Question
The oxidation of glyceraldehyde- 3 -phosphate to 1,3 -bisphosphoglycerate is an endergonic reaction, but the flow through this point in glycolysis proceeds smoothly. How is the unfavorable equilibrium constant overcome?
Step 1
This is represented by the equation: \[ \text{Glyceraldehyde-3-phosphate} + \text{NAD}^+ + \text{P}_i \rightarrow \text{1,3-bisphosphoglycerate} + \text{NADH} + \text{H}^+ \] Show more…
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The oxidation of glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate, catalyzed by glyceraldehyde 3-phosphate dehydrogenase, proceeds with an unfavorable equilibrium constant (K'eq = 0.08; ΔG'° = 6.3 kJ/mol), yet the flow through this point in the glycolytic pathway proceeds smoothly. How does the cell overcome the unfavorable equilibrium?
Free-Energy Change for Triose Phosphate Oxidation The oxidation of glyceraldehyde 3 -phosphate to 1,3 -bisphosphoglycerate, catalyzed by glyceraldehyde 3 phosphate dehydrogenase, proceeds with an unfavorable equilibrium constant $\left(K_{\mathrm{eq}}^{\prime}=\right.$ $\left.0.08 ; \Delta G^{\prime 0}=6.3 \mathrm{kJ} / \mathrm{mol}\right),$ yet the flow through this point in the glycolytic pathway proceeds smoothly. How does the cell overcome the unfavorable equilibrium?
The oxidation of glyceraldehyde 3 -phosphate to 1,3 -bisphosphoglycerate, catalyzed by glyceraldehyde 3 phosphate dehydrogenase, proceeds with an unfavorable equilibrium constant $\left(\boldsymbol{K}_{\mathrm{eq}}^{\prime}=\right.$ $0.08 ; \Delta G^{\prime \circ}=6.3 \mathrm{kJ} / \mathrm{mol}$ ), yet the flow through this point in the glycolytic pathway proceeds smoothly. How does the cell overcome the unfavorable equilibrium?
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