Oligoribonucleotide X was treated with phosphatase (for removal of 3' and 5' - terminal phosphates), then with RNAase T1, which cleaves all phosphodiester bonds located in a 3' position of guanosine in a 5'-specific manner. enter image description here

As a result, oligonucleotides L, M and N were generated in equal amounts. Each of them was further treated with phosphatase and subjected to alkaline hydrolysis. Results are listed in the table below.

enter image description here

Then experiment was modified: oligoribonucleotide X after treatment with phosphatase was hydrolyzed with RNAaseP, which cleaves all phosphodiester bonds in a 3'-position of pyrimidines in a 5' - specific manner.

enter image description here

This hydrolysis yielded five products in approximately equimolar concentrations: uridine monophosphate, cytidine monophosphate and oligonucleotides P, Q and R. After resolution of the mixture and alkaline hydrolysis of these oligonucleotides data listed in the table below were obtained.

enter image description here

My thinking : G can not be at the 5' end otherwise we would get a single Guanosine also in the first experiment. G can also not be at the second last from 3' end as we have not got any singles. We have got A and C in a duplet and so they must have been at the 3' end.

I also have trouble understanding the significance of "mole/mole of oligoribonucleotide".

I am trying to fit in other results but am unable to do so. A GOOD HINT WILL BE APPRECIATED MORE THAN A DIRECT ANSWER.


  • $\begingroup$ @Satwik In the above comment - do you mean UMP and CMP ? $\endgroup$
    – biogirl
    Nov 26 '13 at 16:37
  • $\begingroup$ @SatwikPasani And by doublet u don't mean they are 1 after the other because I don't think that is possible. $\endgroup$
    – biogirl
    Nov 26 '13 at 16:48
  • $\begingroup$ @biogirl It's not clear from what you have written so far that you realise the significance of getting a cytidine rather than a CMP in one of the products - it means, for example that you can deduce the sequence of M, which in turn allows you to deduce the sequence of Q. $\endgroup$
    – Alan Boyd
    Nov 26 '13 at 18:35
  • $\begingroup$ @biogirl Did you solve this? If yes, I wanted to know the answer they(the source of the question) provide, to make sure I had the correct solution. $\endgroup$ Nov 30 '13 at 4:40
  • 1
    $\begingroup$ @SatwikPasani ok i have done it. It's right. Thanks a ton ! $\endgroup$
    – biogirl
    Dec 11 '13 at 18:00

Here are a few hints which might help you:-

1) You get free UMP and CMP (in contrast to UMP + C or CMP + U) in the second experiment before alkaline hydrolysis.

2) There are two CMP in "N" oligoribonucleotide, and a free cytidine (and not CMP) in "M" and "Q".

3)Use the fact that all oligoribonucleotides were treated with phosphatase before being subjected to alkaline hydrolysis.

(Referring to your comment, I dont know why do you think consecutive cytidine is not possible).

These are a few hints that helped me solve this. If you want further clarification or the answer, tell me in the comments.

Answer for reverse derivation:- $5'-UACGCCGAC-3'$


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