Please type answer not write by hend The activation energy of an uncatalyzed reaction is91 kJ/mol. The addition of a catalyst lowers theactivation energy to 57 kJ/mol.Part AAssuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at25 °C?Express the ratio to two significant digits.ke/kuSubmitPart B=ke/kuΠΑΣΦ=Request AnswerAssuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at134 °C?Express the ratio to two significant digits.Review | Constants | Periodic TableVE ΑΣΦ?

The rate of a chemical reaction is generally given by the Arrhenius equation:k = A *…

Answered: The activation energy of an uncatalyzed…

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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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.
Urease enzyme hydrolysed urea at [S]= 0.03 mmol/L with a Km value of around 0.06 mmol/L. The initial velocity observed was 1.5x103 mmol/L.min1. Calculate the initial velocity of the enzymatic reaction when using [S]= 0.12 mmol/L.
An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :
A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).
An enzyme that follows Michaelis-Menten kinetics has a KM value of 3.00 µM and a keat value of 181 s1. At an initial enzyme concentration of 0.0100 µM, the initial reaction velocity was found to be 1.07 x 10-0 µM/s. What was the initial concentration of the substrate, S, used in the reaction ? Express your answer in micromolar to three significant figures. > View Available Hint(s) ? [S] !! µM Submit
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 Michaelis–Menten kinetics Lineweaver–Burk plot 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.
An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.
An enzyme catalyzes a reaction at a velocity of 20 μmol/min when the concentration of substrate (S)is 0.01 M. The Km for this substrate is 1 × 10-5 M. Assuming that Michaelis-Menten kinetics arefollowed, what will the reaction velocity be when the concentration of S is 1 ×10-6 M?
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
From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
The enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose 6-phosphate. After the reactants and products were mixed and allowed to reach equilibrium at 25°C, the concentration of glucose 1-phosphate was 4.5 mM and that of glucose 6-phosphate was 86 mM. Calculate Keq' and AG for this reaction. The reaction coordinate diagram for an enzyme-catalyzed reaction is shown below. How many transition states and intermediates are in the reaction? Is the reaction thermodynamically favorable? Which step is the rate-determining step of the reaction? G Reaction coordinate
An enzyme-catalyzed reaction is allowed to proceed for 3.4 minutes under steady-state conditions (with large excess of substrate). If the enzyme concentration is 7.9 nM, and the concentration of product formed is 6.3 µM, what is the kcat of this enzyme? Please give your answer to two significant figures in s1.

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