If glucose is labeled with 14C at the C1 position only, which of the following would occur if the glucose underwent glycolysis?
Hint: Open chain form of Glucose has an aldehyde group at C1

Question

BlackTom AI · Accepted Answer

We start by restating the scenario and options to ensure clarity of what’s being evaluated.
Question: If glucose is labeled with 14C at the C1 position only, which of the following would occur if the glucose underwent glycolysis?
Answer options:
1) 50% of the pyruvate would contain 14C.
2) 100% of the pyruvate would contain 14C.
3) 14C would be found as part of the carboxyl group of pyruvate.
4) The 14C would be lost as CO2.
5) None of the above.

Option 1 analysis: This is the correct understanding. During glycolysis, one molecule of glucose (a 6-carbon sugar) is split into two molecules of glyceraldehyde-3-phosphate, which are then processed to two molecules of pyruvate (each with 3 carbons). The labeled C1 carbon of glucose ends up in only one of the two resulting pyruvate molecules, not both. Therefore, about half of the pyruvate molecules produced would carry the 14C label, matching 50% of the pyruvate containing 14C.

Option 2 analysis: This would imply that both pyruvate molecules retained the 14C label. Since glycolysis splits the six-carbon glucose into two three-carbon units, the label at C1 cannot appear in both pyruvates. Thus this statement is incorrect.

Option 3 analysis: It’s true that in one of the pyruvate molecules the labeled carbon can end up as the carboxyl carbon, but the statement claims that the 14C would be found as part of the carboxyl group of pyruvate in all pyruvate molecules, which is not accurate. Only one of the two pyruvates would carry the label, and depending on the carbon mapping, it would correspond to a specific carbon position (often the carboxyl carbon in that particular labeled pyruvate). Since the statement does not specify that this occurs in both pyruvates, it is an incomplete/general claim and can be misleading.

Option 4 analysis: The idea that the 14C is lost as CO2 during glycolysis is incorrect. CO2 release from carbon labeling occurs later in metabolism (e.g., decarboxylation steps in the citric acid cycle as the pyruvate is converted to acetyl-CoA and CO2 is released). In glycolysis itself, the label is not released as CO2.

Option 5 analysis: Given that option 1 accurately describes the outcome, this “None of the above” choice is incorrect.

In summary, the key conceptual point is that glycolysis divides the glucose molecule into two three-carbon fragments, distributing the labeled carbon into only one of the resulting pyruvates, which leads to approximately 50% of the pyruvate molecules carrying the 14C label. The other options misstate the distribution or fate of the label during glycolysis or rely on events (like CO2 loss) that occur outside glycolysis.

If glucose is labeled with 14C at the C1 position only, which of the following would occur if the glucose underwent glycolysis? Hint: Open chain form of Glucose has an aldehyde group at C1

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