8 e.Consider the reversible enzyme reactionwwm mF+H2O ==== MWhere F=Fumarate and M=MalonateAt 25 C and Ph 8 the maximum velocities V and Michaellis constants of these reactions are :(0.6 × 10ª s-1)(E).V = (0.2 x 10³ s-1) (E).K, = 7 x 10-6 mol L-1VM =KM = 100 x 10-6 mol L-1%3D%3Dwhere (E), is the total molar concentration of enzymatic sites. Calculate the values of thefour rate constants in the mechanismkaE + F EX E + MkakaShow your work.

Enzyme kinetics is a study of the rate of enzyme catalyzed reactions. In Michaelis-Menten kinetics,…

Answered: Calculate the values of the

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter18: Glycolysis

Section: Chapter Questions

Problem 17P

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Utilising the provided class data generate the following graphs: I) Michaelis Menten; II) Lineweaver-Burk; and III) Hanes-Woolf. Ensure that you clearly label each graph,and add the relevant trendlines with equations. Table 1: Class data demonstrating the Absorbance at 700nm obtained for the alkaline phosphatase enzyme reaction Table 1 tube Abs700mm 1 0.000 2 0.060 2 0.090 4 0.140 5 0.190 6 0.250 7 0.290 The equipment we used are • 20mM Tris Buffer pH 8.5 • 33mM MgCl2 • Alkaline Phosphatase (2mg/ml) in 20mM Tris Buffer pH 8.5 • 4mM Glucose-1-phosphate • Acid Molybdate pH 5.0 • Reducing Agent • Distilled Water • Glass Test tubes • Tube Rack • Cuvette • Pipettes and Tips • Water bath set to 37oC The method we used is Method/Protocol: 1. Read the protocol in its entirety before starting. Take note of any additional information that appears in subsequent steps that may influence how previous steps are performed. 2. Using glass tubes, generate the reactions mixtures…
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
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
a) Calculate the enzyme and specific activity of a reaction with 3 μM Hsp90 using the following information: The rate is measured in a spectrophotometer as 0.028 OD units/min in a 1 ml reaction volume. The absorbance was detected at 340nm and the extinction coefficient for NADH at this wavelength is 6200 L M-1 min-1 and the molecular mass of Hsp90 is 82.7 kDa. The rate of NADH utilisation is equivalent to the rate of ATP utilised by Hsp90. Show all your calculations and the units for your answers. b) Calculate the turnover number for the reaction described in (a) above
Based on the definition of kcat, substitute a value that can be measured and yet still represents the value associated with the original concentration of the R. What would the rate or velocity of the reaction be equal to under these circumstances? How can cells increase Vmax? What variable that we could change would directly impact Vmax? Would the value of KM be affected by the ways you determined that Vma,x could be increased? What does this indicate about KM? Thinking about how catalysts work, about the Michaelis-Menten Equation, and the definition of kcat, what specifically does the enzyme change in the reaction mechanism to increase the rate? If an enzyme follows the 2 step mechanism proposed by Michaelis-Menten, what do you know about this enzyme? Be very specific and comprehensive. Please answer very soon will give rating surely
Calculate the equilibrium concentration of H2O for the following esterification reaction performed in ethanol(C2H5OH) C2H5OH + CH3CO2H ⇌CH3CO2C2H5 + H2O KC= 4.0 At equilibrium: [CH3CO2H] = 0.75 M; [CH3CO2C2H5 ]= 2.2 M
A) Is this reaction ( in picture provided) in equilibrium? B) If it is not then ,what is ∆G' at 25°C if the concentration of Glucose-1-phosphate is 15.04µM and the concentration of Glucose-6-phosphate is 1.62 mM? Answer in Joules. Round to the correct number of significant figures. (There are 103 µM in 1mM.) Thank you so Much!!!
The conversion of glucose-1-phosphate to glucose-6-phosphate by the enzyme phosphoglucomutase has a △G°' of -7.6 kJ/mol. Calculate the equilibrium constant for this reaction at 298 K and a pH of 7. (R = 8.315 J/K-mol) A. 0.003 B. 0.047 C. 1.00 D. 21
With appropriate chemical structures, explain the mechanism (mode-of-action) of fluoroacetate poisoning? Example: Step 1: Fluoroacetate is converted to Product “A”. This reaction is catalyzed by Enzyme __________________________ Structures of fluoroacetate and the product “A”. Name of Enzyme. Step 2: Product “A” from Step 1 is converted to Product “B.” Catalyzed by enzyme 2. Structure of Product B and name of Enzyme 2. etc.
The formation of maltose, a disaccharide, from two glucose molecules, is not energetically favorable. However, if this reaction is coupled with the hydrolysis of ATP, the reaction occurs more favorably. Maltose + H2O = 2 Glucose , ΔG'o = -15.5 KJ/mol or -3.7 kcal/mol a. Determine if the coupled reaction will occur spontaneously at standard state through calculating the Gibbs Free Energy of Reaction. b. Calculate the equilibrium constant for each individual reaction, and for the coupled reaction (using free energy of reaction). Show that the equilibrium constant for the coupled reaction equals the equilibrium constants for the individual reactions multiplied together. c. If the reaction medium contains the following chemical species at their given concentrations (298 K and 1.0 atm, pH = 7.0), will the reaction proceed in the forward or the reverse direction? [Maltose] = [Glucose] = 10.0 mM; [ATP] = 5.0 mM; [ADP] = [Pi] = 20 mM
If a 0.1 M solution of glucose 1- phosphate at 25 °C is incubated with a catalytic amount of phosphoglucomutase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are Calculate Keq and ΔG′° for this reaction.
From your Lineweaver-Burk plot,the vlaues are: Km Vmax Uninhibited 0.09 mmol/L 3.02 min/mmol Inhibited 6.22 mmol/L 9.98 min/mmol By describing the potential changes in the kinetic parameters, identify and justify the type of inhibitor that was inhibiting the acid phosphatase activity.

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