Answer:
50%
Explanation:
Glucose is a simple sugar with a total of 6 carbon atoms in its structure. Pyruvate has a total of three carbon atoms. Two molecules of pyruvate are obtained per glucose by glycolysis. None of the carbon of glucose is released in the form of CO2 during glycolysis. Therefore, the radio-labeled C-1 of glucose will be the component of the carbon skeleton of one of the total of two pyruvate molecules produced during glycolysis. So, 50% of the pyruvate will exhibit radioactivity.
For #1, the answer is A) Ribosomes.
