2. The Km of a particular enzyme is 2.0 x 105 M. If the initial velocity (Vo) is 0.16 mmol/min for [S] =0.15 M, what will be the initial velocity when [S] =2.0 x 10+M?

For a one-substrate enzyme-catalyzed reaction, the Michaelis-Menton equation shows the quantitative…

Answered: 2. The Km of a particular enzyme is 2.0…

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 with concentration of 3.0 uM is added to a flask containing 125 uM of substrate. You know the Km is equal to 0.054 mM, and has a Vmax of 1.4 uM/s. What is the observed velocity? O 0.98 uM/s 0.074 uM/s O 5.2 um/s 1.4 uM/s
Given the following enzyme kinetics data, determine the maximal velocity and the Michaelis constant of the enzyme. Show your solutions
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 :
7.1 The following results were obtained for an enzyme-catalysed reaction Substrate concentration (mmolr'): Initial velocity (umol I' min"): 6.67 182 10.0 20.0 40.0 147 233 323 400 Calculate Km and Vmax
A bacterial enzyme catalyzes the hydrolysis of maltose as shown in the reaction given below: Maltose + H2O -> 2 glucose If the reaction has a Km of 0.135 mM and a V max of 65 m mol/min. What is the reaction velocity when the concentration of maltose is 1.0 mM?
Given an enzyme with a KM= 10 mM and Vmax = 100 mmol/min. If [S] = 100 mM, which of the following will be true? a. A 10 fold increase in Vmax would increase the velocity 10 fold b. A 10 fold decrease in KM would increase the velocity c. A 10 fold increase in Vmax would decrease velocity 20 fold d. Both A) and B) e. 15000 pM
Suppose that the data shown in the margin are obtained for an enzyme -catalyzed reaction. [S] (mM) V (mmol/ml min) 0.1 3.33 0.2 5.0 0.5 7.14 0.8 8.00 1.0 8.33 2.0 9.09 a.) Determine the Km and Vmaxb.) Assuming htat the enzyme present in the system had a concentration of 10-6M, calculate the turn -over number
The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)
An enzyme is present at 100 nM (nanomolar) and has a Vmax value of 25 uM/s (micromolar/second). The Km for the substrate is 5.2 uM. What is the initial velocity (V0) at a substrate concentration of 15.2 uM? Report your answer to three significant figures in units of uM/s.
1. a. Calculate the physiological DG of the reaction shown below at 37°C, as it occurs in the cytosol ofneurons, with phosphocreatine at 4.7 mM, creatine at 1.0 mM, ADP at 0.73 mM, and ATP at 2.6mM. The standard free energy change for the overall reaction is –12.5 kJ/mol. Phosphocreatine + ADP ® creatine + ATP b. The enzyme phosphoglucomutase catalyzes the conversion of glucose 1-phosphate to glucose6-phosphate. Calculate the standard free energy change of this reaction if incubation of 20 mMglucose 1-phosphate (no glucose-6 phosphate initially present) yields a final equilibrium mixtureof 1.0 mM glucose 1-phosphate and 19 mM glucose 6-phosphate at 25°C and pH 7.0. c. If the rate of a nonenzymatic reaction is 1.2 x 10–2 μM s–1, what is the rate of the reaction at 37℃ inthe presence of an enzyme that reduces the activation energy by 30.5 kJ/mol?
Given the initial rate data for the reaction A + B → C, determine the rate expression for the reaction. [A], M [B], M A[C]/At (initial) M/s 0.0418 0.122 2.73 x 10-4 0.0836 0.122 10.9 × 10 4 0.0418 0.488 2.73 x 104 O A[C] At ○ A[C] ΔΙ = 0.439 Ms[A][B]² = 1.27 M²s¹ [A][B] ● A[C] = 6.53 × 10³ s¹ [A] At O A[C] = 0.156 M's¹[A]² At ○ A[C] At = 5.35 x 102 M2s[A][B]
You are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μM

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2. The Km of a particular enzyme is 2.0 x 105 M. If the initial velocity (Vo) is 0.16 mmol/min for [S] = 0.15 M, what will be the initial velocity when [S] =2.0 x 10+M?