Ionic effects on the temperature–force phase diagram of DNA

Abstract

Double-stranded DNA (dsDNA) undergoes a structural transition to single-stranded DNA (ssDNA) in many biologically important processes such as replication and transcription. This strand separation arises in response either to thermal fluctuations or to external forces. The roles of ions are twofold, shortening the range of the interstrand potential and renormalizing the DNA elastic modulus. The dsDNA-to-ssDNA transition is studied on the basis that dsDNA is regarded as a bound state while ssDNA is regarded as an unbound state. The ground state energy of DNA is obtained by mapping the statistical mechanics problem to the imaginary time quantum mechanics problem. In the temperature–force phase diagram the critical force F _c (T) increases logarithmically with the Na⁺ concentration in the range from 32 to 110 mM. Discussing this logarithmic dependence of F _c (T) within the framework of polyelectrolyte theory, it inevitably suggests a constraint on the difference between the interstrand separation and the length per unit charge during the dsDNA-to-ssDNA transition.

Appendices

Appendix A: Ground state eigenenergy of unperturbed Hamiltonian

We derive the expression of the ground state eigenenergy E0(0), (21), of the unperturbed Hamiltonian $\widehat {\textbf {H}}_{0}$. The ground state eigenenergy $E_{0}^{(0)} = < \psi _{0}^{(0)} | \widehat {\textbf {H}}_{0} | \psi _{0}^{(0)} >$ gives

$$\begin{array}{@{}rcl@{}} E_{0}^{(0)} & = & \int d^{3}r \left( \psi_{0}^{(0)}(\textbf{r})\right)^{*} \left[-\frac{ (k_{B}T)^{2}} {K} \left( {\nabla} - \frac{1}{k_{B}T} \textbf{F}\right)^{2} - \frac{e^{2}}{\epsilon r}\right] \psi_{0}^{(0)}(\textbf{r}) \\ & = & E_{k} + E_{p} \end{array} $$

(30)

where we define the kinetic-like energy as

$$ E_{k} \equiv \int d^{3}r \left( \psi_{0}^{(0)}(\textbf{r})\right)^{*} \left( -\frac{ (k_{B}T)^{2}} {K}\right) \left( {\nabla} - \frac{1}{k_{B}T} \textbf{F}\right)^{2} \psi_{0}^{(0)}(\textbf{r}) $$

(31)

and we define the potential energy, due to the attractive Coulomb potential as,

$$ E_{p} \equiv \int d^{3}r \left( \psi_{0}^{(0)}(\textbf{r})\right)^{*} \left( - \frac{e^{2}}{\epsilon r}\right) \psi_{0}^{(0)}(\textbf{r}). $$

(32)

Without loss of generality, let the external constant force F be in the $\widehat {\textbf {r}}$-direction, i.e., $\textbf {F}{\kern -.5pt} ={\kern -.5pt} F\widehat {\textbf {r}}$. The Laplacian term takes the form

$$\begin{array}{@{}rcl@{}} \lefteqn{ \left( {\nabla} - \frac{1}{k_{B}T} \textbf{F}\right)^{2} \psi_{0}^{(0)}(\textbf{r}) = \frac{1}{r^{2}} \frac{\partial}{\partial r} \left( r^{2} \frac{ \partial\psi_{0}^{(0)}} {\partial r}\right) - 2\frac{F}{k_{B}T} \frac{ \partial\psi_{0}^{(0)}} {\partial r} + \left( \frac{F}{k_{B}T}\right)^{2} \psi_{0}^{(0)}} \\ & = & - \frac{1}{a} \left( 1 - \frac{Fa}{k_{B}T}\right) \left( 2 - \frac{r}{a} \left( 1 - \frac{Fa}{k_{B}T}\right)\right) \frac{1}{r} \psi_{0}^{(0)} + \frac{2}{a} \frac{F}{k_{B}T} \left( 1 - \frac{Fa}{k_{B}T}\right) \psi_{0}^{(0)} + \left( \frac{F}{k_{B}T}\right)^{2} \psi_{0}^{(0)} \end{array} $$

(33)

in which to obtain the last line we use (19) for the unperturbed ground state eigenfunction ψ0(0)(r). Substituting (33) in (31) and use the normalization of ψ0(0)(r) gives the kinetic-like energy

$$\begin{array}{@{}rcl@{}} E_{k} & = & -\frac{ (k_{B}T)^{2}} {K} \left[- \frac{1}{a} \left( 1 - \frac{Fa}{k_{B}T}\right) 4\pi{\int}_{0}^{\infty} dr r^{2} \left( 2 - \frac{r}{a} \left( 1 - \frac{Fa}{k_{B}T}\right)\right) \frac{1}{r} | \psi_{0}^{(0)}(\textbf{r}) |^{2}\right. \\ & & \hspace{13mm} \left. + \frac{2}{a} \frac{F}{k_{B}T} \left( 1 - \frac{Fa}{k_{B}T}\right) + \left( \frac{F}{k_{B}T}\right)^{2}\right] \\ & = & \frac{1}{K} \left( \frac{k_{B}T}{a}\right)^{2} \left[2\left( \frac{Fa}{k_{B}T}\right)^{2} - 4\frac{Fa}{k_{B}T} + 1\right]. \end{array} $$

