User:ElNando888/Blog/Conformational 2

From EteRNA WiKi
Jump to: navigation, search

(back to Conformational 1)

 

Of course, you know this already: the minimum free energy structure is only one of the many possible shapes that a given RNA sequence can take. And the Laws of Thermodynamics tell us that it is the state which this sequence will tend to. Though, if we consider a "batch" of identical sequences in a solution (a test tube, some water and a little magnesium salt), not all the sequences will form this "very best shape" called MFE, or even stay there once reached. An equilibrium will be reached, and the only thing that we (in principle) can say for sure, is that the count of the sequences that take the MFE shape will always be greater than the count of any other individual shapes (let's forget for now the issue of energy barriers and kinetic traps)

Taking a simple example, let's assume that a sequence S can take shapes A, B, C and D. In a solution containing 10 such sequences, we find 4 with shape A, 3 with shape B, 2 with shape C and 1 with shape D. Obviously, A is the MFE, and B, C and D are what we call "suboptimals". Notice that the MFE is only 40% of the total, and yet, it is the MFE.

And there is a strict equivalence between the difference in free energy and the relative concentration of the shapes in a solution. This is true also for other molecules than RNA, but we have some numbers and formulas about RNA itself:

https://getsatisfaction.com/eternagame/topics/relative_concentrations_of_rna_shapes#reply_10948182

Thanks Rhiju :)

 


So, what does this tell us about switches? Well, let's call the unbound state A, and its free energy FE(A). We also have the bound state, which I'll call B+ (because it's got a molecule bonus) with free energy FE(B+). And we should not forget B-, which is the same shape as the bound state, but without the molecule bonus, with its FE(B-) free energy. And the minimum conditions for a switch to work (in principle) can be represented by following inequation:

 

   
FE(B-) > FE(A) > FE(B+)

 

Alright. That's the principle. In theory, any free energy values that verify the above condition should work. How does this translate in switch labs?

 

Switch labs consist basically of 2 experiments, once without the ligand, and once with it. In the first case (no bonuses), there will be A and suboptimals, among them B-. What happens when FE(B-) is very close from FE(A)? As we saw above, it means that the difference in free energy is low, which indicates that the concentrations of the shapes will be similar. So there will be almost as many B- than A's. In the context of our EteRNA experiments, with the way the labs are scored, this is bad news.

Similarly, if FE(B+) is only marginally lower than FE(A), then in the presence of the ligand, there will be barely more bound shapes (B+) than unbound ones (A).

Intuitively, it seems that the best combination is when FE(A) is right in the middle between FE(B-) and FE(B+). For our FMN-based labs, this would mean that the theoritical best free energy difference should be 4.86 / 2 = 2.43 kcal/mol...

 

...

 

Wait... isn't there a "wisdom" about switches in EteRNA labs, that says that energy difference between states is best around 1 kcal/mol?

 

Conformational 3

Personal tools
Main page
Introduction to the Game