enter image description hereI have a linear ssDNA oligo that is designed such that the ends are complementary, and conducive to hairpin formation (see attached figure). My question is, which has a higher likelihood of formation, a hairpin, or a self-dimer? Would the likelihood of hairpin formation increase if I decrease the concentration of the ssDNA oligo during the annealing process? If I enter the sequence in IDT oligo analyzer, the gibbs free energy for formation of self-dimer is significantly lower compared to hairpin. The melting temperature of the complementary part of my oligo is 50 deg. The annealing will be done in nuclease free water, without any Mg+.


I have tested a similar design to this. With a duplex protocol the chance of forming a hairpin is much higher than that of forming self-dimer, even at concentration of 25 μM. You can run a 3-5% agarose gel to test the outcome.

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    $\begingroup$ Welcome to SE Biology. Your answer sounds reasonable but we have no way of judging whether it is correct for lack of evidence. Do you have your own gel to illustrate this? $\endgroup$ – David Aug 2 '19 at 17:14

It seems kinetics have more thing to do than thermodynamics.

Intuitively it seems if you take this DNA at a low concentration, then the 2 termini of the same molecule will have more probability to find each other since they are within same molecule, as well the 60 nucleotide intermediate portion will bend. So the hairpin form is likely to form at lower concentration of this ssDNA.

Now from this condition if they have to go to the linear form, they have to first unwind the paired region. That is the "uphill" region of the kinetics, or the energetically unfavourable action.


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