IChO 2018 · Theory problem 1

DNA

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Palindromic sequences are an interesting class of DNA. In a palindromic double-
-stranded DNA (dsDNA) species, the sequence of one strand read in the \(5'\to3'\) direction matches the \(5'\to3'\) reading on the complementary strand. Hence, a palindromic dsDNA consists of two identical strands that are complementary to each other. An example is the so-called Drew–Dickerson dodecanucleotide (1):

Figure 1

1.1 How many different palindromic double-stranded DNA dodecanucleotides (i. e. dsDNA species with twelve base pairs) exist?

1.2 How many different palindromic double-stranded DNA undecanucleotides (i.e., dsDNA species with eleven base pairs) exist?

The melting temperature of dsDNA, \(T_m\) is defined as the temperature at which 50% of the original amount of DNA double strands are dissociated into separate strands.

1.3 Consider the Drew–Dickerson dodecanucleotide (1). Assume that a G–C nucleobase pair contributes to the DNA duplex stability more than an A–T pair does.

What is the probability that its \(T_m\) increases when a single randomly selected base pair is replaced by a G–C pair?

Let us analyze the thermodynamics of formation of double-helical DNA from single strands, and its dependence on the length of the DNA and on the temperature. The equilibrium constant of association of single strands to form dsDNA differs for palindromic and non-palindromic dsDNA.

A solution of dsDNA with the initial concentration of \(c_{\mathrm{init}}=1.00\times10^{-6}\ \mathrm{mol\,dm^{-3}}\) was heated to \(T_m\) and equilibrium was reached.

1.4 Calculate the equilibrium constant of association of single strands at \(T_m\) for both non-palindromic and palindromic DNA.

The mean contributions to the Gibbs energy of association of two single strands to form dsDNA were estimate over a certain range of experimental conditions, and they amount to \(-6.07\ \mathrm{kJ\,mol^{-1}}\) per one G–C pair, and \(-1.30\ \mathrm{kJ\,mol^{-1}}\) per one A–T pair present in a dsDNA.

1.5 How many base pairs are there in the shortest dsDNA oligonucleotide that has \(T_m\) above \(330\ \mathrm K\)? At this \(T_m\), consider the following values of the equilibrium constant of association of single strands to form a dsDNA: \(K_{\mathrm{np}}=1.00\times10^6\) for a non-palindromic dsDNA, \(K_{\mathrm p}=1.00\times10^5\) for a palindromic dsDNA. Is the shortest oligonucleotide palindromic or non-palindromic?

Finally, let us leave the simplified idea of base pairs contributing individually to the association of DNA strands. The Gibbs energy of this process may be considered explicitly dependent on temperature. The dependence of the inverse \(T_m\) of the Drew–Dickerson dodecanucleotide (1) on the logarithm of the initial duplex concentration \(c_{\mathrm{init}}\) is shown below. (Note: a standard concentration \(c_0=1\ \mathrm{mol\,dm^{-3}}\) is introduced.)

\(c_{\mathrm{init}}/10^{-6}\ \mathrm{mol\,dm^{-3}}\) 0.25 0.50 1.00 2.0 4.0 8.0
\(T_m/\mathrm K\) 319.0 320.4 321.8 323.3 324.7 326.2
Figure 2

1.6 Calculate the standard enthalpy \(\Delta H^\circ\) and the standard entropy \(\Delta S^\circ\) of the association of DNA single strands to form the palindromic double-stranded Drew–Dickerson dodecanucleotide (1). Assume that \(\Delta H^\circ\) and \(\Delta S^\circ\) do not vary with temperature.

Official English text of the problem.

