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You are repeating the Hershey-Chase expe...

You are repeating the Hershey-Chase experiment and are provided with two isotopes `.^(32)Pand .^(15)N`(in place of `.^(35)S` in the original experiment ). How does yoe expect your results to be differnet?

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`""^(32)P` is radioactive isotope and `""^(15)N` is not radioactive (it is heavier isotope of nitrogen). Method for detection of `""^(32)P` and `""^(15)N` is different. Thus final conclusion can not be drawn by using `""^(15)N`. Suppose, even `""^(15)N` is considered radioactive, its presence will be shown inside the cell due to a incorporation of `""^(15)N` in nitrogenous bases DNA as well as in supernatant because it will be incorporated into amino groups of proteins also.
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