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

T_m

? (d) A dsDNA solution with

c_{init} = 1.00\times10^{-6}

mol dm

^{-3}

is heated to

T_m

(50% dissociated). Calculate the association equilibrium constant at

T_m

for a non-palindromic dsDNA (

K_{np}

) and a palindromic dsDNA (

K_p

), taking the standard concentration

c^0 = 1

mol dm

^{-3}

. (e) The mean Gibbs energies of association per base pair are

-6.07

kJ mol

^{-1}

(G–C) and

-1.30

kJ mol

^{-1}

(A–T). At

T_m = 330

K, use

K_{np} = 1.00\times10^{6}

and

K_p = 1.00\times10^{5}

: how many base pairs has the shortest dsDNA with

T_m

above 330 K, and is it palindromic? (f) The inverse melting temperature of the dodecanucleotide varies linearly with

\ln(2c_{init}/c^0)

: for

c_{init}/10^{-6}

= 0.25, 0.50, 1.00, 2.0, 4.0, 8.0 mol dm

^{-3}

T_m

= 319.0, 320.4, 321.8, 323.3, 324.7, 326.2 K. Calculate the standard enthalpy

\Delta H^{\circ}

and entropy

\Delta S^{\circ}

of strand association.

Step 4 of 5: Shortest duplex above 330 K

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n_{np} = \dfrac{37.9}{6.07} = 6.2 \Rightarrow 7;\quad n_{p} = \dfrac{31.6}{6.07} = 5.2 \Rightarrow 6\ \text{bp, palindromic}

Analysis

The required association Gibbs energies are $\Delta G^{\circ} = -RT\ln K$ : $-37.9$ kJ mol $^{-1}$ (non-palindromic) and $-31.6$ kJ mol $^{-1}$ (palindromic) at 330 K. Dividing by the strongest per-pair contribution ( $-6.07$ kJ mol $^{-1}$ , all G–C) gives 6.2 and 5.2, rounded up to 7 and 6 base pairs. The shortest dsDNA melting above 330 K is palindromic, with 6 base pairs.