDev/omm/catalase/frames/cattx. htm http://www. beyondtechnology. com/tips016. shtml The above preview is unformatted text This student written piece of work is one of many that can be found in our GCSE Patterns of Behaviour section.