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The constant, 0 v/v % hydrogen peroxide solution, did not react with the catalase, therefore it was recorded as 300+ seconds and had a reaction rate of 0 seconds-1. The 0.1 v/v % hydrogen peroxide concentration recorded a reaction rate of 0.17 seconds-1. The 0.2 v/v % hydrogen peroxide concentration, recorded a lower reaction rate than the 0.1 v/v% at 0.14 seconds-1. The 1 v/v % hydrogen peroxide concentration recorded higher at 0.49 seconds-1. The 3 v/v % hydrogen peroxide concentration recorded the highest reaction rate at 1.57 seconds-1. The 5 v/v % hydrogen peroxide concentration recorded a lower reaction rate than the 3 v/v % at 1.00 seconds-1. The trendline for the data is polynomial increasing to the highest value (1.57 seconds-1) …show more content…
This is due to the absence of the substrate. If the hydrogen peroxide was not present then there were no molecules which fit the catalase enzyme and as such no reaction can take place. The reaction rate between the catalase and the hydrogen peroxide appeared to increase with the amount of available substrate (hydrogen peroxide).
This is because catalase is capable of breaking down 40 million hydrogen peroxide molecules for every 1 catalase molecule and the more molecules of substrate present in the solution means that there are more molecules to fit the enzyme. The reaction results in more products being created. The products for this reaction were water and oxygen. The more oxygen molecules that are created, the more there are to travel to the surface. It is when these molecules get trapped underneath the paper disc that it is lifted to the surface by the …show more content…
The yeast consumes the sugar and turns it into energy and carbon dioxide. Over time the sugar would reduce as the yeast consumes it. The yeast over time may have had less energy and therefore behaved differently. This change in behaviour may have had an effect on the catalase enzyme by perhaps reducing the amount which would decrease the reaction rate.
The hydrogen peroxide concentrations 1, 3, and 5 v/v % were contained in a different beaker. The two beakers used were both 50 mL beakers, however, one was plastic and one was glass. The plastic beaker had a different shape and may have had a different diameter. The variation in diameter would result in a different height of the solution. The change in height would affect the travel time it takes for the disc to travel to the surface because the surface is further away.
The catalase solution should be placed in the 30°C warm water bath for the duration of the experiment to keep the catalase solution at a constant 30°C. Therefore mitigating the error of catalase solution temperature reduction. This keeps the enzyme at the same temperature and allows it to function the
The constant, 0 v/v % hydrogen peroxide solution, did not react with the catalase, therefore it was recorded as 300+ seconds and had a reaction rate of 0 seconds-1. The 0.1 v/v % hydrogen peroxide concentration recorded a reaction rate of 0.17 seconds-1. The 0.2 v/v % hydrogen peroxide concentration, recorded a lower reaction rate than the 0.1 v/v% at 0.14 seconds-1. The 1 v/v % hydrogen peroxide concentration recorded higher at 0.49 seconds-1. The 3 v/v % hydrogen peroxide concentration recorded the highest reaction rate at 1.57 seconds-1. The 5 v/v % hydrogen peroxide concentration recorded a lower reaction rate than the 3 v/v % at 1.00 seconds-1. The trendline for the data is polynomial increasing to the highest value (1.57 seconds-1) …show more content…
This is due to the absence of the substrate. If the hydrogen peroxide was not present then there were no molecules which fit the catalase enzyme and as such no reaction can take place. The reaction rate between the catalase and the hydrogen peroxide appeared to increase with the amount of available substrate (hydrogen peroxide).
This is because catalase is capable of breaking down 40 million hydrogen peroxide molecules for every 1 catalase molecule and the more molecules of substrate present in the solution means that there are more molecules to fit the enzyme. The reaction results in more products being created. The products for this reaction were water and oxygen. The more oxygen molecules that are created, the more there are to travel to the surface. It is when these molecules get trapped underneath the paper disc that it is lifted to the surface by the …show more content…
The yeast consumes the sugar and turns it into energy and carbon dioxide. Over time the sugar would reduce as the yeast consumes it. The yeast over time may have had less energy and therefore behaved differently. This change in behaviour may have had an effect on the catalase enzyme by perhaps reducing the amount which would decrease the reaction rate.
The hydrogen peroxide concentrations 1, 3, and 5 v/v % were contained in a different beaker. The two beakers used were both 50 mL beakers, however, one was plastic and one was glass. The plastic beaker had a different shape and may have had a different diameter. The variation in diameter would result in a different height of the solution. The change in height would affect the travel time it takes for the disc to travel to the surface because the surface is further away.
The catalase solution should be placed in the 30°C warm water bath for the duration of the experiment to keep the catalase solution at a constant 30°C. Therefore mitigating the error of catalase solution temperature reduction. This keeps the enzyme at the same temperature and allows it to function the