Chemistry Labs

Problem 1

DNA. Palindromic double-stranded DNA consists of two identical strands complementary to each other, e.g. the Drew–Dickerson dodecanucleotide 5'-CGCGAATTCGCG-3'. (a) How many different palindromic dsDNA dodecanucleotides (12 base pairs) exist? (b) How many palindromic dsDNA undecanucleotides (11 base pairs) exist? (c) Assume a G–C pair stabilizes the duplex more than an A–T pair; what is the probability that replacing one randomly selected base pair of the dodecanucleotide by a G–C pair increases its melting temperature TmT_m? (d) A dsDNA solution with cinit=1.00×10−6c_{init} = 1.00\times10^{-6} mol dm−3^{-3} is heated to TmT_m (50% dissociated). Calculate the association equilibrium constant at TmT_m for a non-palindromic dsDNA (KnpK_{np}) and a palindromic dsDNA (KpK_p), taking the standard concentration c0=1c^0 = 1 mol dm−3^{-3}. (e) The mean Gibbs energies of association per base pair are −6.07-6.07 kJ mol−1^{-1} (G–C) and −1.30-1.30 kJ mol−1^{-1} (A–T). At Tm=330T_m = 330 K, use Knp=1.00×106K_{np} = 1.00\times10^{6} and Kp=1.00×105K_p = 1.00\times10^{5}: how many base pairs has the shortest dsDNA with TmT_m above 330 K, and is it palindromic? (f) The inverse melting temperature of the dodecanucleotide varies linearly with ln⁡(2cinit/c0)\ln(2c_{init}/c^0): for cinit/10−6c_{init}/10^{-6} = 0.25, 0.50, 1.00, 2.0, 4.0, 8.0 mol dm−3^{-3}, TmT_m = 319.0, 320.4, 321.8, 323.3, 324.7, 326.2 K. Calculate the standard enthalpy ΔH∘\Delta H^{\circ} and entropy ΔS∘\Delta S^{\circ} of strand association.
Step 5 of 5: Extract ΔH∘\Delta H^{\circ} and ΔS∘\Delta S^{\circ}
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1Tm=ΔS∘ΔH∘−RΔH∘ln⁡ ⁣2cinitc0  ⇒  slope=−2.0×10−5 K−1⇒ΔH∘=−416 kJ mol−1, ΔS∘=−1.18 kJ K−1mol−1\dfrac{1}{T_m} = \dfrac{\Delta S^{\circ}}{\Delta H^{\circ}} - \dfrac{R}{\Delta H^{\circ}}\ln\!\frac{2c_{init}}{c^0} \;\Rightarrow\; \text{slope} = -2.0\times10^{-5}\ \text{K}^{-1} \Rightarrow \Delta H^{\circ} = -416\ \text{kJ mol}^{-1},\ \Delta S^{\circ} = -1.18\ \text{kJ K}^{-1}\text{mol}^{-1}
Analysis

Combining ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT\ln K with Km=c0/(2cinit)K_m = c^0/(2c_{init}) gives 1/Tm=ΔS∘/ΔH∘−(R/ΔH∘)ln⁡(2cinit/c0)1/T_m = \Delta S^{\circ}/\Delta H^{\circ} - (R/\Delta H^{\circ})\ln(2c_{init}/c^0). Fitting the six points gives slope =−2.0×10−5= -2.0\times10^{-5} K−1^{-1} and intercept =2.845×10−3= 2.845\times10^{-3} K−1^{-1}, hence ΔH∘=R/slope=−416\Delta H^{\circ} = R/\text{slope} = -416 kJ mol−1^{-1} and ΔS∘=ΔH∘×intercept=−1.18\Delta S^{\circ} = \Delta H^{\circ}\times\text{intercept} = -1.18 kJ K−1^{-1} mol−1^{-1}.