If I have a predicted RNA stem-loop of energy -0.30 kcal/mol, and another of -4.9 kcal/mol, how do I tell whether such a structure is a significantly stable structure or not? Our two competing hypotheses are that there is some structure with sufficient energetic stability to fold inside a cell, versus the one that there is not.

  • $\begingroup$ A predicted stem loop of what? A nucleotide chain presumably, and most likely RNA but you need to tell us. $\endgroup$
    – terdon
    Apr 23 '14 at 0:34
  • $\begingroup$ Yes, it is an RNA. Allow me to update the original question. Many apologies for being unclear. $\endgroup$
    – ericmjl
    Apr 23 '14 at 0:46
  • $\begingroup$ probably the best way is to make it and see if the sequence forms a stem loop. there are lots of energy calculations based on 3d models and sequences if you'd like to know, but i'm not clear on what is best myself. $\endgroup$
    – shigeta
    Apr 23 '14 at 2:41
  • $\begingroup$ I don't know what you mean by significant. More negative the value of folding energy more is the stability. For better predictions you can use the salt correction formulae. $\endgroup$
    Apr 23 '14 at 3:21
  • $\begingroup$ are you working on miRNA or tRNA or just other rnas? $\endgroup$
    – user1357
    Apr 23 '14 at 6:37

You can convert the $\Delta G$ values into equilibrium constants $K$ using the formula

$$ \Delta G = R T \ln K $$

This might help you understand what those energies roughly mean.

But the more important and also more difficult part is judging how good your predicted free energies actually are. I assume you used a secondary structure prediction program like Mfold to calculate this. They typically have a several parameters you can adjust like ion concentrations and temperature, you should make sure that you have some appropriate values here.

Without knowing anything more about your system or data I'd ignore the -0.3 kcal/mole result, the -4.9 kcal/mole structure could be formed, but I wouldn't regard this as certain.

Now, in the in vivo situation the whole thing gets far more complicated. The presence of RNA binding proteins, helicases and the generally rather complex environment in the cell makes it much harder to predict which secondary structures will actually form.

If you wanted to know for sure you could e.g. do SHAPE experiments on your RNA. This is a technique that also works in vivo.

  • $\begingroup$ no thats too much work $\endgroup$
    – user1357
    Apr 23 '14 at 6:37
  • 3
    $\begingroup$ RNA folding is definitely a lot of work if you are not an RNA molecule :) $\endgroup$
    – shigeta
    Apr 23 '14 at 13:20

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