Question

The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the geometric shape of these molecules, adenine bonds with thymine and cytosine bonds with guanine. The figure (Figure 1) shows the thymine-adenine bond. Each charge shown is ±e, and the H−N distance is 0.110 nm.

A) Calculate the net force that thymine exerts on adenine. To keep the calculations fairly simple, yet reasonable, consider only the forces due to the O−H−N and the N−H−N combinations, assuming that these two combinations are parallel to each other. Remember, however, that in the O−H−N set, the O− exerts a force on both the H+ and the N−, and likewise along the N−H−N set.

B) Is the net force attractive or repulsive?

C) Calculate the force on the electron in the hydrogen atom, which is 5.29×10−2 nm from the proton.

D) Compare the strength of the bonding force of the electron in hydrogen with the bonding force of the adenine-thymine molecules.

Answer 1

Concepts and reason

First calculate the potential energy of the bond O-H-N and then potential energy of bond N-H-N by using the expression for potential energy.

Secondly, find the net force by using the fact that the like charges repel each other and unlike charge attract each other’s.

Finally, calculate the potential energy of electron proton in hydrogen atom and then compare it with the net potential energy calculated in the first step.

Fundamentals

The expression to calculate the electric potential energy of two charges is,

$U = \frac{{kqQ}}{r}$

Here, k is the constant, q is the charge 1, Q is the charge 2, r is the distance between these two charges, and U is the electric potential energy.

The total energy is calculated as follows:

$U = {U_{{\rm{O}} - {\rm{H}} - {\rm{N}}}} + {U_{{\rm{N - H - N}}}}$

Here, ${U_{{\rm{N - H - N}}}}$ is the potential energy due to the bond N-H-N, ${U_{{\rm{O}} - {\rm{H}} - {\rm{N}}}}$ is the potential energy due to the bond O-H-N, and U is the total electric potential energy of the adenine and thymine bond.

The potential energy of electron proton in hydrogen atom is calculated as follows:

${U_{e - p}} = k\frac{{{e_e}{e_p}}}{r}$

Here, k is the constant, ${e_e}$is the charge of electron, ${e_p}$ is the charge on proton, r is the distance between the electron and proton, and ${U_{e - p}}$ is the potential energy of electron-proton system.

(a)

The expression to calculate the electric potential energy of two charges is,

$U = \frac{{kqQ}}{r}$

Here, k is the constant, q is the charge 1, Q is the charge 2, r is the distance between these two charges, and U is the electric potential energy.

The expression to calculate the potential energy of O-H-N bond is,

${U_{{\rm{OHN}}}} = \frac{{k\,{q_{\rm{O}}}{q_{\rm{N}}}}}{{{r_{{\rm{O - N}}}}}} + \frac{{k\,{q_{\rm{O}}}{q_{\rm{H}}}}}{{{r_{{\rm{O - H}}}}}}$

Substitute $0.28 \times {10^{ - 9}}\,{\rm{m}}$for ${r_{{\rm{O - N}}}}$, $0.17 \times {10^{ - 9}}\,{\rm{m}}$for ${r_{{\rm{O - H}}}}$, $9.0 \times {10^9}\,{\rm{N \bullet }}{{\rm{m}}^{\rm{2}}}{\rm{/ }}{{\rm{C}}^{\rm{2}}}$for k, $- 1.60 \times {10^{ - 19}}\,{\rm{C}}$ for ${q_{\rm{O}}}\,{\rm{and}}\,{q_{\rm{N}}}$, and $1.60 \times {10^{ - 19}}\,{\rm{C}}$ for ${q_{\rm{H}}}$.

$\begin{array}{c}\\{U_{{\rm{OHN}}}} = \left( \begin{array}{l}\\\frac{{9.0 \times {{10}^9}\,{\rm{N \bullet }}{{\rm{m}}^{\rm{2}}}{\rm{/ }}{{\rm{C}}^{\rm{2}}}\,\left( { - 1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( { - 1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)}}{{0.28 \times {{10}^{ - 9}}\,{\rm{m}}}}\\\\ + \frac{{9.0 \times {{10}^9}\,{\rm{N \bullet }}{{\rm{m}}^{\rm{2}}}{\rm{/ }}{{\rm{C}}^{\rm{2}}}\,\left( { - 1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( {1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)}}{{0.17 \times {{10}^{ - 9}}\,{\rm{m}}}}\\\end{array} \right)\\\\ = - 5.32 \times {10^{ - 19}}\,\,{\rm{J}}\\\end{array}$

The expression to calculate the potential energy of N-H-N bond is,

${U_{{\rm{NHN}}}} = \frac{{k\,{q_{\rm{N}}}{q_{\rm{N}}}}}{{{r_{{\rm{N - N}}}}}} + \frac{{k\,{q_{\rm{N}}}{q_{\rm{H}}}}}{{{r_{{\rm{N - H}}}}}}$

Substitute $0.30 \times {10^{ - 9}}\,{\rm{m}}$for ${r_{{\rm{N - N}}}}$, $0.19 \times {10^{ - 9}}\,{\rm{m}}$for ${r_{{\rm{N - H}}}}$, $9.0 \times {10^9}\,{\rm{N \bullet }}{{\rm{m}}^{\rm{2}}}{\rm{/ }}{{\rm{C}}^{\rm{2}}}$for k, $- 1.60 \times {10^{ - 19}}\,{\rm{C}}$ for $\,{q_{\rm{N}}}$, and $1.60 \times {10^{ - 19}}\,{\rm{C}}$ for ${q_{\rm{H}}}$.

