Why the Ezyme activity deacreses when the MgCl2 concentration increases to 4mM onwards? Discuss
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Q: An enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The ...
A: Enzyme concentration= 100 nM Vmax value= 25 uM/s Km for substrate = 5.2 uM
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Why the Ezyme activity deacreses when the MgCl2 concentration increases to 4mM onwards? Discuss
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- Draw the Michaelis-Menten Plot and Lineweaver-Burke Plot of an enzyme in thefollowing situation. An enzyme, Enzyme A, has an optimal temperature at 34.5oC. The enzyme kinetic parameters were recorded to be Km = 15.0 μM, Vmax = 12.0 μM/s. The condition for an enzyme kinetic experiment has the temperature raised to 36.5oC. What will happen to the results? Provide at least five data sets prior to raising the temperature and after raising the temperature and plot the graphs as overlaying figures for comparison. Explain youranswer using a minimum of 200 words.If the data from an enzyme experiment is plotted as a Lineweaver-Burk plot, and the Vmax is 0.02 mol/sec, and x-intercept is –2.5 mM then what is the KM value? Show yourwork/reasoning.The following data was obtained during kinetic analysis of an enzyme with and without an inhibitor. Substrate concentration (mM) Reaction rate without inhibitor (µM/s) Reaction rate with inhibitor (µM/s) 10 28 12 20 50 23 40 83 42 60 107 58 100 139 83 200 179 125 300 197 150 400 209 167 560 227 197 How do you calculate the KM for the enzyme in the absence of an inhibitor. And how do you calculate kcat with the given enzymatic concentration of 5 µM.
- At what substrate concentration would an enzyme with a Km of 0.005M operate at one quarter of its Vmax? What would be the Km for an enzyme if it is operating at 90% of its Vmax when the substrate concentration is 0.01MDraw three different Lineweaver-Burke plots for an enzyme in the presence or absence of a (1) competitive inhibitor, (2) uncompetitive inhibitor, (3) noncompetitive inhibitor. Indicate on your graphs: Vmax, max (app), Km and Km (app) for each case.Since KM is an intrinsic property of an enzyme, its value does not depend on the enzyme concentration. True or False
- The following were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: Reaction Velocity (mmol/min) Substrate added (mmol/L) 217 0.8 325 2 433 4 488 6 647 20 652 1000 The Km for this enzyme is approximately _____________. (Round to the nearest integer)An enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The Km for the substrate is 5.2 uM. What is the initial velocity (V0) at a substrate concentration of 15.2 uM? Report your answer to three significant figures in units of uM/s.Given the following set of kinetic data, what is the KM of the enzyme? Numerical answer, only (assuming the same units as seen in the table). [S] (micro M) vo (Micro mol/sec) 1 49 2 96 8 249 50 498 100 656 1000 778 5000 999
- Given the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMAn enzyme-catalyzed reaction has a KM of 5 mM and a Vmax of 60 nM/sec. What is the substrate concentration when the initial velocity of the reaction is 30 nM/sec? Show your work/reasoningQ is an analog of substrate A that binds to enzyme X and produces the following kinetics:[A] V0 (µmole/ml/min) [Q] = 0 [Q] = 0.5 µM [Q] = 2 µM1 µM 10 7 43 µM 20 16 1010 µM 35 31 2330 µM 43 41 3680 µM 47 46 43a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of dataon your graph) and determine the following: the KM and Vmax of the enzyme X in theabsence of Q, and the KMapp the Vmaxapp at each concentration of Q. b) What type of inhibition is this?c) Calculate the inhibition constant (Ki) for Compound Q