Qis an analog of substrate A that binds to enzyme X and produces the following kinetics:[A]Vo (umole/ml/min)[Q] = 0[Q] = 0.5 pM[Q] = 2 µM1 µM10743 μΜ20161010 µM35312330 μΜ43413680 µM474643a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of dataon your graph) and determine the following: the KM and Vmax of the enzyme X in tboabsence of Q, and the KMapp the Vmaxapp at each concentration of Q.Vmaxb) What type of inhibition is this?c) Calculate the inhibition constant (K) for Compound Q

Enzymes are protein molecules that increase the rate of reaction by decreasing the activation…

Answered: Qis an analog of substrate A that binds…

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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For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.
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.
An enzyme that follows Michaelis-Menten kinetics has a KM value of 10.0 μM and a kcat value of 206 s−1 . At an initial enzyme concentration of 0.0100 μM , the initial reaction velocity was found to be 1.07×10−6 μM/s . What was the initial concentration of the substrate, [S] , used in the reaction ? Express your answer in micromolar to three significant figures..
Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?
In a alkaline phosphatase kinetics experiments a studnet first made a standard curve with different amounts of PNP. In a reaction carried out with 10nM alkaline phosphatase and 100 uM PNPP substrate the student measured an absorbance value of 0.76 at 410nm after 30 seconds of reaction. Assuming the 30 second time point is in the lienar range of the reaction and the substrate has not been depleted at this time point, calculate the Vo at 100uM PNPP (in units of concentration per time).Beer's law formula given is y=0.0323x (y axis is Absorbance) (x Axis is PNP uM)
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.
An enzyme that follows simple Michaelis–Menten kinetics has an initial reaction velocity of 10 µmol⋅min-1 when the substrate concentration is five times greater than the KM. What is the Vmax of this enzyme in µmol⋅min−1?
(a) At what substrate concentration would an enzyme with a kcat of 30.0s−1 and a Km of 0.0050 M operate at one-quarter of its maximum rate?(b) Determine the fraction of Vmaxthat would be obtained at the following substratecon centrations [S]: ½Km, 2Km, and 10Km.(c) An enzyme that catalyzes the reaction X ⇌ Y 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 μM, and enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 μM. Which curve corresponds to which enzyme?
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.
The following kinetic data were obtained for an enzyme in the absence of an inhibitor (1) and in the presence of two different inhibitors (2) and (3) at 5mM concentration. [S], mM (mmol/mL/sec) (mmol/mL/sec) (mmol/mL/sec) (1) (2) (3) 1 12 4.3 5.5 2 20 8 9 3 29 14 13 8 35 21 16 12 40 26 18 Determine the Vmax and KM of the enzyme Determine the type of inhibition.
Given the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμM
The following were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics: Reaction Velocity (mmol/min) Substrate added (mmol/L) 217 0.8 325 2 433 4 488 6 647 20 652 1000 The Km for this enzyme is approximately _____________. (Round to the nearest integer)

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