3. Relation between Reaction Velocity and Substrate Concentration: Michaelis-Menten Equation (a) At what substrate concentration would an enzyme with a k of 30.0 s¹ cat and a K of 0.0050M operate at one-quarter of its maximum rate? m Vo/Vmax=0.25 (b) Determine the fraction of that would be obtained at the following substrate concentrations [S]: ½ K, 2 K, and 10 K. m max (c) An enzyme that catalyzes the reaction XY is isolated from two bacterial species. The enzymes have the same Vmax, but different Km values for the substrate X. Enzyme A has a Km of 2.0 M, while enzyme B has a Km of 0.5 M. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 [M. Which curve corresponds to which enzyme? [Y] Time

3. Relation between Reaction Velocity and Substrate Concentration: Michaelis-Menten Equation (a) At what substrate concentration would an enzyme with a k of 30.0 s¹ cat and a K of 0.0050M operate at one-quarter of its maximum rate? m Vo/Vmax=0.25 (b) Determine the fraction of that would be obtained at the following substrate concentrations [S]: ½ K, 2 K, and 10 K. m max (c) An enzyme that catalyzes the reaction XY is isolated from two bacterial species. The enzymes have the same Vmax, but different Km values for the substrate X. Enzyme A has a Km of 2.0 M, while enzyme B has a Km of 0.5 M. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 [M. Which curve corresponds to which enzyme? [Y] Time

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 catalyzes a reaction at a velocity of 20 micromole/min when the concentration of substrate (S) is 0.01 M. The Km for this 1 X 10^-5 M. Assuming that Michaelis-Menten Kinetics are followed, what will the reaction velocity be when the concentration of S is 1.0 X 10^-6 M?
1) Is the ratio of the forward rate constant and reverse rate constant changed by the presence of an enzyme catalyst? 2) 4) What is the simplest mathematical relationship between substrate concentrations [S], initial velocity (Vo) of an enzyme catalyzed reaction, the maximal velocity of the reaction (Vmax), and the ½ maximal Vo (i.e. Km) ? (Hint: what is the Michaelis-Menten Equation? 3) Graphically illustrate the most useful derivation of the Michaelis-Menton equation (that darn linearized, double-reciprocal)
The rates at various substrate concentrations for an enzyme-catalysed reaction (b) are recorded as follows: 2.5 5.0 10.0 15.0 (S]/10 M 2.2 3.8 5.9 7.1 Vo/ M min Assuming the reaction obeys Michaelis-Menten kinetics, calculate the following values by using Lineweaver-Burk plot () maximum rate, Vmax- (i) Michaelis constant, Km- (ii) turnover number, kz.
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?
Vmax [S] Vo = Km+ [S] Eadie-Hofstee plot Lineweaver-Burk (L-B) plot v=Vm-Km [S] Km 1 Vm Vm [S] The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For competitive inhibition The Michaelis-Menten equation becomes Vo=Vmax[S]/(aKm+[S]) Put the Michaelis-Menten equation for competitive inhibition in the Eadie-Hofstee format. The Y-intercept 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.
Table below shows the kinetic data obtained for an enzyme in the absence of any inhibitor (1), and in the presence of an inhibitor (2), at a concentration of 7.0 mM. Assume the total enzyme concentration, [E]T, is the same for each experiment. Table 1 [S] (mM) Vo Vo {umol/(mL•s)} {umol/(mL•s)} Without Inhibitor With Inhibitor (1) 3.8 (2) 2.2 1 2 7.3 4.3 4 13.3 8.2 8 22.9 14.9 12 30.0 20.4 a); Construct a Lineweaver -Burk plot using the kinetic data shown in Table 1. Determine Vmax and Km for the uninhibited enzyme and inhibited enzyme.
1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062
1.1)the following data duscribe an enzyme-catalyzed reaction(hydrolysis of cabobenzoxyglycyl-L-tryptophan) Plot these results using a lineweaver-Burk method, and determine values for Km and Vmax. substrate concenrate(mM) Velocity(mM.sec-1) 2,5 0.024 5 0.036 10 0.053 15 0.060 20 0.061 25 0.062 1.2) If the Km of an enzyme for it's substrate remains constant as the concentration of the inhibitor icreaces, what can be said about the mode of inhibition and why? 1.3) calculate the turnover number for an enzyme, assuming Vmax is 0.5M.sec-1 and the concentration of the enzyme used is 0.002M . why is it usefull to know this? 1.4) discuss the mechanism of the bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues. make use of relavant graphs and diagrams to explain your answer.
1. For enzyme catalyzed reaction: ki S + E=X (fast) k2 X -2> P + E (slow) derive the rate law and answer the following: a) What is meant by Michaelis-Menten constant km? (b) Prove that order of reaction in substrate changes from unity to zero at higher [S]. (c) What is the rate and value of km, if k2 >> k1. (d) Draw the potential energy diagram for enzyme catalyzed reaction.
The kinetics of an enzyme were measured as a function of substrate concentration in the presence or absence of 10-4 M inhibitor. The following initial rate data were obtained: Substrate (So µM/L) 3 5 Velocity (Vo, µmol/L-min) Without inhibition 10 30 90 10.4 14.5 4.5 6.8 8.1 Is this a competitive or noncompetitive inhibition? Evaluate KM, KI, and Vmax. (HINT: use linear regression to first estimate kinetic parameters for uninhibited and then for the inhibited) With inhibition 2.1 2.9 22.5 33.8 40.5
The steady-state kinetics of an enzyme are studied in the absence and presence of an inhibitor (inhibitor A). The initial rate is given as a function of substrate concentration in the following table: v[(mmol/L)min- [S] (mmol/L) No inhibitor Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38 (a) What kind of inhibition (competitive, uncompetitive, or mixed) is involved? (b) Determine Vmax and Ky in the absence and presence of inhibitor.