Experimental hypothesis:
I expect to obtain a set of results which show that an increase in the substrate concentration (hydrogen peroxide concentration) will increase the enzyme activity within the four minutes, and therefore increase the rate of reaction. This I expect to be reflected in the volume of oxygen gas collected, where I predict that a largest volume of oxygen will be collected for the 20 volume solution, as this is the highest concentration. This can be explained by the fact that an increase in substrate will result in more hydrogen peroxide molecules surrounding the catalase, compared to a low concentration. Taking into account that there is the same amount of catalase, this means that more hydrogen peroxide molecules will be bound to the active sites of the catalase every second. This will increase the enzyme activity as there will be more active sites being used, and therefore the increase in substrate broken down, will result in a larger volume of oxygen produced.
Testing the reliability of the results
I have decided to use 10 randomly selected results within the class to test my hypothesis. Each individual set of results was obtained by a different person. It is therefore important to test the reliability of the results by using various statistical methods. Statistical methods will give an exact numerical value, measuring the spread and consistency of the data, which will give a good idea on the distribution, and therefore the likelihood the data occurred by chance. A good statistical method to be used in doing this is the calculation of the two averages: mean and median between two randomly selected sets of data. The data selected for this statistical test, has been highlighted on the results table.
Mean is equal to the sum of all the values divided the number of values there are. This can be shown by the equation:
X = Σx/n
Therefore the mean for data set 1 is:
1.4+1.8+2.1+2.4/4
= 1.9
The mean for data set 2 is:
0.8+1.4+2+2.6/4
=1.7
The median can be defined as the middle value of the set of data in numerical order, (from lowest to highest). There are four values for each set of data, and therefore the middle value would be the 2 ½ value, or the mean between the 2nd and 3rd values. Therefore, the median for:
Data set 1 = 1.8+2.1/2
= 1.95
Data set 2 = 1.4+2/2
= 1.7
The standard deviation of each set of data results will give a numerical value showing the spread of the data. This is the last statistical test to be taken, and after comparison of these results, I will conclude whether there seems to be a difference between them.
Standard deviation = √(Variance)
Variance = (Σx2 /n) - x2
The variance can be defined as the square of the mean subtracted from the sum of all values squared, divided by the frequency.
Therefore for data set 1, variance is:
15.53/4 – 3.61
=0.2325
Therefore, the standard deviation = 0.48
For data set 2, the variance is
13.36/4 – 2.89
=0.45
The standard deviation is = 0.67
The statistical tests on the two sets of data show that although they were collected as a result of two individual experiments, they have similarities between them in terms of mean and median. The mean and median are two methods of an average, and it being similar for the two sets of data proves that the data has good consistency. The spread of the data was measured by the standard deviation, and as can be seen from the values, the distributions of the sets of values are different. However, this does not make such a difference in the reliability of the data, as it is the trends and patterns that the data shows that is of importance to me. Overall, the statistical tests have provided me with results showing that they are consistent, and therefore are a good basis for my conclusion.
Statistical significance test
Statistical significance tests are used to show whether the difference between the means of two sets of data are truly significant. I have decided to use the Mann-Whitney U test, as this is the best significance test to use for independent measures design experiments, which does not take the distribution of the data into consideration. It will tell me whether the difference between two randomly selected sets of results is significant, and whether it supports my experimental hypothesis. If the results are significant, it will show that my hypothesis is correct and the concentration of hydrogen peroxide does effect the rate of reaction of catalase.
I have selected two sets of results to use for the Mann-Whitney U test, which I have made clear in the results table. Before, doing the test it is important to make a null hypothesis, which will show what the results imply, if they do not support the experimental hypothesis. The null hypothesis is that the concentration of the hydrogen peroxide does not affect the rate of reaction of the enzyme catalase.
