The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V₁ for anenzyme-catalyzed, single-substrate reaction E + S ⇒ ES →→ E + P. The model can be more readily understood whencomparing three conditions: [S] << Km, [S] = Km, and [S] >> Km.Match each statement with the condition that it describes.Note that "rate" refers to initial velocity Vo where steady state conditions are assumed. [Etotal] refers to the total enzymeconcentration and [Efree] refers to the concentration of free enzyme.[S] << Km[ES] is much lower than [Efree].The rate is directly proportional to [S].[S] = Km[Efree] is equal to [ES].[S] >> KmAlmost all active sites willbe filled.Adding more S will not increasethe rate.Answer BankNot true for any of theseconditionsIncreasing [Etotal] will lower Km.

Those proteins or biological catalysts which help to speed up the chemical reaction are termed…

Answered: The Michaelis-Menten equation models…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate V% for an enzyme-catalyzed, single-substrate reaction E + S=ES → E + P. The model can be more readily understood when comparing three conditions: [S] > Km- Match each statement with the condition that it describes. Note that "rate" refers to initial velocity Vo where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km Not true for any of these conditions [ES] is much lower than [Efree]. Reaction rate is independent of Increasing [Etotal] will lower Almost all active sites will Km- be filled. [S). [Efree] is about equal to [Etotal]. Show All W- 5179933 (3).docx 5179933 (4).docx PCR-MINI RES....docx MacBook Pro
Consider this intermediate in the derivation of the Michaelis-Menten equation. [E] [S] [ES| k-1 + kz km Assume that k is negligible compared to the other rate constants. If the k is very small, it suggests that the enzyme has a Select an option affinity for its substrate, while if the if the km is very large, it suggests that the enzyme has a Select an option. affinity for its substrate. Select an option Submit You have used 0 of high Sav low moderate
The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
For the following two scenarios, sketch the complete reaction free energy diagram for an enzyme-catalyzed conversion of a single substrate (S) into product (P), where the reaction is spontaneous in the forward direction. For each of these, overlay the free energy diagram for the uncatalyzed reaction and indicate AAG" on your sketch: a). Substrate binding is rate limiting b). The chemical step is rate limiting
Can you make a graph in excel
The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. Vmax [S] Km + [S] where v is the velocity, or rate, Vmax is the maximum velocity. K is the Michaelis-Menten constant, and [S] is the substrate concentration. A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (ro) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for Vmax and Km- Vmax= µM/min v (µM/min) 300 275 250 225 200 175 150 125 100 75 50 Km = 25 K 0 10 20 30 V max 40 50 [S] (M) 1/2 V max Michaelis-Menten curve 60 70 80 90 100 HM
Given the following reaction and equation for the initial velocity of the reaction: k₁ k3 E+S ES E + P V=Keat [ES] = k3 [ES] k₂ where keat is the rate constant for the reaction which forms the product from the ES complex. Explain in words why the velocity is directly proportional to theamount of enzyme added in the presence of saturating substrate levels.
k_1 E + SES E + P +0 k₂ S ↓↑ KIS ESS Based on this model, please answer the following questions: i) This is the model for what? Uncompet tive ii) Show the rate equation for an enzyme reaction of this type. iii) Show the Lineweaver-Burk plot for this case. A
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
Consider the enzyme-catalyzed conversion of a substrate S into a product in the presence of an inhibitor I with the following properties (mixed inhibition): (i) The inhibitor competes with S to bind the active site of the enzyme with dissoci- ation constant K¡ for the complex (EI). (ii) The inhibitor is also capable of binding the enzyme at a secondary site of the enzyme with dissociation constant K{ for the complex (IES). The complex (IES) is unable to form P. iii) The dissociation constants of the (IES) and (EI) complexes satisfy the relation K{ = K1/2. What is the value of the ratio vo/(k2 [E]o) when the concentration of the substrate is [S]o = KM/3 and the concentration of the inhibitor is [I]o = K{/4? Recall that KM is the Michaelis constant.
You will perform the protocol below for the calf intestinal alkaline phosphatase (CIP) provided. For each reaction, your final enzyme concentration should be 10 nM CIP. Note: Enzymes purchased are typically labelled with their “units of activity” (U), as this relates to how much enzyme is needed to catalyze a reaction. The 100 nM CIP provided has approximately 3 U/mL and was diluted 1 in 1,000 from a 500 U/mL purchased enzyme. 1) Create a table (similar to the one below) to help you determine and keep track of what to add to each of the cuvettes in which your reactions will be measured. The five different concentrations of PNPP should be: 25, 50, 100, 200, 300 μM. Each reaction will be in a final volume of 1 mL and contain 10 nM alkaline phosphatase. Concentrations of stock solutions: 1.0 mM PNPP, 100 nM calf intestinal phosphatase
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.

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