[S] << Km [S] = = Km [S] >> Km Not true for any of these conditions Almost all active sites are empty. [ES] is much higher than [Efree]. [Efree] is equal to [ES]. Reaction rate is independent of [S. [Efree] is about equal to [Etotai]- Increasing [Etotal] will lower Km-

In an enzyme kinetics study, three inhibitors resulted to the following results: Inhibitor ABC Inhibitor XYZ Inhibitor PQR Without Inhibitor Vmax 40.2 mM/ sec 40.3 mM/ sec 12.32 mM/ sec 65.43 mM/ sec Km 24.3 mM 28.5 mM 24.3 mM 15.7 mM The inhibition occurred at 37.0°C. Assuming that the temperature increased to 37.5°C, what will happen to the Vmax and Km in the chemical system involving Inhibitor XYZ? Why do you say so? Provide illustrations using appropriate dimensions in the x and y axes. C.

In a alkaline phosphatase kinetics experiments a studnet first made a standard curve with different amounts of PNP. In a reaction carried out with 10nM alkaline phosphatase and 100 uM PNPP substrate the student measured an absorbance value of 0.76 at 410nm after 30 seconds of reaction. Assuming the 30 second time point is in the lienar range of the reaction and the substrate has not been depleted at this time point, calculate the Vo at 100uM PNPP (in units of concentration per time).Beer's law formula given is y=0.0323x (y axis is Absorbance) (x Axis is PNP uM)

The total concentration of enzyme in a reaction, [E], is made up of the concentration of enzyme bound to substrate, [ES], and the concentration of enzyme still free in solution, [Ef]. Similarly, the total amount of substrate is made up of [Sf] and [ES]. We can assume that the concentration of enzyme is much less than that of the substrate, [E] << [S]. Assuming the steady state condition and the relationships between [E], [Ef] and [ES], and similar ones for S, given in lectures, derive an expression for the saturation factor,  , in terms of [S] and . (Note that [E] and [S] denote the total amounts of enzyme and substrate added to the reaction, respectively. You may assume that [S]>>[E].)

Which of the following statements are true about the relationships of [S], KM, and Vmax?  (Choose all that are true) As the [S] is increased, vo approaches the limiting value, Vmax KM = Vmax/2 The rate of product formed, vo, is at Vmax when [S] <<< KM KM and Vmax assist in finding the rate of the enzyme catalyzed reaction only if the reaction is considered irreversible.

You are working on an enzyme that obeys standard Michaelis-Menten kinetics. Based on the following reaction expression, what is the Km value for this enzyme? E+SESE + P . . . k₁ = 880.8 M-¹5-1 k.₁ = 42.18 S-1 k₂ = 56.29 S-1

The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 104 M. At a substrate concentration of 0.20 M, v = 43 μmol/min. Calculate the rate of reaction when the substrate concentration is tenfold lower, 0.020M. 43 μmol min rate of reaction → Vmax Км substrate concentration →

From a series of flasks with a constant concentration of enzyme the following initial velocities weretaken, they were obtained as a function of the concentration of the substrate.a) Calculate the KM and Vmax kinetic parameters of the three forms (Lineweaver-Burk, Eadie-Hofstee, Dixon).b) Analyze which are the atypical data that cause a low correlation, which can be eliminated and explain youranswer.

Given the assumption used to derive the Michaelis-Menten equation for enzyme kinetics that the back reaction from P + E back to ES does not occur, why is it important in measuring experimental data for Michaelis-Menten analysis to determine the initial rate of the reaction just after adding substrate to the sample being measured? O Because faster rates can be measured more accurately than slower rates later in the process. Because the Michaelis-Menten equation also contains the position of the equilibrium of the reaction. ● By using this approach, the amount of P at the start of the experiment is extremely low, causing the back reaction to be essentially absent. Because in this way the [ES] concentration has not yet reached steady state.

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 O a. The rate of the enzyme catalyzed reaction is faster than the uncatalyzed reaction O b. The change in free energy for the reaction is greater in the catalyzed reaction, compared to the uncatalyzed reaction O c. The enzyme stabilizes the transition state for the reaction Od. The reaction is exergonic е. The reaction is now spontaneous due to the addition of enzyme Released Energy