Solve correctly please.Should correct 20) The below energy diagram shows that enzymes catalyze reactions by lowering the activation energy.Uncatalyzed reaction (AG'u) is larger than that for the enzyme-catalyzed reaction (AG*e).AGa) 3.3X1010b) 4.5X109EXIReaction coordinateE+SES EXE+P1.9X10¹2d) 2.5X10¹3e) 4.4X1010AGIf the energy of the ES complex is 15 kJ/mol lower than the energy of the E + S, the value of AG* e' is 20kJ/mol, and the value of AGu u is 90 kJ/mol. What is the rate enhancement achieved by an enzyme in thiscase? Universal gas constant, R = 8.314 x 10-3 kJ mol-1 and temperature, T=25 °C<29

Energy ES complex is 15 kJ/mol lower than the energy of the E + S∆Ge= 20KJ/mol∆Gu= 90KJ/molR = 8.134…

Answered: If the energy of the ES complex is 15…

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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1. You are studying the enzyme catalyzed reaction below, and you find the KM is 3.3x10-4 M,. You also find that k1 is 4.3x106 M-'s-1. What is the dissociation constant (KD) for the enzyme/substrate complex? k1 k2 E+S ES E+P k.1
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]
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 Oa. The reaction is now spontaneous due to the addition of enzyme b. The rate of the enzyme catalyzed reaction is faster than the uncatalyzed reaction O C. The reaction is exergonic O d. The change in free energy for the reaction is greater in the catalyzed reaction, compared to the uncatalyzed reaction e. The enzyme stabilizes the transition state for the reaction Released Energy pes
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.
Comment on Figure 2a and Figure 2b regarding stabilization of the transition state and destabilization of ES complex during catalysis. (b) AG AG EX EX E+S E+P ES EP AG AG,+ AG- TAS ES EP Figure 2
1). The initial velocity of an enzyme-catalyzed reaction has been determined at several substrate concentrations with and without an inhibitor I. The total enzyme concentration [E]ot present in each experiment is 1.20 x 10 M. [1] = 5.00 x10 M. (Note: v initial rate or velocity.) [S], M 1.25E-3 2.50E-3 5.00E-3 10.0E-3 v, M/min, no inhibitor 0.480E-3 0.660E-3 0.810E-3 0.920E-3 v, M/min, 5.00x10 M 9.43E-5 1.41E-4 1.70E-4 2.10E-4 b). What kind of inhibition occurred. c). Calculate the value of Ki = a). Use the Lineweaver-Burk plot to determine the values of Vmax and KM at [1] = 0.00 mM, and the values of Vmax and KM at 0.500 mM of I. it in 'How many molecules of
An inhibitor "I" is added to the enzymatic reaction at a concentration of 1.0 g/L. The data obtained are shown in the table below. No Inhibitor With Inhibitor [S](g/L) v ([E0] = 0,015 g/L) (g/L.min) v ([E0] = 0,015 g/L)(g/L.min) 40 2,28 1,818 20 1,74 1,316 13,4 1,4 0,986 10 1,18 0,76 8 1 0,62 (a)What kind of inhibition occurred?(b)Determine Ki (in g/L) for the inhibition of the previous exercise.
e) Two 20th century scientists, Leonor Michaelis and Maud Leonora Menten, proposed the model known as Michaelis-Menten Kinetics to account enzymatic dynamics. The model serves to explain how an enzyme can cause kinetic rate enhancement of a reaction and explains how reaction rates depends on the concentration of enzyme and substrate. k1 k2 E + S= ES → E + P k-1 Equation 1 i. Prove Equation 2 by referring to Equation 1. Vmax[S] V = [S] + KM Equation 2 ii. Give three characteristic features of catalyst.
a particular enzyme catalyzes a single reactant S to a single product P, following michaelis-menten kinetics rp=(VmaxCs) / (Km + Cs) 1. A reaction with this enzyme is carried out at very low substrate concentrations. Draw and label a curve on the plot that describes the reaction kinetics under those conditions.
An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-s
From the following data of an enzymatic reaction, determine the type of inhibition, the KM for the substrate and Ki for the enzyme-inhibitor complex. [S] mM 2.0 3.0 4.0 10.0 15.0 [P] mM/h (without inhibitor) 139 179 213 313 370 [P] mM/h ([I]=6mM) 88 121 149 257 313
An enzymes catalyzed reaction is studied in the presence and absence of an inhibitor. The following data was obtained in the image provided. Plot 1/[S] as abscissa and 1/V as ordinate for both catalyzed reactions and reaction with inhibitor. Use the same graph for both plots Calculate the following: Km of enzyme in the reaction without inhibitor Km' of the enzyme in the reation with inhibitor Vmax of the uninhibited reaction Vmax of the inhibited reaction What kind of inhibitor was added to the enzyme catalyzed reaction? Explain your answer in terms of changes in Km and Vmax.

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