Question

Question: (III) The two strands of the helix-shaped DNA molecule are held together by electrostatic forces as shown in Fig. 16–39. Assume that the net average charge (due to electron sharing) indicated on H and N atoms have magnitude 0.2e and on the indicated C and O atoms is 0.4e. Assume also that atoms on each molecule are separated by \({\bf{1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\;{\bf{m}}\). Estimate the net force between (a) a thymine and an adenine; and (b) a cytosine and a guanine. For each bond (red dots) consider only the three atoms in a line (two atoms on one molecule, one atom on the other). (c) Estimate the total force for a DNA molecule containing \({\bf{1}}{{\bf{0}}^{\bf{5}}}\) pairs of such molecules. Assume half are A–T pairs and half are C–G pairs.

Short answer

(a) The net force between the thymine and the adenine is \({\rm{5}} \times {\rm{1}}{{\rm{0}}^{ - 10}}\;{\rm{N}}\).

(b) The net force between a cytosine and a guanine is \({\rm{7}} \times {\rm{1}}{{\rm{0}}^{ - 10}}\;{\rm{N}}\).

(c) The total force of the DNA molecule containing \({10^5}\) pairs is \(6 \times {10^{ - 5}}\;{\rm{N}}\).

Step-by-step solution

Formula of electrostatic force

The expression for the electrostatic force is as follows:

\(F = \frac{{k{q_1}{q_2}}}{{{r^2}}}\)

Here,\(k\)is the proportionality constant,\({q_1}\)and\({q_2}\)are the charges, and\(r\)is the distance between the charges.

Evaluation of the net force between a thymine and an adenine

(a)

Let the net charge on H and N atoms be 0.2e and C and O atoms be 0.4e. The net force between the thymine and the adenine is due to the following forces.

The force of attraction between O and H atoms is:

\(\begin{aligned}{l}{F_{{\rm{OH}}}} &= \frac{{k\left( {0.4e} \right)\left( {0.2e} \right)}}{{{{\left( {1.80 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\\{F_{{\rm{OH}}}} &= \frac{{0.08k{e^2}}}{{{{\left( {1.80 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\end{aligned}\)

The force of attraction between O and N atoms is:

\(\begin{aligned}{l}{F_{{\rm{ON}}}} &= \frac{{k\left( {0.4e} \right)\left( {0.2e} \right)}}{{{{\left( {2.80 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\\{F_{{\rm{ON}}}} &= \frac{{0.08k{e^2}}}{{{{\left( {2.80 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\end{aligned}\)

The force of attraction between N and N atoms is:

\(\begin{aligned}{l}{F_{{\rm{NN}}}} &= \frac{{k\left( {0.2e} \right)\left( {0.2e} \right)}}{{{{\left( {3.00 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\\{F_{{\rm{NN}}}} &= \frac{{0.04k{e^2}}}{{{{\left( {3.00 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\end{aligned}\)

The force of attraction between H and N atoms is:

\(\begin{aligned}{l}{F_{{\rm{HN}}}} &= \frac{{k\left( {0.2e} \right)\left( {0.2e} \right)}}{{{{\left( {2.00 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\\{F_{{\rm{HN}}}} &= \frac{{0.04k{e^2}}}{{{{\left( {2.00 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\end{aligned}\)

The net force between the thymine and the adenine can be calculated as:

Thus, the net force between the thymine and the adenine is \({\rm{5}} \times {\rm{1}}{{\rm{0}}^{ - 10}}\;{\rm{N}}\).

Evaluation of the net force between a cytosine and a guanine

(b)

The net force between a cytosine and a guanine is due to the following forces.

The force of attraction between O and H atoms is:

\(\begin{aligned}{l}{F_{{\rm{OH}}}} &= \frac{{k\left( {0.4e} \right)\left( {0.2e} \right)}}{{{{\left( {1.90 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\\{F_{{\rm{OH}}}} &= \frac{{0.08k{e^2}}}{{{{\left( {1.90 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\end{aligned}\)

The force of attraction between O and N atoms is:

\(\begin{aligned}{l}{F_{{\rm{ON}}}} &= \frac{{k\left( {0.4e} \right)\left( {0.2e} \right)}}{{{{\left( {2.90 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\\{F_{{\rm{ON}}}} &= \frac{{0.08k{e^2}}}{{{{\left( {2.90 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\end{aligned}\)

The force of attraction between H and N atoms is:

\(\begin{aligned}{l}{F_{{\rm{HN}}}} &= \frac{{k\left( {0.2e} \right)\left( {0.2e} \right)}}{{{{\left( {2.00 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\\{F_{{\rm{HN}}}} &= \frac{{0.04k{e^2}}}{{{{\left( {2.00 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\end{aligned}\)

The force of attraction between N and N atoms is:

\(\begin{aligned}{l}{F_{{\rm{NN}}}} &= \frac{{k\left( {0.2e} \right)\left( {0.2e} \right)}}{{{{\left( {3.00 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\\{F_{{\rm{NN}}}} &= \frac{{0.04k{e^2}}}{{{{\left( {3.00 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}^2}}}\end{aligned}\)

The net force between a cytosine and a guanine can be calculated as:

Thus, the net force between a cytosine and a guanine is \({\rm{7}} \times {\rm{1}}{{\rm{0}}^{ - 10}}\;{\rm{N}}\).

Evaluation of the total force of the DNA molecule

(c)

The total force of the DNA molecule containing \({10^5}\) pairs can be calculated as:

\(\begin{aligned}{l}{F_{{\rm{net}}}} &= \frac{1}{2}{10^5}\left( {{F_{{\rm{AT}}}} + {F_{{\rm{CG}}}}} \right)\\{F_{{\rm{net}}}} &= \frac{1}{2}{10^5}\left( {\left( {{\rm{5}} \times {\rm{1}}{{\rm{0}}^{ - 10}}\;{\rm{N}}} \right) + \left( {{\rm{7}} \times {\rm{1}}{{\rm{0}}^{ - 10}}\;{\rm{N}}} \right)} \right]\\{F_{{\rm{net}}}} &= 6 \times {10^{ - 5}}\;{\rm{N}}\end{aligned}\)

Thus, the total force of the DNA molecule containing \({10^5}\) pairs is \(6 \times {10^{ - 5}}\;{\rm{N}}\).