Biochem#2: a) Phosphoenolpyruvate is one of the two phosphate donors in the synt

Question

Biochem#2:

a) Phosphoenolpyruvate is one of the two phosphate donors in the synthesis of ATP during glycolysis. In human erythrocytes (red blood cells) the steady state concentration of ATP is 2.24 mM, that of ADP is 0.25 mM, and that of pyruvate is 0.051 mM. Calculate the concentration of phosphoenolpyruvate at 25oC, assuming that the pyruvate kinase reaction is at equilibrium in the cell. b) The physiological concentration of phosphoenolpyruvate in human erythrocytes is 0.023 mM. Compare this with the value obtained in (a). What is the significance of this difference (if any)? c) Please draw the reaction you listed, including all reactants and products and list cofactors.

Explanation / Answer

a)

Phosphoenol pyruvate (PEP) + H2O ----> Pyruvate + Pi   Go'= -61.9 KJ/mol

ADP+Pi -----> ATP + H2O Go'= -30.5KJ/mol

Go'=-61.9 + 30.5 = -31.45KJ/mol

-31,400= -8.315(298)ln Keq

Keq= 3.19x105

Keq=[pyruvate] [ATP]/ [PEP][ADP]

By putting the above values, we have:

3.19x105= [0.051x10-3][2.24x10-3]/ [PEP] [0.25x10-3]

[PEP]= 1.43x10-9

b)

There is significant difference in calculated valus of PEP in erythrocytes and physiological value of PEP. Physiological value is far higher than the calculated value.

Hence, it is implied that the reaction is not in equilibrium.

c)

Phosphoenol pyruvate (PEP) + H2O ----> Pyruvate + Pi   Go'= -61.9 KJ/mol

ADP+Pi -----> ATP + H2O Go'= -30.5KJ/mol

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