Turner 1999
Watson-Crick Helices Example 1

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Turner 1999

Watson-Crick Helices

(Example 1)

 

 

        

∆G°₃₇ = ∆G°_{37 intermolecular initiation} + 2×∆G°_{37 AU end penalty}+ ∆G°_{37 symmetry} + ∆G°₃₇(AU followed by GC) + ∆G°₃₇(GC followed by CG) + ∆G°₃₇(CG followed by GC) + ∆G°₃₇(GC followed by CG) + ∆G°₃₇(CG followed by UA)

 

∆G°₃₇ = 4.09 kcal/mol + 2×0.45 kcal/mol + 0.43 kcal/mol –2.08 kcal/mol – 3.42 kcal/mol – 2.36 kcal/mol – 3.42 kcal/mol – 2.08 kcal/mol

 

∆G°₃₇ = –7.94 kcal/mol

 

 

 

Note that, for example, the parameter for (AU followed by GC) is the same as (CG followed by UA) because the correct directionality of the strands is preserved.

Page Revised 8/14/09