321 a thermodynamic treatment allows predictions of the

Question

3.21 A thermodynamic treatment allows predictions of the stability of DNA. The table below lists the standard Gibbs energies, enthalpies,

and entropies of formation at 298 K of short sequences of base pairs as two polynucleotide chains come together:

Sequence                              5′-A-G                  5′-G-C                      5′-T-G

                                               3′-T-C                  3′-C-G                      3′-A-C

D_seqG^o/(kJ mol^-1)                -5.4                         -10.5                      -6.7

D_seqH^o/(kJ mol^-1)                -25.5                    -46.4                        -31.0

D_seqS^o(kJ mol^-1)                 -67.4                    -118.8                       -80.8

To estimate the standard Gibbs energy of formation of a double stranded piece of DNA, DDNAG3, we sum the contributions from the

formation of the sequences and add to that quantity the standard Gibbs energy of initiation of the process, which in the case treated

in this exercise may be set equal to D_initG^o = +14.2 kJ mol-1: D_DNAG^o = D_initG^o + D_seqG^o(sequences)

Similar procedures lead to D_DNAH^o and D_DNAS^o.

 

 

(a) Provide a molecular explanation for the fact that D_initG^o is positive and D_seqG^o negative.

(b) Estimate the standard Gibbs energy, enthalpy, and entropy changes for the following reaction:

                5′-A-G-C-T-G-3′ + 5′-C-A-G-C-T- 3′ →    5′-A-G-C-T-G-3′

                                                                                          3′-T-C-G-A-C-5′

            Use D_initH^o = +2.5 kJ mol^-1

               and D_initS^o = -37.7 J K^-1mol^-1

(c) Estimate the 'melting' temperature of the piece of DNA shown in part (b).