(34)

The potential energy is

$$ E_{p} = - \frac{e^{2}}{\epsilon} 4\pi{\int}_{0}^{\infty} dr r^{2} \left( \psi_{0}^{(0)}(\textbf{r})\right)^{*} \frac{1}{r} \psi_{0}^{(0)}(\textbf{r}) = - \left( 1 - \frac{Fa}{k_{B}T}\right) \frac{e^{2}}{\epsilon a}. $$

(35)

Adding the kinetic-like energy E _k with the potential energy E _p yields the ground state eigenenergy of the unperturbed Hamiltonian

$$\begin{array}{@{}rcl@{}} E_{0}^{(0)} & = & \frac{1}{K} \left( \frac{k_{B}T}{a}\right)^{2} \left[2\left( \frac{Fa}{k_{B}T}\right)^{2} - 4\frac{Fa}{k_{B}T} + 1\right] - \left( 1 - \frac{Fa}{k_{B}T}\right) \frac{e^{2}}{\epsilon a} \\ & = & \frac{2}{K} F^{2} + \left( \frac{e^{2}}{\epsilon k_{B}T} - 4\frac{k_{B}T}{Ka}\right) F + \frac{1}{K} \left( \frac{k_{B}T}{a}\right)^{2} - \frac{e^{2}}{\epsilon a} \\ & = & \frac{2}{K} F^{2} + \left( 2\frac{k_{B}T}{Ka} - 4\frac{k_{B}T}{Ka}\right) F + \frac{1}{2} \frac{e^{2}}{\epsilon a} - \frac{e^{2}}{\epsilon a} \\ & = & \frac{2}{K} \left( F^{2} - \frac{k_{B}T}{a} F - \frac{1}{4} \frac{Ke^{2}}{\epsilon a}\right) \\ & = & \frac{2}{K} \left( F - \frac{1}{2} \frac{k_{B}T}{a}\right)^{2} - \frac{3}{4} \frac{e^{2}}{\epsilon a} \end{array} $$

(36)

which is (21) in the main text.

Appendix B: First-order correction to ground state eigenenergy of full Hamiltonian

We derive the first-order correction, (23), to the ground state eigenenergy of full Hamiltonian $\widehat {\textbf {H}}$. Using the perturbation $\widehat {\textbf {H}}^{\prime }$, (13), and the ground state eigenfunction of the unperturbed Hamiltonian ψ0(0)(r), (19), we obtain

$$\begin{array}{@{}rcl@{}} < \psi_{0}^{(0)} | \widehat{\textbf{H}}^{\prime} | \psi_{0}^{(0)} > & = & \frac{e^{2}}{\epsilon\lambda_{D}} \frac{1}{\pi} \left( \frac{ 1 - \frac{Fa}{k_{B}T}} {a}\right)^{3} \sum\limits_{n=0}^{\infty} \frac{ (-1)^{n}} { (n+1)! {\lambda_{D}^{n}}} 4\pi{\int}_{0}^{\infty} dr r^{n+2} \exp\left[-2\left( 1 - \frac{Fa}{k_{B}T}\right) \frac{r}{a}\right] \\ & = & \frac{1}{2} \frac{e^{2}}{\epsilon\lambda_{D}} \sum\limits_{n=0}^{\infty} (-1)^{n} (n+2) \left[\frac{a}{ 2\lambda_{D} \left( 1 - \frac{Fa}{k_{B}T}\right)}\right]^{n} \\ & = & \frac{1}{2} \frac{e^{2}}{\epsilon\lambda_{D}} \left( \sum\limits_{n=0}^{\infty} (-1)^{n} nC^{n} + 2\sum\limits_{n=0}^{\infty} (-1)^{n} C^{n}\right) \end{array} $$

(37)

where defining C ≡ a/2λ _D (1 − (Fa/k _B T)). The second summation in parentheses $S \equiv {\sum }_{n=0}^{\infty } (-1)^{n} C^{n}$, which is a geometric series equal to 1/(C + 1). The first summation in parentheses ${\sum }_{n=0}^{\infty } (-1)^{n} nC^{n} = C\partial S/\partial C$ equals − C/(C + 1)². Hence, we can write explicitly, upon substituting the definition of C,

$$ < \psi_{0}^{(0)} | \widehat{\textbf{H}}^{\prime} | \psi_{0}^{(0)} > = \frac{1}{2} \frac{C+2}{ (C+1)^{2}} \frac{e^{2}}{\epsilon\lambda_{D}} = \left( 1 - \frac{Fa}{k_{B}T}\right) \frac{e^{2}}{\epsilon} \frac{ 4\left( 1 - \frac{Fa}{k_{B}T}\right) \lambda_{D} + a} { \left[2\left( 1 - \frac{Fa}{k_{B}T}\right) \lambda_{D} + a\right]^{2}} $$

(38)

which is (23) in the main text.