The first step in glycolysis is catalyzed by the enzyme hexokinase: glucose + AT

Question

The first step in glycolysis is catalyzed by the enzyme hexokinase: glucose + ATP doubleheadarrow glucose-6-? + ADP This reaction can be thought o f as two coupled reactions (P, is phosphate): glucose + P doubleheadarrow glucose-6-? ATP doubleheadarrow ADP + P, For the overall reaction, AG degree = -16. 7 kJ/mol. For the hydrolysis o f ATP (equation 2), AG degree = -31. kJ/mol. All reactions are performed at 37 degree C. What is delta G degree for reaction 1 ? Whal is [glcose- 6 - P]/ [gl ucose] [ phosphate] if G = 0? Starting with 1 M glucose and I M phosphate, what are the equilibrium concentrations o f glucose, phosphate, and glucose-6-phosphate if only reaction 1 occurs?

Explanation / Answer

a)
overall reaction can be obtained by adding reaction 1 and 2
delta G(over all) = delta G(rxn 1) + delta G(rxn 2)
-16.7 = delta G(rxn 1) - 31
delta G(rxn 1) = 14.3 KJ/mol
Answer: 14.3 KJ/mol

b)
[glucose 6P] / [glucose] [phosphate] is Kc for reaction1

since delta G= 0
delta Go (rxn 1) = -R*T* ln Kc
14300 J = -8.314*298* ln Kc
Kc=1.69*10^-6
Answer: 1.69*10^-6

c)
from reaction 1
glucose + phosphate --->glucose 6P
1                      1                             0    (initial)
1-x                 1-x                          x (at equilibrium)
Kc = [glucose 6P] / [glucose] [phosphate]
1.69*10^-6 = x / (1-x)^2
since KC is very small, x will be very small and it can be ignored as compared to 1
so above expression becomes:
1.69*10^-6 = x / 1
x= 1.69*10^-6 M

At equilibrium, glucose and phosphate will not change much as x is small
so,
[glucose 6P]= x= 1.69*10^-6 M
[glucose] =1M
[phosphate] =1M

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