The answer is false. <span>In a </span>lytic infection, the virus enters the cell, reproduces viral offspring, and then releases the offspring as it destroys the cell.<span> that is the reason why it is false. i am 100 percent sure my answer is correct.
Hope this helps.
can you please make this the brainliest answer it would really help me. thank you :) </span>
Answer:
D. The methyl group of acetyl CoA becomes radio-labeled
Explanation:
During the steps in glycolysis, the carbon at position 1, becomes C-1 in dihydroxyacetone phosphate during the cleavage of fructose-1,6-bisphosphate to dihydroxyacetone phosphate and glyceraldehyde-3-phosphate. Subsequently on isomerization of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate, C-1 of dihydroxyacetone phosphate becomes C-3 of glyceraldehyde-3-phosphate.
Furthermore, in pyruvate, the end product of glycolysis, C-3 is converted to a methyl group which then becomes the methyl group in the acetyl-CoA molecule produced from the oxidative decarboxylation of pyruvate.
Since the radioactive 14-C of radio-labeled glucose occupies position 1, it will become the methyl group of acetyl-CoA.
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