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Turner 2004

Coaxial Stacking

Example 2

 

 

example structure        

∆G°37 coaxial stack = ∆G°37(Continuous Backbone Stack) + ∆G°37(Discontinuous Backbone Stack)

 

∆G°37 coaxial stack = ∆G°37(AU pair followed by GG mismatch) + ∆G°37(Discontinuous Backbone Stack)

 

∆G°37 coaxial stack= -0.8 kcal/mol – 2.1 kcal/mol

 

∆G°37 coaxial stack= -2.9 kcal/mol

 

∆H°coaxial stack = ∆H°(Continuous Backbone Stack) + ∆H°(Discontinuous Backbone Stack)

 

∆H°coaxial stack = ∆H°(AU pair followed by GG mismatch) + ∆H°(Discontinuous Backbone Stack)

 

∆H°coaxial stack = –3.5 kcal/mol – 8.46 kcal/mol

 

∆H°coaxial stack = –12.0 kcal/mol

 

 

Note that at this interface, the terminal AU pair penalty would still apply when calculating the helix stability.