Turner 2004
Watson-Crick Helices Example 1

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Turner 2004

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

 

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

 

∆H°= 3.61 kcal/mol + 2×3.72 kcal/mol – 10.48 kcal/mol – 14.88 kcal/mol – 10.64 kcal/mol – 14.88 kcal/mol – 10.48 kcal/mol

 

∆H°= –50.31 kcal/mol

 

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

Page Revised 8/14/09