Answers

  • 1.1: \(\boxed{4096}\).
  • 1.2: \(\boxed{0}\).
  • 1.3: \(\boxed{\frac13\approx33.3\%}\).
  • 1.4: \(\boxed{K_{c,\mathrm{np}}=2.00\times10^6\ \mathrm{dm^3\,mol^{-1}}}\) and \(\boxed{K_{c,\mathrm p}=5.00\times10^5\ \mathrm{dm^3\,mol^{-1}}}\). The dimensionless standard-state constants have the same numerical values when \(c_0=1\ \mathrm{mol\,dm^{-3}}\).
  • 1.5: \(\boxed{6\text{ base pairs; palindromic, with all six pairs G–C}}\).
  • 1.6: \(\boxed{\Delta H^\circ\approx-416\ \mathrm{kJ\,mol^{-1}}}\) and \(\boxed{\Delta S^\circ\approx-1.18\times10^3\ \mathrm{J\,mol^{-1}\,K^{-1}}}\), per mole of duplex formed.

Solution

1.1

A palindromic DNA strand equals its reverse complement. Thus, if its bases are \(b_1,\ldots,b_{12}\),

\[ b_i=\overline{b}_{13-i}, \]

where the bar denotes the complementary base: A pairs with T, and G pairs with C.

The first six positions can each contain any of four bases. The last six positions are then fixed:

\[ N=4^6=\boxed{4096}. \]

There is no division by two: the two strands of each palindromic duplex already have identical sequences when read \(5'\to3'\).

1.2

For an eleven-base strand, the central position would have to satisfy

\[ b_6=\overline{b}_6. \]

None of the four standard DNA bases is complementary to itself. Consequently, an odd-length DNA duplex cannot be palindromic:

\[ \boxed{N=0}. \]

1.3

The Drew–Dickerson duplex contains eight G–C pairs and four A–T pairs. Under the stated stability model, replacing an A–T pair with a G–C pair increases stability and hence \(T_m\); replacing an existing G–C pair does not increase its contribution.

All twelve positions are equally likely to be selected, so

\[ P(T_m\text{ increases}) =\frac{4}{12} =\boxed{\frac13\approx33.3\%}. \]

1.4

Let \(D\) denote a duplex and \(S_1,S_2\) its complementary single strands. These are symbols for whole DNA strands, rather than elemental formulas.

For non-palindromic DNA, the strands are distinct:

\[ \ce{S_1 + S_2 <=> D}. \]

At \(T_m\), half the original duplex has dissociated:

\[ [D]=\frac{c_{\mathrm{init}}}{2}, \qquad [S_1]=[S_2]=\frac{c_{\mathrm{init}}}{2}. \]

Therefore,

\[ K_{c,\mathrm{np}} =\frac{[D]}{[S_1][S_2]} =\frac{c_{\mathrm{init}}/2}{(c_{\mathrm{init}}/2)^2} =\frac{2}{c_{\mathrm{init}}}. \]

Substitution gives

\[ \boxed{K_{c,\mathrm{np}} =2.00\times10^6\ \mathrm{dm^3\,mol^{-1}}}. \]

For palindromic DNA, both strands are the same species \(S\):

\[ \ce{2S <=> D}. \]

Each dissociated duplex produces two molecules of \(S\). Hence,

\[ [D]=\frac{c_{\mathrm{init}}}{2}, \qquad [S]=2\left(\frac{c_{\mathrm{init}}}{2}\right)=c_{\mathrm{init}}. \]

Thus,

\[ K_{c,\mathrm p} =\frac{[D]}{[S]^2} =\frac{c_{\mathrm{init}}/2}{c_{\mathrm{init}}^2} =\frac{1}{2c_{\mathrm{init}}}, \]

giving

\[ \boxed{K_{c,\mathrm p} =5.00\times10^5\ \mathrm{dm^3\,mol^{-1}}}. \]

The dimensionless equilibrium constants used in thermodynamic logarithms are

\[ K_{\mathrm{np}}=\frac{2c_0}{c_{\mathrm{init}}}, \qquad K_{\mathrm p}=\frac{c_0}{2c_{\mathrm{init}}}. \]

These equal \(c_0K_c\), so their numerical values agree with those above when \(c_0=1\ \mathrm{mol\,dm^{-3}}\).

1.5

In the stated additive model,

\[ \Delta G^\circ =-\left(6.07n_{\mathrm{GC}}+1.30n_{\mathrm{AT}}\right) \ \mathrm{kJ\,mol^{-1}}. \]

At the melting temperature,

\[ \Delta G^\circ=-RT_m\ln K. \]

For a fixed number of base pairs, the greatest stabilization—and therefore the greatest \(T_m\) in this model—is obtained when every pair is G–C. For an all-G–C duplex of \(n\) base pairs,

\[ T_m=\frac{6070n}{R\ln K}, \]

using \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\).

