In the first step of glucose breakdown in our bodies, hexokinase catalyzes the f

Question

In the first step of glucose breakdown in our bodies, hexokinase catalyzes the following reaction; glucose + Pi glucose-6-phosphate Delta G degree ' at 25 degree C for this reaction is 13.8 KJ/mol. Does the Delta G degree ' indicate that formation of glucose-6-phosphate is favored at 25 degree C ? Explain your answer; no calculations required. In certain conditions, the formation of glucose-6-phosphate can be coupled to ATP hydrolysis. What is Keq' for the coupled reaction? ATP ADP + P_i Delta G degree ' at 25 degree C for this reaction is -30.5 kJ/mol. If ATP and ADP are present at 0.8 mM and 0.2 mM, what is the equilibrium ratio of glucose-6-phosphate to glucose? In the first step of glucose breakdown in our bodies, hexokinase catalyzes the following reaction; glucose + Pi glucose-6-phosphate Delta G degree ' at 25 degree C for this reaction is 13.8 KJ/mol. Does the Delta G degree ' indicate that formation of glucose-6-phosphate is favored at 25 degree C ? Explain your answer; no calculations required. In certain conditions, the formation of glucose-6-phosphate can be coupled to ATP hydrolysis. What is Keq' for the coupled reaction? ATP ADP + P_i Delta G degree ' at 25 degree C for this reaction is -30.5 kJ/mol. If ATP and ADP are present at 0.8 mM and 0.2 mM, what is the equilibrium ratio of glucose-6-phosphate to glucose?

Explanation / Answer

a] Given delta G is value positive one

Then reaction is spontaneous only when delta G is negative .

delta G = -RTlnK ; 13.8KJ = -RTln K ===> K = 3.81*10^-3

K =equilibrium constant

When delta G is negative ; then K is > 1 and the forward reaction favours

when delta G is positive ; then K is < 1 and the backward reaction favours

b] delta G -30.5KJ/mol

delta G = -RT ln K

T = 25C = 298 K

R = 8.314

then K = 222001

c] glucose + Pi = Glucose -6 Phosphate

ATP ----> ADP + Pi

Add both reactions ====>

Glucose + ATP ------> Glucose-6 Phosphate + ADP

K of first * K of second = Glucose - 6 Phospahte * ADP / ATP * Glucose

Glucose -6 Phosphate / Glucose = 3383.3

d] Ratio =338.3 as 10 fold decreased

delta G = -RT ln K

Initial delta G = -RT ln K1*K2 = 16.699 KJ /mol

K = 338.3 * 0.2 /0.8 = 84.575

delta G = -10.999KJ /mol

Change in delta G = 5.7 KJ /mol

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