The Hershey-Chase experiment was designed to prove that DNA is the genetic material in organisms. In this experiment, two batches of viruses were grown in two separate media A and B, with A having an abundance of nutrients involving $\ce{^32_15P}$, and B having a lot of $\ce{^35_16S}$.

Leaving solution B aside, we have $\ce{^32_15P}$ undergo a beta decay:

$$\ce{^32_15P -> ^32_16S + e- + \bar{\nu_e}}$$

This process releases $\pu{1.709 MeV}$ of energy, which is quite sufficient to break apart the phosphodiester linkages.

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With all the phosphorus in the viral DNA being radiolabelled, this would very well break apart all the phosphodiester linkages. and give us a soup of sulfur-based nucleosides. In such a case, how come this hasn't affected the virus' mechanism to transfer the DNA to a host cell?

  • $\begingroup$ The decay phenomena of Phosphorous would be , most likely, included in the experimental design since 14 days half life means, that if experiment is finished in 3 -4 days the loss is less than 20 %. $\endgroup$
    – ankit7540
    Commented Jun 9, 2017 at 5:50
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    $\begingroup$ @ankit7540 Agreed, but even if a few phosphorus disintegrated, it could create highly unstable species that form as the sulphur formed recoils out. $\endgroup$ Commented Jun 9, 2017 at 5:59

2 Answers 2


Of course it would break, just like you said; also, a high-energy $\beta$ particle would kill quite a lot of bystander molecules. Also, if not for other reason, the resulting molecule would no longer be DNA , since the decayed atom would no longer be $\rm P$. Also, the product would no longer be radioactive, so we wouldn't be able to detect it anyway.

The same applies to any other case of radiolabeling, in biochemistry or elsewhere, and you seem to be missing the very point of it: All this is not important.

Say, we have some $\ce{A + B -> C + D}$, and we label some atom in $\ce{A}$ so as to find out whether it will go to $\ce{C}$ or to $\ce{D}$. Then we run the reaction as usual. Now, radioactive atoms do not decay all at once; some decay earlier and some later. Those which decay earlier (that is, before and during the reaction) are lost to us, and we don't care about them one bit. Then we separate $\ce{C}$ from $\ce{D}$ and measure the radioactivity of each, and it is those atoms which decay now that matter.

Oh, and it is not like we label all atoms in $\ce{A}$. No, a very tiny fraction would suffice. Those viruses would probably get but a few radioactive atoms each. Some will be lucky enough to reproduce. Some will be killed by an early decay and remembered as martyrs of science.

So it goes.

  • $\begingroup$ Thanks for your answer. I have a few doubts. If we radiolabel $\ce{A}$, and it disintegrates as the reaction is run, how come the unstable species produced as a result of recoil of radioatoms not interfere with the reaction? Couldnt such a random unstable intermediate combine with $\ce{B}$ and form some other product $\ce{E}$ instead? $\endgroup$ Commented Jun 9, 2017 at 6:51
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    $\begingroup$ Sure, they would form all kinds of side products. It is just that we don't care, because we have extremely few of those. Again, the fraction of labeled molecules in $\rm A$ is so small that from chemical point of view they don't exist at all. Only the decay event makes them visible. $\endgroup$ Commented Jun 9, 2017 at 7:16

Why the disintegration of $\ce{^32_15P}$ does not affect the experiment. Following reasons:

  1. As mentioned since the experiment is relatively short relative to the half-life of the radioactive element under question. From the original paper (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2147348/) it appears that 1-2 days would be maximum for needed for a set of experiment. And this work involved several sets of experiments. In such case, the theoretical amount of remaining radio-active element is > 88%.

  2. The typical number of cells in this experiment was $\ce{10^9}$ cells per $\ce{mL}$. And starting cell culture broth usually involves several $\ce{mL}$s. Hence, the total phages would be very large in number. This means that of the < 12% cells can have $\ce{^32_15P}$ disintegration (assuming the maximal time of experiment). Even if going by this maximal approach this experiment would be accurate to give correct interpretation since the radio-active assay used has $\ce{\pm}20$ % error (check paper). Again, as mentioned by Ivan Neretin, the disintegration is random and and can happen in any cell absolutely randomly, so considering one phospho-diester breakage in a cell would not cause the phage to die. I am not sure but there might be correction mechanisms too.

Why this disintegration not affects the virus' mechanism to transfer the DNA to a host cell ?

Consider a phage which has undergone such disintegration. If the phage(virus) survives and somehow can still replicate its genetic material it can continue to act on bacteria. But nobody knows what happens !

If it dies, then when centrifugation is done it either be sticking to the bacteria or be in the broth(culture solution). Then either it can go with bacteria or be separated as lower weight fraction. This might skew the result a little though !

In this work, before releasing phage on bacteria initial separation of phages for labelling was done.

The $\ce{^35_16S}$ is a stable species and can only affect the numerical accuracy of this experiment by < 12% by the maximal disintegration as discussed above.


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