Use the following information ( T = 37 oC) ADP + H2O ---> AMP + Pi + H+ ΔGo’ = -35.7 kJ/mol ATP + H2O ---> ADP + Pi + H+ ΔGo’ = -30.5 kJ/mol The steady state concentrations in mildly fatigued muscle cell are: [ATP] = 6.07 mM [AMP] = 1.80 mM [ADP] = 1.33 mMCalculate ΔG for the reaction under these cellular conditions (T= 37 oC): AMP + ATP ---> 2ADP Use the following information (T = 37°C)ADP + H2OATP + H2OAMP+ Pi + H+ADP+ Pi + H+AG°=-35.7 kJ/molAGO-30.5 kJ/mol[ADP] = 1.33 mMThe steady state concentrations in mildly fatigued muscle cell are:[ATP] = 6.07 mM [AMP] = 1.80 mMCalculate AG for the reaction under these cellular conditions (T=37 °C):2ADPAMP + ATP --->Type your answer...

Answered: Image /qna-images/answer/ea57564f-59bc-4142-8f3a-692acc0b5d33.jpg

Use the following information (T = 37°C) ADP + H2O ATP + H2O AMP+ Pi + H+ ADP+ Pi + H+ AG°=-35.7 kJ/mol AGO-30.5 kJ/mol [ADP] = 1.33 mM The steady state concentrations in mildly fatigued muscle cell are: [ATP] = 6.07 mM [AMP] = 1.80 mM Calculate AG for the reaction under these cellular conditions (T=37 °C): 2ADP AMP + ATP ---> Type your answer...

Use the following information (T = 37°C) ADP + H2O ATP + H2O AMP+ Pi + H+ ADP+ Pi + H+ AG°=-35.7 kJ/mol AGO-30.5 kJ/mol [ADP] = 1.33 mM The steady state concentrations in mildly fatigued muscle cell are: [ATP] = 6.07 mM [AMP] = 1.80 mM Calculate AG for the reaction under these cellular conditions (T=37 °C): 2ADP AMP + ATP ---> Type your answer...

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter16: Thermodynamics: Directionality Of Chemical Reactions

Section: Chapter Questions

Problem 10QRT

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Consider the coupled reactions: kcal AGO'= -16.3 mol 1,3-ВPG 3-PG + Pi AG°'= -7.3 kcal mol ATP → ADP + P; What is the net reaction? 3-PG + ADP + Pi→ 1,3-BPG + ATP 3-PG + ATP → 1,3-BPG + ADP + P¡ O 1,3-BPG + ADP + P¡ –→ 3-PG + ATP O 1,3-BPG + ATP → 3-PG + ADP + P¡
The value of ΔGÂ° for the conversion of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate (GAP) is 7.53 kJ/mol. If the concentration of dihydroxyacetone phosphate at equilibrium is 2.85 mM, what is the concentration of glyceraldehyde-3-phosphate?Assume a temperature of 25.0 Â°C.[GAP] = ???? mM
Calculate the standard free energy change for the reactionFADH2 + 1/2O2 → FAD + H2Ogiven that the standard reduction potential for the reduction of oxygen to water is +0.82 V and for the reduction of FAD to FADH2 is +0.03 V.
CH₂OH T C=O CH₂OPO²- DHAP Triose phosphate isomerase H HCOH CH₂OPO3²- G-3-P Under standard conditions, AG° = +7.53 kJ/mol. What is AG when the initial concentration of DHAP is 2 x 10-4 M and the initial concentration of G-3-P is 3 x 10-6 M? R = 8.314 x 10-3 kJ*mol-¹*K-1 T = 310 K
Calculate the equilibrium constant for the following reaction at 37 oC. Fumarate + H2O → Malate ΔGo’ = -3.13 kJ/mol
Calculate the actual, physiological deltaG for the reaction Phosphocreatine + ADP > creatine + ATP at 35.84oC, as it occurs in the cytosol of neurons, with phosphocreatine at 4.70 mM, creatine at 1.39 mM, ADP at 0.89 mM, and ATP at 2.39 mM. Give the answer in kJ/mol to one decimal place. The standard deltaG hydrolysis of phosphocreatine with H2O is -42.49 kJ/mol and the standard deltaG for the phosphorylation of ADP with Pi is +35.99 kJ/mol. Give the answer in kJ/mol to one decimal place.
Given the reaction: 6CO2+ 6H2O→ C6H12O6 + 6O2 ΔG°f CO2 = −394.39 kJ/mol ΔG°fH2O = −237.14 kJ/mol ΔG°fC6H12O6 = -2870 kJ/mol ΔG°fO2 = 0 What will be the ΔG at equilbrium?
Which of the reactions are spontaneous (favorable)? DHAP = glyceraldehyde-3-phosphate AG = 3.8 kJ/mol L-malate + NAD+ → oxaloacetate + NADH + H+ AG = 29.7 kJ/mol C,H¸O; C,H,O, + H,0 AG= 3.1 kJ/mol Rh(I) C,H, + H, C,H, AG = -150.97 kJ/mol C,H1,0,P O̟P + ATP C,H14012P2 + ADP AG = -14.2 kJ/mol glutamate + NAD+ + H,O → NH† + a-ketoglutarate + NADH + H* AG = 3.7 kcal/mol
The Keq for the isomerization of glucose-6-phosphate to fructose-6-phosphate is 0.504. What is the ΔG for the reaction if the concentration of glucose-6-phosphate is 0.01 M and the concentration of fructose-6-phosphate is 0.05 M? This reaction is in the cell (this is your temperature = 30 degree celcius ). the answer should be 0.588 kcal/mol - please show all steps I am not sure how to get that answer
What is the product of the CNO cycle? What do you think it is called CNO cycle?
Energy from ATP hydrolysis drives many nonspontaneouscell reactions:AT P⁴⁻(aq) +H₂O(l) ⇌ADP³⁻(aq) +HPO₄²⁻(aq) +H(aq) ΔG°'=-30.5 kJ Energy for the reverse process comes ultimately from glucose metabolism:C₆H₁₂O₆(s)+ 6O₂g) →6CO₂(g) +6H₂O(l)(a) Find K for the hydrolysis of ATP at 37°C.(b) Find ΔG'°ᵣₙₓ for metabolism of 1 mol of glucose. (c) How many moles of ATP can be produced by metabolism of1 mol of glucose? (d) If 36 mol of ATP is formed, what is the actual yield?
A reaction has a ΔG‡ = 15 kcal/mol at 280 K. (b) At what temperature will the reaction be accelerated by a factor of 10? (give answer with units in kelvin)

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