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 experimen

Question

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 experimen

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?

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Use of `.^(15)N` was redioactive then its pressure would have been detected, of `.^(32)P and .^(15)N` differnet `(.^(32)`P being a radioactive isotope while`.^(15)N` is non-radioactive but is the heavier isotope of nitrogen).
Even if `.^(15)`n was radioactive then its presence would have been detected, both inside the cell (`.^(15)N` incorporated as introgenous base in DNA) as well as in the supernatant, because `.^(15)N` would also get incorporated in amino group of amino acids in proteins. Hence, the use of `.^(15)`N would not give any conclusive results.