Using the method for the Mann Whitney U test for the two sets of results, I have gained that the S values are:
312.5 for data set 1
304.5 for data set 2
Using the equation
N1.N2 + (NX(NX+1)/2) – SX = U value
where N1 = number of values in data set 1
N2 = number of values in data set 2
NX= highest number of values either data set
SX= highest s value
The U value can be calculated as
64 + (272/2) – 312.5
= 79.5
Looking in a table of critical values, I have found that for a 16 by 16 value data size, this is 81. Therefore, the U value obtained is less than the critical value, and therefore I can reject the null hypothesis.
Using this it can be said that where p=0.05, that 95% of my results support my experimental hypothesis. This means that the results are truly significant and that the concentration of hydrogen peroxide does affect the rate of reaction of catalase.
Graphical analysis
In general, the results obtained by the class show a large variation, not only within the measurements, but also within the distribution. Therefore, to be able to closely analyse the data, it is necessary to display the results in the form of graphs. These will aid me in my explanation, and provide a clear presentation of the data.
On conducting this experiment, it was obvious that the speed at which the bubbles of oxygen was being produced by the catalase activity on the hydrogen peroxide increased considerably as the concentration concerned increased. The results obtained show evidence for this, from which I have created the column graph below:
This column graph shows the results obtained by experiment number 2, for all four hydrogen peroxide concentrations over four minutes. It can be seen clearly that the total amount of oxygen collected for each concentration, (at four minutes) exceeds one another as the concentration increases. As both the time and the amount of catalase were controlled variables, and thus kept roughly the same for each concentration, this can only lead to the suggestion that the increase of the total oxygen collected each time was the result of an increased rate of reaction. An increase in the rate of the breakdown of hydrogen peroxide by the catalase present in the potato disks increases the rate at which the products oxygen and water is produced. As the time of four minutes allocated for the reaction was not sufficient time for all the hydrogen peroxide to be broken down, the volume of oxygen collected after this was a reflection of speed at which the reaction was occurring, or the rate of the reaction. As the volume of oxygen collected increased with the concentration, the rate at which the enzyme catalase was interacting with the hydrogen peroxide increased with the concentration. This implies that my hypothesis was correct, and that the increase in the hydrogen peroxide substrate, allows it to be broken down quicker, as an increase in the use of more active sites of the catalase results in a faster rate of reaction. However, the graph uses only one set of results, from which I cannot accept that my hypothesis is correct, as this is not a solid base for my conclusion.
It is therefore necessary to compare the class results, to see whether they too show a similar pattern. To do this I have created a line graph showing the total amount of oxygen collected with an increasing concentration of the hydrogen peroxide solution.
It can be seen that the class results also support the fact that the volume of oxygen increased with the increase in peroxide solution, shown by the positive correlation of each line. This therefore supports the biological theory that increased substrate increases the rate of enzyme activity upon it. In the 5 volume concentration, the concentration of hydrogen peroxide was low, and therefore there were a low number of hydrogen peroxide particles to be broken down. A low number of particles, in a high number of water particles means only a small number of particles will combine with the catalase to be broken down at any one time. In comparison, a high concentration, like the 20 volume concentration contains many more hydrogen peroxide particles available for the catalase to combine with. This results in more particles combined with the active sites of the catalase, causing a larger number of particles being broken down at any one time. Therefore, a higher number of oxygen and water particles are produced at a given time, which results in a larger volume of oxygen given off by the reaction with a higher concentration of the hydrogen peroxide.
In conclusion, the increase in oxygen with the concentration supported the idea of an increase in rate of reaction of the breakdown of hydrogen peroxide stated within my hypothesis. This showed that the increase of the amount of hydrogen peroxide substrate caused the enzyme catalase to act upon it at a faster rate increasing the amount of the oxygen product produced. Fundamentally, this supports the biological theory that increasing an amount of substrate will increase the enzyme activity upon it, providing that the amount of enzyme active sites available for that substrate is more than the substrate particles involved.
Shakila Chowdhury Analysis Biology Paul Smith
