Turner 1999 Hairpin Example 4

example structure

∆G°₃₇ = ∆G°₃₇(Watson-Crick Pairs) + ∆G°₃₇(Hairpin Loop)

∆G°₃₇ = ∆G°₃₇(Watson-Crick Pairs) + ∆G°₃₇(terminal mismatch) + ∆G°_{37 Hairpin initiation}(6) + ΔG°_{37
penalty}(all C loops)

∆G°₃₇ = ∆G°₃₇(CG followed by AU) + ∆G°₃₇(AU followed by CG) + ∆G°₃₇(CG followed by AU) + ∆G°_{37 AU end penalty}+ ∆G°₃₇(AU followed by CC) + ∆G°_{37 Hairpin initiation}(6) + 6×A + B

∆G°₃₇ = –2.11 kcal/mol – 2.24 kcal/mol – 2.11 kcal/mol + 0.45 kcal/mol – 0.7 kcal/mol + 5.4 kcal/mol + 6×0.3 kcal/mol + 1.6 kcal/mol

∆G°₃₇ = +2.1 kcal/mol

Turner 1999
Hairpin Example 4

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