Do 4' to 5' Phosphodiester Bonds Exist in Oligonucleotides?

[Soaring Crane]

Can 4' to 5' phosphodiester bonds, rather than 5' to 3' bonds, in reference to oligonucleotides exist? If so, then what would the structure look like (especially since carbon 4 on the sugar lacks a hydroxyl group)?


Thank you.

 

[Ygggdrasil]

I suppose such a bond might be possible if ribose is in the straight-chain form rather than the cyclic form (in the cyclic form the OH on carbon 4 reacts with carbon 1 to form the ring structure). However, I don't know how likely such a reaction is or if it is even possible. Certainly I'm not familiar with any examples of structures with a 4' to 5' phosphodiester bond.

 

[Soaring Crane]

Assuming that the oligonucleotide is comprised of nucleotides with deoxyribose (D isomer), then how would the bonds concerning the sugar’s noncyclic form, or representation other than the Haworth projection, be depicted?

Would the base still be attached to carbon 1? If so, what happens to the carbonyl group?

Normally, when I see the 5’ to 3’ phosphodiester linkage, the phosphate group on 5’ (cyclic sugar) comes first, and, then, it is followed by the sugar. The second phosphate group follows on 3’. For the straight chain sugar, how are the phosphate groups organized in relation to one another? For 4’ to 5’, would the phosphate groups be on the same side (be in the same positions where their respective OH groups were)?

Thank you for any response.

 

[Ygggdrasil]

You make an excelent point that I overlooked earlier. Because the base is attached to the 1' carbon of the deoxyribose sugar, the sugar would not be able to convert to its straight chain form.