Turnip Peroxidase Lab Report PDF

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Turnip Peroxidase Lab Report...

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Post-Lab Questions Part B: Following the Activity of Turnip Peroxidase 1. Explain the control in this experiment (the what and why).  Tube 1 (the blank) was used as the control in this experiment because hydrogen peroxide (the substrate) was not in this tube. The Spectronic 200 can only be zeroed when no reaction is taking place in the control test tube. In this case, Tube 1 had turnip extract and no hydrogen peroxide. If hydrogen peroxide was present it would have bind to the active sites of turnip peroxidase (an enzyme found in the extract) and a reaction would have occurred. 2. Can you be absolutely sure that you are testing peroxidase? What would you need to do to be sure turnip peroxidase is responsible for the color change of guaiacol and not some other turnip enzyme?  No, you cannot be absolutely sure that you are testing peroxidase.  A brown product called tetraguaiacol is produced when hydrogen peroxide is reduced by the turnip peroxidase while guaiacol is oxidized. The formation of tetraguaiacol can be measured by following the absorbance at a wavelength of 470 nm. 3. What volume of turnip extract (or range of volumes) yielded a linear curve for your turnip peroxidase activity?  200 μl yielded a linear curve for the turnip peroxidase activity. 4. How might you determine how much peroxidase is ACTUALLY in the volume of extract used to create your linear curve? Think about the experiments you might need to do. Refer to your biochemistry textbook if you need ideas.  Determine the amount of substrate needed to reach maximum saturation for the enzyme (peroxidase) by adding certain amounts of substrate and immediately measuring equilibrium after addition. The concentration of peroxidase can be determined by varying the substrate concentration until a suitable concentration is found.

Part C: Varying the Amount of Enzyme in the Reaction 1. Does the reaction go to completion for any of the enzyme concentrations (absorbance reaches a maximum threshold)?  All reactions go to completion except for the enzyme concentration of 200 μl. 2. Calculate the ∆A/min over a time period where the assay is linear for each level of H2O2 tested.  ∆A/min was not able to be calculated because the data and graphs were taken on the lab computer. Also, no graphs for Parts B and C were made because data from other groups was not recorded. However, all values should be less than 1.0. If the post-lab questions were read before conducting the experiment data collected from every lab group for the graph would have been recorded. 3. Are all the turnip peroxidase active sites saturated in any of the reactions? Which one and how do you know?  The enzyme concentration of 200μl because the absorbance increased slowly and then leveled off. On the other hand, the absorbances of other enzyme concentrations were not constant and did not level off. Part D: Varying the Amount of Substrate in the Reaction 1. What conclusions can you draw from your data about the impact of substrate concentration on reaction rate?

Increasing substrate concentration increases the reaction rate until the enzyme is fully saturated. When maximum threshold is reached the reaction rate cannot be increased any further. 2. At what point does increasing the substrate concentration no longer have an effect on the reaction rate? Why is this so?  Increasing the substrate concentration no longer has an effect on the reaction rate once the enzyme is fully saturated. The reason is because there are no more available active sites for anymore substrates and as such the reaction rate has reached its maximum. 3. Using the Michaelis-Menten curve you created, what is the Vmax? What is the KM?  The Vmax is about 14.0 μmol/min and the KM is about 0.001. 4. Using the Lineweaver-Burke plot you created, what is the slope of each line? What is the KM? What is the Vmax? The slope of the line can also be expressed as (slope = KM/Vmax). Substitute your value of the slope for one of your experimental conditions. Plug in your estimate of the KM for that condition based on where your line crosses the x-axis. Do you get the same value of Vmax this way as you do by estimating where the line crosses the yaxis? Why might your values be a little different?  The slope is 0.00006199, the KM is 0.000843, and the Vmax is 13.6 μmol/min. The calculated Vmax based on the instructions in the question is 13.598 μmol/min and this value is extremely close to the Vmax acquired from the Lineweaver-Burke plot. Since the percent error is very small Part D was carried out efficiently with little mistakes incurred. 

Part E: The Effect of Temperature on the Reaction Rate 1. What can you conclude about the effect of temperature on the rate of turnip peroxidase activity?  As the temperature is increased the reaction rate is also increased. However, once the temperature is increased to a very high value (such as 100 °C) the protein (peroxidase) will denature and the rate will become zero. 2. What effect might some of your temperature treatments (such as freezing or boiling) have on enzyme structure?  A freezing temperature wouldn’t affect enzyme structure; however, a very low temperature in turns means less kinetic energy and less substrate binding to the enzyme. As such, the reaction rate would be low at low temperatures. On the other hand, at a boiling temperature the enzyme would denature and the reaction rate will become zero. 3. How would you determine the ideal temperature for turnip peroxidase activity? Can you tell from your data the range where the ideal temperature for this enzyme may be? Explain.  Based on the data, room temperature (about 21 °C) would be the ideal temperature for the turnip peroxidase because only at this temperature is the maximum reaction threshold reached. Furthermore, turnips are raised at room temperature so this also supports the aforementioned notion.

Part F: The Effect of pH on Reaction Rate 1. Why must the blank be changed each time a new buffer is used?  The blank must be changed each time a new buffer is used because each buffer has its own unique absorbance. Not changing the blank will screw the data collected. 2. How does pH affect the rate of the reaction?

Low and high pH are far from being optimal conditions for enzyme activity because reaction rates are close to zero (very little absorbance). pH 7 is the best pH based on the data for enzyme activity. 3. How would you determine the optimum pH for turnip peroxidase activity?  The optimum pH for turnip peroxidase activity is when the reaction rate is maximized. 

Part G: The Effect of Hydroxylamine, an Inhibitor of Peroxidase, on Reaction Rate 1. Explain the action of hydroxylamine on the enzyme’s activity.  Hydroxylamine is a competitive inhibitor which competes with hydrogen peroxide for binding to peroxidase’s active sites. 2. What is the KM and the Vmax with and without inhibitor.  With hydroxylamine, the KM is 0.0126178 and the Vmax is 46.0 μmol/min. A reaction without hydroxylamine to set the inhibition rate at 50% was attempted but not able to be achieved. Other groups were also not able to achieve 50% inhibition without hydroxylamine. 3. What type of competition is this? Competitive, Uncompetitive, Mixed, Mixed Noncompetitive?  This is competitive inhibition because the line intersects the y-axis. Had the reaction without hydroxylamine been done the line would intersect the inhibitor line to imitate Figure 7 on page 122 of the lab manual (a figure of competitive inhibition because the two lines would be not parallel)....