Why re-label the strand ends in 3' DNA labelling?

Question

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?

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.