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I have a problem with a molecular biology question; I don’t understand how DNA 3’ labelling works. I took a diagram from my lesson and tried to understand with it; this is what I understood. If I’m wrong, please tell me and correct me.

  • We take a restriction enzyme which will cut both strands. The bottom strand will be a 5’ end. So, the top strand is shorter than the bottom strand.

  • We will add a Klenow fragment which, with its polymerase activity, polymerizes in the 5’->3’ direction the missing part of the top strand, with dNTP α 32P (labelled by a radioactive phosphate in the alpha position).

  • We get two strands which have the same size, of which one will be strongly labelled.

  • We finish with a terminal transferase which will catalyse the addition of dNTP α 32P on the 3’OH ends of both DNA strands (in order to label these ends).

But, if I’m right, could you explain the necessity of the last step; why label both ends since we have already labelled the top strand?

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Welcome! It will help if you explain the purpose of labeling the DNA according to the lesson: Are you creating a radioactive probe for in situ hybridization? For Northern blotting? Is this a RACE assay? –  Drosophila Feb 26 '13 at 22:25
    
The goal of this label is not defined ; it is just in the part "how to label DNA " ? I see that "in situ hybridization", "Northem blotting" are quoted after this part. –  Math's Feb 27 '13 at 8:41
    
Are you asking why it is necessary to label both strands in the final step versus just labeling the bottom strand, or are you asking why it is necessary to label both strands versus labeling only the top strand? –  dd3 Mar 20 '13 at 18:00

1 Answer 1

There are many ways to label nucleic acids for probes (Summary of Methods).

While there are cases where Klenow fragment-mediated fill-in should sufficiently label both strands of your probe, there are many others where this would not be enough. Here are two illustrative, although somewhat contrived, examples:

Example 1: Double digest with 5' and 3' overhangs

Imagine you did a double digest with BamHI and KpnI to release your probe sequence from the plasmid in which you propagate the sequence. BamHI will give you a 5' overhang, but KpnI gives a 3' overhang.

Resulting BamHI & KpnI double digest fragment:
  5'-    GATCCNNNNNNNG      - 3'
             |||||||||
  3'-        GNNNNNNNCCATG  - 5'

Subsequent treatment with the Klenow fragment + $\alpha$P32-dATP + cold dNTPs will both fill-in the 5'-overhang (Klenow's 5'->3' polymerase activity) and remove the 3' overhang (Klenow's 3'->5' exonuclease activity).

Post-Klenow reaction (bold = new bases, * = radio-labeled nucleotide):
  5'-    GATCCNNNNNNNG      - 3'
         |||||||||||||
  3'-    CTAGGNNNNNNNC      - 5'
           *

At this point only one strand is radio-labeled so the terminal transferase step is needed to add a label to the top strand.

When using $\alpha$P32-dNTPs, you often pick only one base to be radioactive (eg dATP as above). Sometimes you also just spike the reaction with radioactive dNTP leaving a population of cold dNTPs to be used most of the time. In either case you now have the chance of not incorporating only non-radioactive bases.

Example 2: Fill-in reaction that fails to incorporate $\alpha$P32-dATP:

Imagine a digest with PspI and want to fill-in the 5' overhangs using $\alpha$P32-dATP + cold dNTPs. PspI recognizes CCWGG sequences where W is either A or T so that some fill-ins will incorporate the $\alpha$P32-dATP, but others will not.

Possible digests and fill-in reactions:
   5'-  CCTGGNNNNNNNNNNCCTGG - 3'
        ||||||||||||||||||||
   3'-  GGACCNNNNNNNNNNGGACC - 5'
                 |
         Digest with PspI
                 |
                 V
   5'-  CCTGGNNNNNNNNNN      - 3'
             ||||||||||
   3'-       NNNNNNNNNNGGACC - 5'
                 |
        Fill-in with Klenow + hot dATP + cold dNTPs
                 |
                 V
   5'-  CCTGGNNNNNNNNNNCCTGG - 3'
        ||||||||||||||||||||
   3'-  GGACCNNNNNNNNNNGGACC - 5'
          *
 

Again you are left with only one strand being labeled (or no strand labeling if the sequence was 5'- CCAGGNNNNNCCTGG - 3') and a need for the terminal transferase step.

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