An enzyme is present at a concentration of 1 nM and has a Vmax of 2 µM s-'. The Km for its primary substrate is 4 µM.Calculate kcat-kcatCalculate the apparent Vmax and apparent Km of this enzyme in the presence of sufficient amounts of an uncompetitive inhibitorto generate an a' of 2. Assume that the enzyme concentration remains at 1 nM.apparent VmaxµM s-1apparent Km =µM

K cat is nothing but turnover number of enzyme which is equal to number of substrate molecules…

Answered: An enzyme is present at a concentration…

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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An enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.
A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.
Staphylococcal nuclease has a ΔΔG‡ of -84.1 kJ mol-1 at 25.0 °C. If the uncatalyzed rate is 0.630x10-13 µmol s-1, calculate the enzyme-catalyzed rate in µmol s-1. (Use R = 8.3145 J mol-1 K-1)
The equil ibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200.In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1.What is the rate constant for the loss of the unreacted substrate from the active site?
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
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
you are trying to come up with a drug to inhibit the activity of an enzyme thuogth to have a role in a liver disease. in the lab the enzyme was shown to have a Km of 1x10-6 M and Vmax of 0.1 micromoles/min.mg measruing at room temperature. you developed a mixed non-competitve inhibitor with a ki=0.4x10-6M and a Ki` pf 0/2 x 10-5 What will be the apparent Km in the presence of 1.0x10-6 M?
Consider an enzyme that follows standard Michaelis-Menten kinetics and has the following kinetic constants: %3| k2 = 1.5 x 10? s1 Еo 3D 1 х 104 М = 1 x 104 M a. What is the value of the maximum rate VM? b. Prepare a hand-drawn quantitative plot on graph paper (not a simple sketch nor an EXCEL-generated graph) of the enzymatic reaction rate versus substrate concentration (like Fig 3-3) using the kinetic parameters given above. Be sure to label the axes and include numeric values on the axes. C. Based upon your hand drawn saturation plot, at what substrate concentration is the enzymatic reaction rate 75% of Vm?
The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.
Vmax [S] Km + [S] Vo = Eadie-Hofstee plot Lineweaver-Burk (L-B) plot v=Vm-Km [S] Km 1 Vm Vm [S] 1 The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For uncompetitive inhibition The Michaelis-Menten equation becomes Vo-Vmax(S)/(Km+a'[S)) Put the Minchaelis-Menten equation for uncompetitive inhibition in the Lineweaver-Burk format. The slope of this plot will be
when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.
Shown below are Km, and Vmax values obtained for an enzyme A which catalyze the transformation of the following substrates. Enzyme concentration used was 0.01 M. Km, mM 0.02 Vmax, mM/min 5.3 Substrate 1 2 1.5 13.7 3 2.6 100 4 0.1 25 0.05 62 1. Which substrate have the highest affinity for the enzyme? Explain. 2. Which will show higher efficiency of converting the substrate to the product? Show solutions and еxplain.

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