Question about the bonds between adjacent nucleotides

This question is about the bond(s) between adjacent consecutive nucleotides, for example, let’s use nt #10 = A and nt #11 = G.

Is there a specific A-then-G bond that exists in the strand that is used during synthesis? Are all the A-then-G bonds in the synthesis strand the same?

Are there possibly different bonds between the A-then-G that exist in nature?

I am asking because sometimes the G needs to be in a different position relative to the preceding A, for example when the G is first “loose” nt at the opening to a loop.

If there are multiple options for A-then-G bonds that exist in nature, but only one in the synthesis strand, it might help to explain why the synthesis results are different from nature. Hope this makes sense.

I think I get what you’re saying. Here’s my two cents worth… for what it’s worth lol. I just got back from a super bowl party, so blame any inaccuracies on the Patriots.

The interaction between two consecutive bases in a nucleic acid is usually along the lines of the forces that cause two droplets of oil to merge when they’re floating in a glass of water. RNA bases are flat and vaguely coin-shaped. The top/bottom faces are “greasy,” while the edges are closer to water in their ability to form H-bonds. When the RNA molecule is solvated by water, the greasy parts clump together. The strength of this interaction varies with the orientation of the base pairs and identity of the bases.

Sometimes, the strand twists in a way that allows H-bonding between adjacent nucleotides. This happens when the FMN aptamer binds FMN, for example… but I doubt it’s possible to look at a sequence and accurately predict when this will happen.

I’m not sure why the G residue needs to be at the lowest-numbered (5’-most) position of a hairpin to lower energy, but from what I’ve learned about hairpin loops they often contain unusual patterns of H-bonding and stacking. Maybe the well-studied, naturally-occurring loops used to create energy parameters tend to have G in this position. It may be that G has to be in the 5’ most position (intstead of the 3’-most position) in order to expose a part of the residue that allows it to H-bond with other bases in the loop.

Hi Quasi,

Thanks for your response. Your description is interesting, and I will be thinking about it.

Also, thanks for your player projects. I am following several of them,