Question

The experiment conducted by Hershey and Chase was replicated on a mutant bacteriophage which lacked methionine, cysteine, homocysteine and taurine. What would be the expected result?

• A. Radioactivity will be observed only outside the bacterial cell
• B. No radioactivity will be observed inside the cell
• C. No radioactivity will be observed outside the cell
• D. Radioactivity will be observed both inside and outside the cell

Answer 1

The correct option is C No radioactivity will be observed outside the cell

Methionine, cysteine, homocystein and taurine are sulfur containing amino acids. Aminoacids are important part of the protein coat of the bacteriophages. If the mutant bacteriophage lacking these aminoacids are grown in radioactive sulphur then no radioactivity will be observed outside the cell.

Viruses grown in the presence of radioactive phosphorous contains radioactive DNA. So radioactivity will be observed inside the cell.