11. Strategy for Overcoming an ATP-Dependen t Chemical Coupling The Unfavorable
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11. Strategy for Overcoming an ATP-Dependen t Chemical Coupling The Unfavorable Reaction: phosrhorylation of glucose to glucose 6-phosphate is the tnital step in the catabolism of glucobe. The direct. glucose by P, is described by the equation Glucose + P, glucose 6-phosphate H20 AG 13.8 kd/mol Calculate the equilibrium constant for the above reaction at 37 °C. In the rat hepatocyte the physiological concentrations of glucose and P, are maintained at approximately 4.8 mM. What is (a) the equilibrium concentration of glucose 6-phosphate obtained by the direct phosphorylation of glucose? Explairl. the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and glucose by P? Does this reaction represent a reasonable metabolic step for the catabolism of (b) In principle, at least, one way to increase the concentration of glucose 6-phosphate is to drive P. Assuming a fixed concentration of P, at 4.8 mM, how high would the intracellular concentra- tion of glucose have to be to give an equilibrium concentration of glucose 6-phosphate of 250 M (the normal physiological concentration)? Would this route be physiologically reasonable, given that the maximum solubility of glucose is less than 1 M? (e) The phosphorylation of glucose in the cell is coupled to the hydrolysis of ATP; that is, part of the free energy of ATP hydrolysis is used to phosphorylate glucose: () Glucose +P glucose 6-phosphate + H2o AG (2) ATP +H20ADP +P 13.8 kJ/mol AG-30.5 kJ/mol Sum: Glucose ATP glucose 6-phosphate + ADF Calculate Keq at 37 for the overall reaction. For the ATP-dependent phosphorylation of glu cose, what concentration of glucose is needed to achieve a 250 intracellular concentration of glucose 6-phosphate when the concentrations of ATP and ADP are 3.38 mM and 1.32 mM, respec tively? Does this coupling process provide a feasible route, at least in principle, for the phospho- f - rylation of glucose in the cell? Explain (d) Although coupling ATP hydrolysis to glucose phosphorylation makes thermodynamic sense, we have not yet specified how this coupling is to take place. Given that coupling requires a common intermediate, one conceivable route is to use ATP hydrolysis to raise the intracellular concentra tion of P, and thus drive the unfavorable phosphorylation of glucose by Pt. Is this a reasonable route? (Think about the solubility products of metabolic intermediates.) (e) The ATP-coupled phosphorylation of glucose is catalyzed in hepatocytes by the enzyme glucoki- nase. This enzyme binds ATP and glucose to form a glucose-ATP-enzyme complex, and the phos phoryl group is transferred directly from ATP to glucose. Explain the advantages of this route.Explanation / Answer
11.a) delta G 'o= -RT ln K'eq
ln K'eq= -delta G'o/RT
=-(13.8kJ/mol)*(2.48 kJ/mol)
Therefore,K'eq=e-5.56
WHICH EQUALS TO 3.85 * 10-3 M-1.
This equation contains the unit M-1 because the expression of K(eq) contains water.
K'eq=[G6P]/[Glc][Pi]
[G6P]=K'eq[G1c][Pi]
=(3.85*10-3 M-1)(4.8*10-3M)2
=8.9*10-8M
This probably is the reasonable track fir glucose catabolism because the cellular (G6P) is likely to be much higher tham 8.9* 10-8M.
11.b) Because K'eq=[G6P]/[Glc][Pi]
then [Glc]= [G6P]/K'eq[Pi]
=250* 10-6M/(3.85*10-3M-1)(4.8*10-3M)
=14M.
This would not be a reasonable pathway since maximum solubility of glucose is less than 1M.
11.c) 1. Glc + Pi -------> G6P + H2O (delta G'10=13.8kJ/mol)
2. ATP + H2O --------> ADP + Pi ( delta G'20=-30.5kJ/mol)
Sum: Glc + ATP -----> G6P +ADP (delta G'0sum = -16.7 kJ/mol)
ln K'eq= delta G '0/RT
=-(-16.7kJ/mol)(2.48 kJ/mol)
=6.73
therefore,K'eq= 837
From the equation obtained,
[Glc]= [G6P][ADP]/K'eq[ATP]
= (250*10-6M)(1.32*10-3M)/837*(3.38*10-3M)
=1.2*10-7 M.
This pathway shows feasibility because glucose concentration is reasonable.
11.d) No;this ain't reasonable.When glucose is at its physiological level ,amount of Pi would be so high that salts of phospate with divalent cations would precipitate which is not wanted .
11.e) Direct transfer of the phosphoryl group from ATP to glucose takes advantage of the high potential of ATP phosphoryl group and does not demand that the concentration of intermediates be extremely high,unlike the mechanism dicussed in previous answer.
Moreover,the usual benefits of enzymatic catalysis apply,including binding interactions between the enzyme and its substrates; induces fit leading to the extraction of water from the active surface or site;so only gkucose gets phosphorylated;and stabilisation of transition state occurs.
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