The enzyme phosphoserine phosphatase catalyses the reaction:phosphoserine + H2O → serine + PiKinetics measurements were performed using a range of Mg* concentrations andin the presence of different fixed concentrations of Ca-". The enzyme requiresMg²* for activity. The following data were recorded:[Ca²*] / mM0.020.040.0[Mg*]/ mMInitial velocity / µmol of product min0.3320.409.265.990.5025.6512.668.471.0033.3020.0014.282.0040.0028.5722.23What type of inhibitor is Ca2* and what is its Kf?Suggest what role Mg** performs in the catalytic mechanism of phosphoserinephosphatase. Suggest how Ca2* inhibits the reaction.

Inhibition: A substance that decreases or slows down the activity of other substances is known as an…

Answered: phosphoserine + H2O → serine + Pi…

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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What is the Vmax of an enzyme at 1 µM with an activation free energy (AG#) of 55000 J/mol at 298K? -AG # RT k cat = 6.2× 10¹² (Ideal gas constant R = 8.314 J K¯1 mol-¹1) 0.14 μM s-1 -1 S-1 1.4 μM s 14 μM s-1 0.14 mM s-1 1.4 mM s-1
Do not give handwriting solution.
At what substrate concentration would an enzyme with a kcat of 33.0 s-1 and a Km of 0.0046 M operate at one-tenth of its maximum rate? Assume the enzyme obeys Michaeli s-Menten kinetics.
at what substrate concentration is v= 5.0 mM s^-1 for the enzyme catalyzed hydrolysis of trehalose?
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?
You are working on an enzyme that obeys standard Michaelis-Menten kinetics. You have determined the Vmax to be 0.1 mol/sec and the Km to be 2.5 mM. What would the rate of the reaction be when the substrate concentration is 20 mM? 0.09 MS-1 O 0.133 Ms-1 O 0.18 Ms ¹ 9 Ms-1 O 0.018 Ms-1 0.2 MS-1
Vo 60 50 40 30 20 mm/sec 0100 0 O 0 0 50 100 150 200 250 300 Substrate (mM) Use the Michaelis-Menten plot to answer this question. What is the estimated value of Vmax of the enzyme catalyzed reaction (square data points)? (choose the one best answer) 10
Consider the following equilibrium at 25ºC :Glucose-1-Phosphate Glucose-6-PhophateUsing the equilibrium concentrations of [Glucose-1-Phosphate] = 0.35 M and [Glucose-6-Phosphate] = 1.65 M, calculate BOTH K′eqand Gº′ for this reaction. Is this reaction exergonicor endergonic? R = 8.314 J/K·mol
The enzyme triosephosphate isomerase (TIM) catalyzes the following reaction in glycolysis, where it converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP). CH₂OH C=O CH₂OPO²- DHAP triose phosphate isomerase [DHAP] (MM 1.00 2.00 3.00 6.00 O V. (mM/s) 3.700 6.727 9.250 14.800 = H HCOH CH₂OPO²- Kinetics experiments were performed on TIM. Enzyme activity (initial velocity, V.) was measured at varying concentrations of DHAP. The enzyme kinetics were also measured in the presence of two inhibitors, A and B. The enzyme concentration used in all experiments was 25.00 μM. The data are shown below. GAP V. (mM/s) + A 1.452 2.794 4.038 7.281 V₁ (mM/s) + B 0.755 1.379 1.905 3.077
ver is given. A +B C+ D 1) Calculate AG for the above react un and indicáte whether the reaction is favorable or unfavorable [A] = 0.9 M 20°C AG° = 4 KJ/mol %3D [B] = 15mM [C] = 4mM [D] = 3 M R= 8.314 J/moleK %3D %3D 4.7 and explain how you know.
Calculate AG for this reaction under the following conditions: 37°C, pH 7, [Pyruvate] = [CO₂] = 4.0 mM, [OAA] = 2.0 mM, [ATP] = 3.5 mM, [P;] = 5.0 mM, and [ADP] = 1.8 mM. Use 2 signficant figures. AG= i kJ.mol-1 Under these conditions, the reaction is (Tolerance is +/- 2%)
Calculate the Km of the enzyme with these parameters. kcat = 130s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µM

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