For non-palindromic DNA, use the value specified in this part, \(K_{\mathrm{np}}=10^6\):

\[ T_m>330\ \mathrm K \quad\Longrightarrow\quad n>\frac{(8.314)(330)\ln(10^6)}{6070} =6.245. \]

Thus, the shortest possible non-palindromic duplex has seven base pairs.

For palindromic DNA, \(K_{\mathrm p}=10^5\):

\[ n>\frac{(8.314)(330)\ln(10^5)}{6070} =5.204. \]

The smallest allowed even integer is six. Such an all-G–C palindrome exists: choose any three G/C bases as the first half of a strand, then use their reverse complement as the second half.

Its predicted melting temperature is

\[ T_m=\frac{6(6070)}{(8.314)\ln(10^5)} \approx380\ \mathrm K>330\ \mathrm K. \]

Moreover, a six-base-pair palindrome containing any A–T pairs must contain at least two of them, because complementary positions occur in pairs. Its maximum stabilization would then be

\[ 4(6.07)+2(1.30)=26.88\ \mathrm{kJ\,mol^{-1}}, \]

below the required

\[ R(330)\ln(10^5)\approx31.59\ \mathrm{kJ\,mol^{-1}}. \]

Therefore,

\[ \boxed{\text{The shortest duplex is palindromic and contains six G–C base pairs.}} \]

1.6

For the palindromic duplex at its melting temperature, part 1.4 gives

\[ K_{\mathrm p}=\frac{c_0}{2c_{\mathrm{init}}}. \]

Combining

\[ \Delta G^\circ=\Delta H^\circ-T_m\Delta S^\circ \]

with

\[ \Delta G^\circ=-RT_m\ln K_{\mathrm p} =RT_m\ln\left(\frac{2c_{\mathrm{init}}}{c_0}\right) \]

yields

\[ \frac{\Delta H^\circ}{T_m}-\Delta S^\circ =R\ln\left(\frac{2c_{\mathrm{init}}}{c_0}\right). \]

Therefore,

\[ \boxed{ \frac1{T_m} = \frac{R}{\Delta H^\circ} \ln\left(\frac{2c_{\mathrm{init}}}{c_0}\right) + \frac{\Delta S^\circ}{\Delta H^\circ} }. \]

Thus, for the plotted variables

\[ x=\ln\left(\frac{2c_{\mathrm{init}}}{c_0}\right), \qquad y=\frac1{T_m}, \]

the slope and intercept are

\[ m=\frac{R}{\Delta H^\circ}, \qquad b=\frac{\Delta S^\circ}{\Delta H^\circ}. \]

A least-squares straight-line fit to all six tabulated points gives

\[ m=-1.99660\times10^{-5}\ \mathrm{K^{-1}}, \qquad b=2.84523\times10^{-3}\ \mathrm{K^{-1}}. \]

Hence,

\[ \Delta H^\circ =\frac{R}{m} =\frac{8.314}{-1.99660\times10^{-5}} \ \mathrm{J\,mol^{-1}} \approx-416.4\ \mathrm{kJ\,mol^{-1}}, \]

and

\[ \Delta S^\circ =b\Delta H^\circ \approx (2.84523\times10^{-3})(-416.4\times10^3) \ \mathrm{J\,mol^{-1}\,K^{-1}} \approx-1184.8\ \mathrm{J\,mol^{-1}\,K^{-1}}. \]

To sensible precision,

\[ \boxed{\Delta H^\circ\approx-416\ \mathrm{kJ\,mol^{-1}}}, \qquad \boxed{\Delta S^\circ\approx-1.18\times10^3\ \mathrm{J\,mol^{-1}\,K^{-1}}}. \]

The negative enthalpy indicates exothermic duplex formation; the negative entropy is consistent with association and ordering of the strands.

The solution was checked against the official solution.