Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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The effectiveness of allosteric effectors in regulating
interact with multiple substrate-binding sites on the target enzyme(s). |
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interfere with competitive inhibitors. |
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alter the concentration of the target enzyme(s). |
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denature the target enzyme(s). |
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change the conformation of the target enzyme(s). |
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- For each of the statements below about the dUTPase enzyme, mark whether it is true or false. If it is false, change the language to make the statement true. _______If the dUTPase enzyme is the rate determining step in a larger metabolic pathway, the reaction likely lies near to equilibrium and is irreversible. _______For the dUTPase reaction, it is not likely that DG’ = DG°’ _______ dUTPase has a higher affinity for its substrate than for its product, but both of these are higher than the affinity for the transition state _______Kinetic analysis of dUTPase using Michaelis-Menten methods assumes that the formation of the E*S complex has a large negative value for DGarrow_forwardA newly developed drug is suspected to act by inhibiting its enzymatic target. To test this, the rates of reaction in the presence and absence of the new drug were determined. The results of these experiments have been plotted as a double reciprocal plot shown in figure 1 below. Using this figure, calculate the values of Vmax and Km for the enzyme in the presence and absence of the drug. 2. Based on the figure and your answer, what type of inhibition is the new drug displaying? 3. Briefly explain how this type of inhibition exerts its action on an enzyme. Figure 1: Double reciprocal plot of reaction rate against substrate concentration in the presence and absence of a new drug.arrow_forwardYou design an enzyme assay and choose a substrate concentration equal to the Km of the enzymatic reaction. You measure the rate of the reaction at this substrate concentration to be 75.0 μmol/min. Calculate the Vmax of this reaction in μmol/min. Express your answer with one decimal place. Your previous attempt suffered from low signal-to-noise ratio when you used a substrate concentration equal to the Km. So you decide to increase the substrate concentration to 0.45 mM. You remeasure the rate of the reaction at this new substrate concentration and find it to be 135.0 μmol/min. Using the same Vmax value found above, calculate the Km value of this enzymatic reaction. Express your answer in μM with one decimal place. Considering the previous two questions, what would be the rate of the reaction at a substrate concentration of 0.01 mM? Express your answer in μmol/min with one decimal place.arrow_forward
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