AG" for reactionAG" for reactionA⇒ B=-10 kJ/molBC = -8 kJ/molAG' for reactionCD = 5 kJ/molAG" for reaction D⇒ E= 12 kJ/molCalculate the AG' for reaction A+D → B+C.A. -15 kJ/molB.- 5 kJ/molC.5 kJ/molD. 15 kJ/mol For reaction A ⇒B, AG¹° = -22 kJ/molFor reaction BC, AG" = 0 kJ/molFor reaction C ⇒ D, AG" = 31 kJ/molWhich reaction has the largest K'eq?A. A BB. BAC. B CD. CIE. D➜ C

The standard free energy change (∆G°) of a chemical reaction is the free energy change at the…

Answered: AG" for reaction A B = -10 kJ/mol AG"…

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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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.
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
3.An enzyme catalyzed reaction is studied and the following kinetic analysis is obtained: |[S), mM 0.050 0.075 v, (µM min") 0.93 1.264 1.77 2.14 3.7 0.125 0.175 0.935 a. Using Excel, make a fully labeled Lineweaver-Burk plot and determine the Km and Vmax for the enzyme b. The reactions were set up dissolving 1 mg of the enzyme (MW= 100000 Da) in 100 ml of final reaction buffer. Determine the turnover number for the enzyme assuming 1 active site exists per enzyme molecule? c. Determine the catalytic efficiency for the enzyme d. The same reactions are performed in presence of an inhibitor A and the resulting velocities determined: v plus inhibitor, (µM min) 0.272 0.37 0.518 0.626 |1.08 |[S), mM 0.050 0.075 0.125 0.175 0.935 Plot these data on the same graph as above and determine the new Km and Vmax and the type of inhibitor (competitive, non-competitive). e. Can the effects of the inhibitor be over-ridden by adding more substrate? Why?
Carbonic anhydrase catalyzes the hydration of CO. CO2 + H2O ¬ H½CO3 The Km of this enzyme for CO, is 1.20×104 µ.M. When [CO,] = 3.60×104 µM, the rate of reaction was 4.50 umol·mL! sec-1 a What is Vmax for this enzyme? umol·mL-!sec-!
a. What is the equilibrium constant for a reaction that is carried out at 25 °C (298 K) with H° = 20 kcal/mol and S° = 5.0 * 10 - 2 kcal mol -1 K-1? b. What is the equilibrium constant for the same reaction carried out at 125 °C?
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
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 :
1. Determine the Michaelis-Menten parameters of Vmax and Kỵ for the reaction S+E E.S E.S+W P+E The rate of reaction is given as a function of substrate: [S] 0.2 0.02 0.01 kmol/m³ -I'S kmol/m³.s E.S k₂E+S 0.54 0.28 0.19 0.005 0.1 0.002 0.045
ch Select all statements that are correct. Note there might be more than 1 correct statement. From the Lineweaver-Burk plot the equilibrium constant (Keq) can be obtained The Lineweaver-Burk plot gives a more accurate prediction for Vmax than the Michaelis- Menten plot The Lineweaver-Burk plot assumes that products and reactants are present at equal concentrations during the entire time of the reaction The Lineweaver-Burk plot shows velocity of reaction vs substrate concentration The Lineweaver-Burk plot shows 1/velocity of reaction vs 1/substrate concentration O 20°C D) // E
I I 1 1 Testa tabes dm 3 4 5 6 distilled Water 15 10 5 O 6 amylase enzyme absorbance 0.00 0.031 0.043 0:094 0151 0.147 5 5 10 5 O O 5 5 5 3. In this experiment, what does reaction rate represent (what is it a proxy for)? What does a high absorbance value indicate? What does a low absorbance value indicate? What does an ? absorbance value of zero indicate I 1 1 1 I I 1 I I I 1 I I
An enzymatic reaction with KM = 4.4 x 10-5 M, is carried out in 400 μL of of solution containing 0.20 nmoles of enzyme. It is observed that Vmax = 6.6 x 10-3 M/s. What is the kcat value for the enzyme? (HINT: Keep an eye on the units)!!! a. 1.50 x 102 s-1 b. 7.56 x 10-5 min-1 c. 1.32 x 104 s-1 d. 3.3 x 106 s-1
Which rate law has a molecularity of zero? Select one: O a. Rate = k[A]° O b. Rate = k[A][B] %3D O c. Rate = k[A][B][C] %3D O d. Rate k[A] %3D K[AJ? e. Rate = Determine the value of K. for the following reaction if the equilibriu

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AG" for reaction A B = -10 kJ/mol AG" for reaction BC = -8 kJ/mol AG" for reaction CD = 5 kJ/mol AG" for reaction D⇒ E= 12 kJ/mol Calculate the AG' for reaction A+D → B+C. A. -15 kJ/mol B. - 5 kJ/mol C. 5 kJ/mol D. 15 kJ/mol