The first reaction in glycolysis is the phosphorylation of glucose: P_i + glucos
ID: 483680 • Letter: T
Question
The first reaction in glycolysis is the phosphorylation of glucose: P_i + glucose rightarrow glucose - 6 - phosphate + H_2O This is a thermodynamically unfavorable process, with delta G degree' = +13.8 kJ/mol. In a liver cell at 37 degree C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What would be the equilibrium concentration of glucose-6-phosphate, according to the above data? Express your answer to two significant figures and include the appropriate units. This very low concentration of the desired product would be unfavorable for glycolysis. In fact, the reaction is coupled to ATP hydrolysis to give the overall reaction ATP + glucose rightarrow glucose - 6 - phosphate + ADP + H^+ what is delat G degree' for the coupled reaction? Express your answer to three significant figures and include the appropriate units. If, in addition to the constraints on glucose concentration listed previously, we have in the liver cell ATP concentration = 3 mM and ADP concentration = 1 mM, what is the theoretical concentration of glucose-6-phosphate at equilibrium at pH = 7.4 and 37 degree C? Express your answer to three significant figures and include the appropriate units.Explanation / Answer
G = Gº’ + RTln (([G6P]/([Pi][G]))
0 = 13800 J/mol + (8.31 J/molK)(310 K)ln ([G6P]/(0.005M*0.005M)
-5.43 = ln [G6P]/0.000025 0.0044 = [G6P]/0.000025
[G6P] = 1.1x10^-7 M
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(b) 13.8 kJ/mol – 30.5 kJ/mol = -17.5 kJ/mol
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Glucose + ATP->Glucose-6-Phosphate + ADP -16.5 KJ/mol
If in addition to the constraints on glucose cocnnetration listed previously we have in the liver cell ATP concentration =3 mM and adp……..
16500 J/mol = (-8.314 J/ mol-1K-1x 310 K) ln Keq
6.4019 = ln KeqKeq= e6.4019= 603.019 = [G6P][ADP]/ [ATP][Glucose] = [G6P](0.001)/(0.003 x 0.005)
[G6P] = (603.019 x 0.003 x 0.005)/0.001= 9.045 M
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