Synthesis of one molecule of glucose from two molecules of pyruvate via gluconeogenesis requires how many high-energy phosphate bonds, and at which two steps is ATP versus GTP consumed?
- A 6 total; 2 ATP at pyruvate carboxylase and 2 GTP at phosphoenolpyruvate carboxykinase plus 2 ATP at 3-phosphoglycerate kinase reversal ✓
- B 4 total; 2 ATP at pyruvate carboxylase and 2 GTP at phosphoenolpyruvate carboxykinase
- C 6 total; all six consumed at the pyruvate carboxylase step
- D 2 total; 1 ATP each at pyruvate carboxylase and PEP carboxykinase
Correct answer: A. 6 total; 2 ATP at pyruvate carboxylase and 2 GTP at phosphoenolpyruvate carboxykinase plus 2 ATP at 3-phosphoglycerate kinase reversal
Explanation
Per glucose made from two pyruvates: pyruvate carboxylase consumes 2 ATP, phosphoenolpyruvate carboxykinase consumes 2 GTP, and reversal of phosphoglycerate kinase consumes 2 ATP, totaling 6 high-energy phosphate bonds. This energy cost explains why gluconeogenesis is not simply the reverse of glycolysis and why it is energetically expensive compared with the 2 net ATP glycolysis yields.
Reference: Harper's Illustrated Biochemistry, 32nd ed.
High-yield for: NEET PGINI-CETNExTFMGEUSMLEPLABMRCP
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