If glucose is labeled with 14C in position C-1, where does this label end-up if the glucose goes through the oxidative phase of the pentose phosphate pathway?
To determine where the label ends up if glucose labeled with 14C at position C-1 goes through the oxidative phase of the pentose phosphate pathway, we need to understand the key reactions involved.
The oxidative phase of the pentose phosphate pathway includes two reaction steps: glucose-6-phosphate dehydrogenase (G6PDH) and 6-phosphogluconate dehydrogenase (6PGDH).
1. Glucose-6-phosphate dehydrogenase (G6PDH): In this reaction, glucose-6-phosphate (G6P) is oxidized to 6-phosphoglucono-δ-lactone (6PGL), generating NADPH in the process. This reaction does not change the carbon skeleton of glucose, so the label remains at C-1.
2. 6-phosphogluconate dehydrogenase (6PGDH): In this reaction, 6-phosphogluconate (6PG) is further oxidized to ribulose-5-phosphate (Ru5P), producing another NADPH molecule. During this reaction, the carbon skeleton of glucose is rearranged, resulting in the label being distributed throughout the product molecules. The label can be found on various carbon positions within the pentose sugar Ru5P.
In summary, when glucose labeled with 14C at position C-1 goes through the oxidative phase of the pentose phosphate pathway, the label ends up in various carbon positions of the pentose sugar Ru5P.