If a michaelis-type enzyme has an apparent Km of 2*10^-8 and apparent Vmax of 1*10^-4 moles/min in the presence of 1*10^-3 M uncompetitive inhibitor (Ki=1*10^-7 M), what is the true Km of the enzyme
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A: Given Values: Vmax = 12 u s-1ml-1 Total enzyme concentration = 0.016 U ml-1
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A: The answer is given in the image given below.
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A: Here the graph shows, km value of 0.1mM decreased when compared to 0.05mM. Affinity for the enzyme…
Q: a.) Determine the Km and Vmax b.) Assuming htat the enzyme present in the system had a concentration…
A: The Michaelis- Menten Equation for Enzyme kinetics is - V= Vmax[S]/(km+[S]) By putting the above…
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Q: [Substrate, uM] Vo with DEDS (µM/min) Vo without DEDS (µM/min) 3.333 0.774 1.196 4.000 0.877 1.316…
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Q: Vo with Vo without [Substrate, µM] DEDS DEDS (µM/min) (µM/min) 3.333 0.774 1.196
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A: Vmax km Competitive remains same increases Non Competitive Decreases remains same…
Q: The following data were obtained in a study of an enzyme known to follow Michaelis-Menten Kinetics.…
A: In-order to get an accurate value of KM , we need to plot the Lineweaver Burk (LB) plot. The X and Y…
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A: Hello. Since your question has multiple sub-parts, we will solve first three sub-parts for you. If…
Q: 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the…
A: If Km remains unchanged even after increasing the inhibitor concentration it shows that…
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- Using the ActiveModel for aldose reductase, describe the structure of the TIM barrel motif and the structure and location of the active site.What is the likely keq for an enzyme that has a deltaG =-18.3kj/mol?For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.
- Explain why each of the following data sets from a Lineweaver–Burk plot are not individually ideal for determining KM for an enzymecatalyzed reaction that follows Michaelis–Menten kinetics.Calculate the mass of invertase (in mg) and concentration of invertase (in mM) contained in a 25.0mL sample of yeast extract that has 3,000 total units of activity, assuming that pure invertase has a specific activity of 1,000 units/mg with a mass of 270kD.You obtain a calculated Vmax of 4.26uM/s and a Km of 122.5uM from a kinetics experiment performed using 0.5uM enzyme. What is the catalytic efficiency of this enzyme?
- For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K MAssume that the experiments performed in the absence of inhibitors were conducted by adding 5 μL of a 2 mg/mL enzyme stock solution to an assay mixture with a total volume of 1 mL. Take into account that XYZase is a monomeric enzyme with a molecular mass of 45,000 Daltons. Hint: To calculate the ???? in units of per second (s−1), you must first determine the ???? in micromoles per second (μmol/sec). Please explain step by stepThe KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 × 10−2 M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 × 10−4 M. (a) Which substrate has the higher apparent affi nity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax?
- The enzyme, fumarate, has the following kinetic constants: k 1 k 2 k -1 where k 1 = 10 9 M -1 s -1 k -1 =4.4 x 10 4 s -1 k 2 = 10 3 s -1 a. What is the value of the Michaelis constant for this enzyme? b. At an enzyme concentration of 10 -6 M, what will be the initial rate of product"If the higher value of KM resulting in the new plot ( red curb ) is due to the presence of an enzyme inhibitor is inhibitor reversible or irreversible? And why?For the following aspartase reaction in the presence of the inhibitor hydroxymethylaspartate, determine Km and whether the inhibition is competitive or noncompetitive. You have to plot thegraph on the graph paper and also by using excel.[S] V, No Inhibitor V, Inhibitor Present(molarity) (arbitrary units) (same arbitrary units) 1 x 10-4 0.026 0.0105 x 10-4 0.092 0.0401.5 x 10-3 0.136 0.0862.5 x 10-3 0.150 0.1205 x 10-3 0.165 0.142