If C-1 of glucose is radiolabeled and the glucose then enters glycolysis, what percentage of the pyruvate molecules will carry the radioactive label? (Answer with a number only.)

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.

Explanation:

Viruses depend on the host cells that they infect to reproduce. Outside of host cells, viruses exist as a protein coat or capsid, sometimes enclosed within a membrane. The capsid encloses either DNA or RNA which codes for the virus.