For an enzyme with a substrate Km of 0.001 M. a) What substrate concentration for an initial rate equal to one-half maximaly velocity? and b) for an initial rate equal to 0.9 maximal velocity> *         a) [S} = 0.001 M, b) [S] = 0.009 M       a) [S} = Km, b) [S] = 0.09 M       a) [S] = 0.01 M, b) [S] = 0.9 M       a) [S] = 1 M, b) [S] = 9 M

For some Enzyme, the Vmax is 18 micromols/min, Km is 400 microM. If the substrate concentration is 100 microM, what is the velocity of the reaction?

FAD is a coenzyme for dehydrogenation.(a) When a molecule is dehydrogenated, is FAD oxidizedor reduced?(b) Is FAD an oxidizing agent or a reducing agent?(c) What type of substrate is FAD associated with, and whatis the type of product molecule after dehydrogenation?(d) What is the form of FAD after dehydrogenation?(e) Use the curved-arrow symbolism to write a generalequation for a reaction involving FAD.

I need help with parts d-g  a.) For Ado: catalytic efficiency = 4.18 x 106 M-1s-1 For ATP: catalytic efficiency = 12.1 x 107 M-1s-1 b.) Both specificity constants are below the diffusion limit so neither substrate is limiting the reaction rate due to diffusion c.) Line weaver Burk equation for AdoK enzyme with Ado substrate: y = 0.68x + (2.04 x 104), where y = 1/vo and x = 1/[Ado]

Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the above enzyme (lactase). (enzyme lactase has a Vo of 0.111111111111 mM per minute when [S] = 1.0 mM, and a Vo of 0.20 mM per minute when [S] = 5.0 mM) 0.125 per minute 0.25 per minute 0.50 per minute 1.25 per minute 5.0 per minute

You run an experiment using 0.5 µM of an enzyme. The kinetic parameters for your enzyme are kcat = 1800 s-1 and KM = 25 µM. (a) If k-1 is 240 s -1, what is the value of k1? (Report your answer in units of M-1 s -1) (b) What is Vmax for the enzyme (Don't forget to include units in your answer)? (c) You set up an experiment with the following initial concentrations of substrate. Calculate the rate of the reaction at each substrate concentration (report your answer in units of µM sec-1 [S](μM) 1 100 1000 [S][μM) 1 100 1000 v (μMsec ¹) (d) You now add 10 nM of a competitive inhibitor with a KI = 2 nM. What is the rate of the reaction (v) for the following substrate concentrations and how does it compare to the rate without inhibitor (VO)? Vinhib (uMsec-¹) Vinhih/V

Please answer in full.

You begin to study enzyme Z, which catalyzes a simple reversible reaction that interconverts compound S and compound P. You observe that the ∆G´° for the S to P conversion to be –6 kJ/mol, and that compound S has ∆G´° for binding to enzyme Z of –15 kJ/mol, while compound P has a ∆G´° for binding to enzyme Z of –13 kJ/mol. Please explain the effect of enzyme Z on conversion of S to P. (Your answer should include a graph qualitatively  showing energy versus reaction progress; however, you still need to explain youranswer in words!) not sure how to make the correct graph.

The initial velocities of two different enzyme-catalyzed reactions were measured over a series of substrate concentrations. The following results were obtained: Enyme A: KM = 1.5 mM, Vmax = 10 μM s-1 Enyme B: KM = 5.0 mM, Vmax = 85 µM s-1 (a) Which enzyme binds to its substrate more tightly (assume k.1 >> k₂ in the Michaelis-Menten model)? (b) Calculate the initial velocities of each reaction when the substrate concentration is 2.5 mM. (c) Calculate the Kcat of each enzyme if the total enzyme concentration is 100 nM. (d) Which enzyme is the more efficient catalyst? Explain your answer. The enzyme carbonic anhydrase is strongly inhibited by the drug acetazolamide. A plot of the initial reaction velocity (as a percentage of Vmax) in the absence and presence of the inhibitor is shown below. What type of inhibition is taking place? Explain your reasoning. V (% of Vmax) 100 50 0.2 0.4 No inhibitor Acetazolamide [S] (MM) 0.6 0.8 1

What percentage of max is obtained when the substrate is present at 80% of the Km? Use two digits in your answer.