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I'm reading an article about MLPA (Multiplex ligation-dependent probe amplification) and I got stuck on this sentence:

The advantage of splitting the probe into two parts is that only the ligated oligonucleotides, but not the unbound probe oligonucleotides, are amplified.

The principle I don't understand is why won't the unligated oligonucleotides be amplified? As is my current understanding of the PCR, the primers should normally anneal and produce fragments which would end at the end of an oligonucleotide (at the "ligation site" - so they would be shorter, but they should be there), am I wrong?

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  • $\begingroup$ Please edit your question and add a citation to the paper you are reading so we may see the sentence in context. $\endgroup$ – MattDMo Jan 10 '16 at 15:35
  • $\begingroup$ It's the wikipedia article which is linked to MLPA word. $\endgroup$ – mpribis Jan 10 '16 at 20:01
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I work for MRC-Holland, the developer of MLPA, and what Kacper explains is indeed the case. Any RPO (right probe oligo) that is not ligated is also amplified in the MLPA reaction, as the reverse primer can bind to the primer binding site of the RPO. HOWEVER, since this unligated RPO oligo LACKS the binding site for the (fluoresently labelled) forward primer, the amplification ends here. LIGATED probes, in contrast, contain the binding site for BOTH primers, and hence are amplified exponentially. Furthermore, the fluorescent label is only incorporated in the amplicons that results from ligated probes.

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I suppose unligated oligonucleotide would also be "amplified" but on much lesser scale(?only one part of probe?). For instance after 6th run of PCR you would get 2^6=64 amplicons of ligated probe but only 6 for one unligated part(?second starter is designed to anneal to second part of antisense oligonucletide of probe?). After 30 runs there is massive diference that allow you to only detect properly ligated probe. I'm not sure if I'm right but this is how I understood it. Peace:)

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  • $\begingroup$ Why is the down-vote please? I find this pretty reasonable. À propos, thanks for the answer. $\endgroup$ – mpribis Jan 30 '16 at 23:26

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