Glucose Concentation Experiment (Lab Report)

And finally into test tube 3, I pipetted 1.0 ml turnip extract and 4.0 ml of water. The contents of test tube 1 was poured into a spectrometer tube and labeled it “B” for blank. “B” tube was now inserted it into the spectrometer. An adjustment to the control knob was made to zero the absorbance reading on the spectrometer since one cannot continue the experiment if the spectrometer is not zeroed. A combination of two people and a stop watch was now needed to not only record the time of the reaction, but to mix the reagents in a precise and accurate manner. As my partner recorded the time, I quickly poured tube 3 into tube 2. I then poured tube 2 into the experiment spectrometer tube labeled “E” and inserted it into the spectrometer. A partner then recorded the absorbance reading for every 20 seconds for a total of 120 seconds. After the experiment, a brown color in the tube should be observed to indicate the reaction was carried out. Using sterile techniques, any excess liquid left was disposed

The hypothesis is that catalase activity will increase exponentially with higher concentrations of hydrogen peroxide until all catalase active sites are filled, in which case the

at that temperature, there was the most amount of catalase activity. On the pH graph, the

The null hypothesis will be that the test tubes with an increase in temperature, pH values, enzyme concentrations, and substrate concentration will have a very small color change or no color change at all. The alternate hypothesis is that the test tubes containing an increase in temperature, pH values, enzyme concentrations, and substrate concentration will all have an intense color change; the more the change, the more intense the color change will be.

The topic of this lab is on biochemistry.This experiment was conducted to show how cells prevent the build of hydrogen peroxide in tissues. My group consisted of Lekha, Ruth, and Jason. There were used two different concentrations of hydrogen peroxide through this experiment , 1.5% and 3%. By testing two different types it is easier to understand how the H2O2 and catalase react with one another. To do this both the yeast, which was our catalase, and H2O2 were mixed together in a beaker. Each concentration was tested out twice for more accurate results . 1.5% concentrated H2O2 had an average reaction rate of 10.5 seconds while 3% concentrated H2O2 had an average reaction rate of 7.5 seconds. From this experiment we learned that by increasing the concentration of H2O2 and chemically combining it with a catalase it will speed up the reaction. Enzymes speed up chemical reactions . The independent variable in this experiment was the concentration of the H2O2. Some key vocabulary words are Catalase, enzyme, hydrogen peroxide ( H2O2), and concentration.

Tables 2,3,4,5 and 6 show that as duration increased the absorbance also increased for each pH. The solution in the conical flask became darker (yellow) in time this is because the substrate, p-nitrophenyl phosphate was catalysed by acid phosphatase, releasing Nitrophenolate anion. It was the Nitrophenolate anion giving off the yellow colour; the presence of this feature increases the absorbance rate. The addition of sodium hydroxide distorted the shape of the enzyme making it no longer effective in its function.

Time) because it had a correlation closest to 1. All three orders were graphed and a linear regression was used to see which graphed order was closest to 1. The order was determined by comparing the concentration and time to the mathematical predictions made using the integrated rate laws. Analyzing each graph and finding each correlation helped determine which graph was closest to 1. The more concentrated a solution is, the higher the absorbance of that solution. This is due to Beer’s Law. The law measures the absorbance of a solution by determining how much light passes through a solution. As the concentration of a solution increases, fewer wavelengths of light are able to pass through the concentrated solution. The absorbance at 60 seconds was 0.573 (Figure 1: Table1). To calculate the concentration (molarity), the Beer’s Law equation was used, Abs = slope(m)+b. Plugging in what is known into the Beer’s Law equation resulted in 0.573 = 3.172e+004 + 0, where the concentration is determined by M = 0.573-0/ 3.172e+004. So, the concentration at 60 seconds using the equation (M = 0.573-0 / 3.172e+004) was 1.824e-5 M. The 1st order graph resulted in k=0.006152 (Figure 1: Graph 1). Other groups also resulted in their decolorization of CV to be the 1st rate

6. In 1 minute intervals, we observed the color changes of the reactions and compared them to the color palette. We did this for 5 minutes.

The reaction will be complete after approximately 60 seconds. Observe where color develops, and consider what these results indicate. Record your results in Table 1.

Figure one depicts the reaction rate of peroxidase enzyme over time. The y-axis shows the absorbance of the assay solutions, and the x-axis depicts time it took for the reaction to occur in seconds.

In data table 1, record the color and the results from the glucose test. Save the test tube to use as a comparison.

Turnips and horse radish roots are rich source of this enzyme. In this experiment, we would carry out a reaction between hydrogen peroxide and guaiacol which is colorless dye, using peroxidase as a catalyst, to produce water and an oxidized form of guaiacol which is brown. The formation of brown color would serve as an indicator that the breakdown of Hydrogen Peroxide took place. The enzyme activity would be directly proportional to the brown color intensity. The color intensity would be measured using a spectrophotometer and standardized to find the corresponding concentration for each absorbance unit.

All values on graph are just shown by height representing amount ignore the y-axis numbers, water having no CO2, glucose having the most, sucrose second most and fructose the least.

The purpose of this lab was to study some characteristics of enzyme action by investigating how the enzyme catalase decomposes hydrogen peroxide. It connects to our unit about enzymes because it illustrates how enzymes decompose their substrates and their rate of decomposition. In the graph of the more accurate experiment, it shows that the amount of product (in mL) gradually increased over time until at about 120 seconds. After that, it would just gradually begin to plateau meaning enzyme activity would no longer increase. Overall, both sets of data show that the amount of H2O2 decomposed gradually increased over time, and after a certain time, the number would stop climbing. The purpose of the potassium permanganate, or KMnO4, was to titrate the H2O2 in the solution, which we could then use to figure out how much of it was decomposed.

The results support the hypothesis by revealing that there is a correlation between the higher the concentration of Peroxidase, the faster the Enzyme Activity rate is. The more intense the color is, the faster the Enzyme Activity rate is, and base on the graph, since one minute of the experiment, there is already differences in color intensity between all three groups. Group two with the highest concentration of 3.0mL, during the first minute was at the color intensity of four, while the control group with the second highest concentration of 1.5mL was at the color intensity of two, and group one with the lowest concentration of 0.75mL was only at the color intensity of one. As time continue, all three groups rise in color intensity; however,