$\begin{array}{c}\\{U_{{\rm{NHN}}}} = \left( \begin{array}{l}\\\frac{{9.0 \times {{10}^9}\,{\rm{N \bullet }}{{\rm{m}}^{\rm{2}}}{\rm{/ }}{{\rm{C}}^{\rm{2}}}\,\left( { - 1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( { - 1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)}}{{0.30 \times {{10}^{ - 9}}\,{\rm{m}}}}\\\\ + \frac{{9.0 \times {{10}^9}\,{\rm{N \bullet }}{{\rm{m}}^{\rm{2}}}{\rm{/ }}{{\rm{C}}^{\rm{2}}}\,\left( { - 1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( {1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)}}{{0.19 \times {{10}^{ - 9}}\,{\rm{m}}}}\\\end{array} \right)\\\\ = - 4.45 \times {10^{ - 19}}\,\,{\rm{J}}\\\end{array}$

The total potential energy is calculated as follows:

$U = {U_{{\rm{O}} - {\rm{H}} - {\rm{N}}}} + {U_{{\rm{N - H - N}}}}$

Here, ${U_{{\rm{N - H - N}}}}$ is the potential energy due to the bond N-H-N, ${U_{{\rm{O}} - {\rm{H}} - {\rm{N}}}}$ is the potential energy due to the bond O-H-N, and U is the total electric potential energy of the adenine and thymine bond.

Substitute $- 5.32 \times {10^{ - 19}}\,\,{\rm{J}}$ for ${U_{{\rm{O}} - {\rm{H}} - {\rm{N}}}}$ and $- 4.45 \times {10^{ - 19}}\,\,{\rm{J}}$ for ${U_{{\rm{N - H - N}}}}$ in expression $U = {U_{{\rm{O}} - {\rm{H}} - {\rm{N}}}} + {U_{{\rm{N - H - N}}}}$.

$\begin{array}{c}\\U = - 5.32 \times {10^{ - 19}}\,\,{\rm{J}} + - 4.45 \times {10^{ - 19}}\,\,{\rm{J}}\\\\{\rm{ = }} - {\rm{9}}{\rm{.77}} \times {\rm{1}}{{\rm{0}}^{ - 19}}\,{\rm{J}}\\\end{array}$
‎

(b)

The potential energy of electron proton in hydrogen atom is calculated as follows:

${U_{e - p}} = k\frac{{{e_e}{e_p}}}{r}$

Here, k is the constant, ${e_e}$is the charge of electron, ${e_p}$ is the charge on proton, r is the distance between the electron and proton, and ${U_{e - p}}$ is the potential energy of electron-proton system.

Substitute $0.0529 \times {10^{ - 9}}{\rm{m}}$ for r, $- 1.60 \times {10^{ - 19}}\,{\rm{C}}$ for ${e_e}$, $1.60 \times {10^{ - 19}}\,{\rm{C}}$ is for ${e_p}$, and $9.0 \times {10^9}\,{\rm{N \bullet }}{{\rm{m}}^{\rm{2}}}{\rm{/ }}{{\rm{C}}^{\rm{2}}}$ for k.

$\begin{array}{c}\\{U_{e - p}} = \left( {9.0 \times {{10}^9}\,{\rm{N \bullet }}{{\rm{m}}^{\rm{2}}}{\rm{/ }}{{\rm{C}}^{\rm{2}}}} \right)\frac{{\left( { - 1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( {1.60 \times {{10}^{ - 19}}\,{\rm{C}}} \right)}}{{\left( {0.0529 \times {{10}^{ - 9}}{\rm{m}}} \right)}}\\\\ = - 43.55 \times {10^{ - 19}}\,{\rm{J}}\\\end{array}$

Calculate the ratio $\frac{U}{{{U_{e - p}}}}$.

Substitute $- 9.77 \times {10^{ - 19}}\,{\rm{J}}$ for U and $- 43.55 \times {10^{ - 19}}\,{\rm{J}}$ for ${U_{e - p}}$ in expression $\frac{U}{{{U_{e - p}}}},$ and solve for U.

$\begin{array}{c}\\\frac{U}{{{U_{e - p}}}} = \frac{{ - 9.77 \times {{10}^{ - 19}}\,{\rm{J}}}}{{ - 43.55 \times {{10}^{ - 19}}\,{\rm{J}}}}\\\\U = 0.22{U_{e - p}}\\\end{array}$

From the above ratio, It is clear that the electric potential energy between the adenine-thymine bonds is 0.22 times the electric potential energy of the electron-proton in hydrogen atom.

Ans: Part a

The electric potential energy of the adenine-thymine bond is $- 9.77 \times {10^{ - 19}}\,{\rm{J}}$.

Part b

The electric potential energy between the adenine-thymine bonds is 0.22 times the electric potential energy of the electron -proton in hydrogen atom.