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Turner 2004

Dangling Ends

Example

 

 

Example Structure        

∆G°37 = ∆G°37(Watson-Crick Helix) + ∆G°37(3' dangling C adajacent to GC) + ∆G°37(5' dangling A adjacent CG)

 

∆G°37 = –6.04 kcal/mol – 0.4 kcal/mol – 0.2 kcal/mol

 

∆G°37 = –6.6 kcal/mol

 

 

∆H° = ∆H°(Watson-Crick Helix) + ∆H°(3' dangling C adajacent to GC) + ∆H°(5' dangling A adjacent CG)

 

∆H° = –43.75 kcal/mol – 2.8 kcal/mol – 1.6 kcal/mol

 

∆H° = –48.2 kcal/mol

 

Note that this example contains both a 5' and a 3' dangling end (at opposite ends of the duplex).