Q1 Under standard conditions, the free energyneeded to breakdown ATP to ADP and P; is -30.5kJ/mol. Calculate the free energy (AG') required tobreakdown ATP in bacteria, where theconcentrations of ATP, ADP, and P; are 4.0 mM, 2.5mM, and 8.0 mM, respectively.Hint:[products]AG' = AG" +RTln[reactants]where, at 25 °C, RT = 2.479 kJ/molIs this reaction exergonic or endergonic?

Gibbs free energy is the amount of energy that is available for a system to do useful work (∆G').…

Answered: Q1 Under standard conditions, the free…

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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24. Determine equilibrium constants for the hydrolysis of phosphoenolpyruvate (AG" is 61.9 kJ/mol) at pH 7 and 25°C: nariw (d)
A one-to-one protein (P)-ligand (L) complexation (P + L PL) has a dissociation equilibrium constant (Kd) value of 100 nM at 25°C, and the Kd remains the same at 37°C. 1) What is AS of binding at 25°C? Assume ACp of the binding is 0 over the temperature range. AS = 1.34E2 kJ/(mol*K) (note the unit!!) (sig. fig =3) 2) What is the concentration of the PL complex formed at equilibrium when you mix 0.20 uM (microM) of Protein and 1.0 uM of Ligand together at 37°C? PL at equilibrium = 8.1E-1 uM (note the unit!!) (sig. fig =2)
bo Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at a certain ratio of [A] inside to [A] outside, the AG for the transport of substance A from outside the cell to the inside, Macmillan Leaming A outside → Ainside, is 10.7 kJ/mol at 25°C. What is the ratio of the concentration of substance A inside the cell to the concentration outside? [A]inside [A]outside Choose the true statement about the transport of A under the conditions described. Because AG is positive, the ratio [A] inside/[A]outside must be less than one. Increasing [A]outside will cause AG for movement of Aoutside to Ainside to become a larger positive number. Movement of Ainside to Aoutside will be spontaneous. Decreasing the concentration of the uniport protein in the membrane will cause AG to become a smaller positive number.
Bailey and Ollis Biochemical Engineering (Write your solution)
answer only question g
PTP1B Substrate kcat Km. kcat/Km UM 10-7 x (s-1 M) DADEPYLIPQQG DADAPYLIPQQG DAAEP YLIPQQG AAAAPYLIPQQG 44.6 + 1.8 39.8 + 0.32 3.9 + 0.9 13.7 + 0.46 1.1 + 0.25 0.29 + 0.01 35.3 + 0.22 6.6 + 0.22 0.53 + 0.02 34.7 + 0.25 52.7 + 0.7 0.066 + 0.001 ) The units for kcat/KM in the above are given according to standard scientific notation. On this (d) ( basis what is the value of this kinetic parameter for the DADEPYLIPQQG substrate?
Enzyme Kinetics question Enzyme used is 10uL of a 10 ng/uL solution to a reaction mix in a final volume of 2.0 mL. (Enzyme used is 20kDa monomeric enzyme if matters) Based off lineweaver burk where 300 uM of inhibitor is used noninhibited formula is y= 4x + 0.1 (x axis is 1/S 1/mM) (y axis is 1/Vo sec/mM) inhibited formula is y = 4x + 1 I found Km as 1 mM for inhibited Vmax as 1mM/sec for inhibited How would I find Kcat? How would I find Ki?
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

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