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 3 of 5: Association constants at TmT_m
Knp(Tm)=c012cinit=2.0×106;Kp(Tm)=c02cinit=5.0×105K_{np}(T_m) = \dfrac{c^0}{\tfrac{1}{2}c_{init}} = 2.0\times10^{6};\quad K_{p}(T_m) = \dfrac{c^0}{2c_{init}} = 5.0\times10^{5}
Analysis

For a non-palindromic duplex ssDNAX1+ssDNAX2⇌dsDNA\ce{ssDNA1 + ssDNA2 <=> dsDNA}, at TmT_m each species is at cinit/2c_{init}/2, giving Knp=c0/(cinit/2)=2.0×106K_{np} = c^0/(c_{init}/2) = 2.0\times10^6. For a palindromic duplex 2 ssDNA⇌dsDNA\ce{2ssDNA <=> dsDNA}, [dsDNA]=cinit/2[\ce{dsDNA}] = c_{init}/2 but [ssDNA]=cinit[\ce{ssDNA}] = c_{init}, giving Kp=c0/(2cinit)=5.0×105K_p = c^0/(2c_{init}) = 5.0\times10^5.

Common pitfall. For palindromic DNA the two strands are identical, so dissociating half the duplex yields [ssDNA]=cinit[\ce{ssDNA}] = c_{init}, not cinit/2c_{init}/2 — a factor of 4 difference in KK.