Consider a reaction in a cell that is catalyzed by the enzyme hypotheticalase. Under normal cellular conditions, this reaction has AG < 0. You add a drug that causes the cell to produce 1,000× more hypotheticalase. How will this change AG? Please justify your answer.
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- The purified OXA-M290 enzyme can now be tested to determine which β-lactamase inhibitor is most effective. This inhibitor could be prescribed in combination with a β-lactam antibiotic to treat the infection caused by the E. coli KGH1 strain. Before testing inhibitors against OXA-M290, the kinetic activity of this enzyme must first be measured. The activity of OXA-M290 is measured using nitrocefin, a chromogenic β-lactam antibiotic. When nitrocefin is hydrolyzed by a β-lactamase, it changes from yellow to red in colour. The nitrocefin hydrolysis product has an extinction coefficient of 20,500 M-1 cm-1 at 486 nm. The hydrolysis of 60 μM nitrocefin by 1 nM OXA-M290 is monitored using a microplate reader. The absorbance of the wells in the plate is measured at 486 nm every 30 seconds. This experiment is carried out with three replicates, generating the following data: Time (min) Absorbance of Replicate 1 Absorbance of Replicate 2 Absorbance of Replicate 3 0.5 0.0984…Select the graph that correctly illustrates the effect of a positive modifier (effector) on the velocity curve of an allosteric enzyme. Place the correct graph in the set of axes. The solid blue curve represents the unmodified enzyme. The dashed green curve represents the enzyme in the presence of the effector. R is the highly active form of the enzyme and T is the less active form of the enzyme. Assume that this is a positively cooperative enzyme, meaning that the affinity for substrate increases with increasing substrate concentration.Data from enzyme inhibition are used to determine a Kmapp and Vmax PP. Comparison of these values with assays run without inhibitor are used to understand how the inhibition is occurring. This is useful for better understanding the active site as well as the practical aspect of pharmaceutical drugs. Below are idealized Line-Weaver Burke plots of different types of inhibitors. Comnetitive Uncomnetitive Mixed +Inh +Inh 4Inh Anh Inh Anh [S] [S] [S] a. How does the value of Vmax for the enzyme compare to the Vmax PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed b. How does the value of Km for the enzyme compare to the Km PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed c. For each situation in Model 1, consider an inhibitor that is better than the one shown on the graph. Answer the following questions for each type of inhibition: i. How would the KmPP change? ii. How would the Vmax PP change?
- J. C. Servaites, in Plant Physiol. (1985) 78:839–843, observed that Rubisco from tobacco leaves collected before dawn had a much lower specific activity than the enzyme collected at noon. This difference persisted despite extensive dialysis, gel filtration, or heat treatment. However, precipitation of the predawn enzyme by 50% (NH4)2SO4 restored the specific activity to the level of the noon-collected enzyme. Suggest an explanation.In enzyme catalysed reactions, the energy level of the enzyme/substrate (or ES) complex is higher (or raised) compared to the uncatalyzed reaction. List 4 factors that contribute to this raised energy level and explain how each of these factors contribute to the higher energy level of the ES complexThe enzyme urease increases the rate of urea hydrolysis at pH 8.0 and 20 °C by a factor of 10¹4. Suppose that a given quantity of urease can completely hydrolyze a given quantity of urea in 3 minutes at pH 8.0 and 20 °C. How long would it take for this amount of urea to be hydrolyzed in the absence of urease at the same temperature and pH in sterile conditions? time uncatalyzed = years
- You generate mutants in the metabolic pathway for starlase. You conduct some complementation tests (after testing for dominance of course) and come up with the following results: 1 2 4 6 1 + + + 2 + 3 + 4 + 5 6 a. How many complementation groups are there? [Select] b. You conduct some additional experiments to elucidate the starlase metabolic pathway. Your results are shown below. Use this information alongside information from the complementation table above to place the intermediates in the correct order on the pathway. (HINT: use the complementation groups from the table above to help you consolidate information on the tables below. Reference practice question 3 from today's lecture for help). Addition to minimal medium Mutant None starlase P 1 + + 2 + + 4 + + 5 + 6 Precursor --> [ Select ] [ Select ] [ Select ] --> starlase c. Mutant 4 has a loss-of-function mutation for which enzyme in the starlase synthesis pathway? [ Select ] E1 E2 ЕЗ Е4 Precursor > Intermediate 1→ Intermediate…Hello, can someone please help me answer 1 and 2 while focusing on the interaction that the enzyme, the substrate and the product have. Thank youA 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.
- Is the following statement correct or incorrect? It is only possible to employ enzymes that are present naturally in yeast or bacteria for biocatalysis.The figure below shows the dependence of the enzyme's rate, v (in µM/min), as a function of substrate concentration, S (in mM). Also shown is the dependence of the rate in the presence of an inhibitor, present at a concentration of 0.2 mM. Based on this information, which of the following does this inhibitor most likely interact with? 1/v 1- 05 1/[S) O A. Michaelis complex O B. [E O C. free enzyme O D. free substrate O E. Both "A" and "B."Consider the following free energy diagram for an uncatalyzed and enzyme-catalyzed reaction. Select all the statements that are true. Without enzyme With enzyme A+B Time AB Oa. The reaction is now spontaneous due to the addition of enzyme b. The rate of the enzyme catalyzed reaction is faster than the uncatalyzed reaction O C. The reaction is exergonic O d. The change in free energy for the reaction is greater in the catalyzed reaction, compared to the uncatalyzed reaction e. The enzyme stabilizes the transition state for the reaction Released Energy pes