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Am I correct in thinking that in a standard probe synthesis reaction where primer/template are heated up and then cooled down for Klenow to extend, the theoretical maximum amplification is doubling of the template?

Given that there is essentially only one melting/primer hybridization step, there can't be further amplification after doubling?

I understand that there could be edge cases. IE. random hexamers are used where initial amplification product is displaced by a further upstream hexamer. But given an idea situation where we have just one specific primer and one specific dsDNA template that primers can bind to, the maximum yield will be double?

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  • $\begingroup$ Which version of Klenow — i.e. are you asking about the fragment with or without 3'→5' exonuclease activity? $\endgroup$
    – tyersome
    Sep 26, 2019 at 17:57

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Klenow seems to lack exonuclease activity ("proof-reading"), but otherwise works like normal DNA polymerase, so I wouldn't expect anything different from PCR. On each step, as long as there is enough template and primers, you'll get doubling of current template.

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