This question investigates glucose metabolism in liver cells under aerobic condi

Question

This question investigates glucose metabolism in liver cells under aerobic conditions. The cells are given glucose labeled with radioactive carbon (C-14) in positions 3 and 4. 90% of the radioactivity is released as radioactive carbon dioxide within 3 minutes. a) Account for the formation of radioactive carbon dioxide. What metabolic pathways and reactions are responsible? b) The experiment is repeated in the presence of arsenate. How would you expect the amount of radioactive carbon dioxide produced in 3 minutes to compare with the previous experiment? Why? c) The same experiment is repeated a third time but now in the presence of arsenite. How would the production of radioactive carbon dioxide now compare with the earlier two experiments? Why? Be as specific as possible in accounting for any difference. d) Which ion, arsenate or arsenite, would you expect to have a more devastating effect on the liver cells and why? Think about this ONLY in terms of carbohydrate metabolism.

Explanation / Answer

a) Since the glucose is labelled at position 3 and 4, the pyruvate which is formed during glycolysis from glucose will also be labelled. During oxidative decarboxylation by pyruvate decarboxylase to form acetyl CoA (before it enters TCA cyle), the carbon is lost as CO2. Since the carbon is labelled, radioactive CO2 will be produced. The responsible reactions are glycolysis followed by oxidative decarboxylation.

b)

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