Chapter 16, Problem 51P

(a)

To determine

To Calculate: The net force between the thymine and adenine.

Answer to Problem 51P

The net force is $\text{4}{\text{.6×10}}^{\text{-10}}\text{N}$ .

Explanation of Solution

Given:

The given helix-shaped DNA molecule

The net average charge on hydrogen and nitrogen atom is $0.2e$ , on carbon and oxygen atom is $0.40e$ and the separation distance between each molecule is $\text{1}{\text{.0×10}}^{\text{-10}}\text{m}$ .

Formula used:

The net force formula is given as,

  $F=k\frac{q_1{q}_2}{r^2}$

Where, k is the proportionally constant, $q_{1}and{q}_2$ are testing charges and r is the separation distance between the testing charges.

The net force is calculated as,

  $F_{C.G}=2F_{O-H}+2F_{O-N}+F_{H-N}+F_{N-N}$

Calculation:

Draw the figure for the thymine and adenine.

  

Here, in this bonds which are responsible are NHO and NHN.

For NHO bonds:

For the O-H attraction, $F_{O-H}=k\frac{\left(0.4e\right)\times \left(0.2e\right)}{{\left(1.80\times {10}^{-10}\text{m}\right)}^2}=\frac{0.08k{e}^2}{{\left(1.90\times {10}^{-10}\text{m}\right)}^2}$

For the O-N repulsion, $F_{O-N}=k\frac{\left(0.4e\right)\times \left(0.2e\right)}{{\left(\text{2}{\text{.80×10}}^{\text{-10}}\text{m}\right)}^2}=\frac{0.08k{e}^2}{{\left(\text{2}{\text{.80×10}}^{\text{-10}}\text{m}\right)}^2}$

For NHN bond:

For the H-N attraction $F_{H-N}=k\frac{\left(0.2e\right)\times \left(0.2e\right)}{{\left(2.00\times {10}^{-10}\text{m}\right)}^2}=\frac{0.04k{e}^2}{{\left(2.00\times {10}^{-10}\text{m}\right)}^2}$

For the N-N repulsion, $F_{N-N}=k\frac{\left(0.2e\right)\times \left(0.2e\right)}{{\left(3.00\times {10}^{-10}\text{m}\right)}^2}=\frac{0.04k{e}^2}{{\left(3.00\times {10}^{-10}\text{m}\right)}^2}$

The net force is given as,

  $\begin{array}{l}{F}_{C.G}=F_{O-H}-F_{O-N}-F_{N-N}+F_{H-N}\\ =\left(\frac{0.08k{e}^2}{{\left(1.80\times {10}^{-10}\text{m}\right)}^2}-\frac{0.08k{e}^2}{{\left(2.80\times {10}^{-10}\text{m}\right)}^2}-\frac{0.08k{e}^2}{{\left(3.00\times {10}^{-10}\text{m}\right)}^2}+\frac{0.08k{e}^2}{{\left(2.00\times {10}^{-10}\text{m}\right)}^2}\right)\times \left(\frac{1}{1.0\times {10}^{-10}\text{m}}\right)\frac{k{e}^2}{d^2}\\ =\left(0.02004\right)\times \left(\frac{\left(8.988\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\times {\left(1.6\times {10}^{-19}C\right)}^2}{{\left(1.0\times {10}^{-10}\text{m}\right)}^2}\right)=4.6\times {10}^{-10}\text{N}\end{array}$

Conclusion:

The net force is $\text{4}{\text{.6×10}}^{\text{-10}}\text{N}$ .

(b)

To determine

To calculate: The net force between the cytosine and guanine.

Answer to Problem 51P

The net force is $7.1\times {10}^{-10}\text{N}$ .

Explanation of Solution

Given:

The given helix-shaped DNA molecule

The net average charge on hydrogen and nitrogen atom is $0.2e$ , on carbon and oxygen atom is $0.40e$ and the separation distance between each molecule is $\text{1}{\text{.0×10}}^{\text{-10}}m$

Formula used:

The net force formula is given as,

  $F=k\frac{q_1{q}_2}{r^2}$

Where, k is the proportionally constant, $q_{1}and{q}_2$ are testing charges and r is the separation distance between the testing charges.

The net force is calculated as,

  $F_{C.G}=2F_{O-H}-2F_{O-N}-F_{H-N}+F_{N-N}$

Calculation:

Draw the structure explaining the bond between the cytosine and guanine.

  

Here, the bond which are responsible are OHN, NHN, and NHO.

For OHN bond:

For the O-H attraction, $F_{O-H}=k\frac{\left(0.4e\right)\times \left(0.2e\right)}{{\left(1.90\times {10}^{-10}\text{m}\right)}^2}=\frac{0.08k{e}^2}{{\left(1.90\times {10}^{-10}\text{m}\right)}^2}$

For the H-N attraction $F_{H-N}=k\frac{\left(0.2e\right)\times \left(0.2e\right)}{{\left(2.00\times {10}^{-10}\text{m}\right)}^2}=\frac{0.04k{e}^2}{{\left(2.00\times {10}^{-10}\text{m}\right)}^2}$

For NHN bond:

For the H-N attraction $F_{H-N}=k\frac{\left(0.2e\right)\times \left(0.2e\right)}{{\left(2.00\times {10}^{-10}m\right)}^2}=\frac{0.04k{e}^2}{{\left(2.00\times {10}^{-10}m\right)}^2}$

For the N-N attraction, $F_{N-N}=k\frac{\left(0.2e\right)\times \left(0.2e\right)}{{\left(3.00\times {10}^{-10}\text{m}\right)}^2}=\frac{0.04k{e}^2}{{\left(3.00\times {10}^{-10}\text{m}\right)}^2}$

For NHO bond:

For the O-N repulsion, $F_{O-N}=k\frac{\left(0.4e\right)\times \left(0.2e\right)}{{\left(2.90\times {10}^{-10}\text{m}\right)}^2}=\frac{0.08k{e}^2}{{\left(2.90\times {10}^{-10}\text{m}\right)}^2}$

For the H-N attraction $F_{H-N}=k\frac{\left(0.2e\right)\times \left(0.2e\right)}{{\left(2.00\times {10}^{-10}\text{m}\right)}^2}=\frac{0.04k{e}^2}{{\left(2.00\times {10}^{-10}\text{m}\right)}^2}$

The net force is given as,

  $\begin{array}{l}{F}_{C.G}=2F_{O-H}+F_{O-N}+F_{H-N}+F_{N-N}\\ =\left(2\frac{0.08k{e}^2}{{\left(1.90\times {10}^{-10}\text{m}\right)}^2}-2\frac{0.08k{e}^2}{{\left(2.90\times {10}^{-10}\text{m}\right)}^2}-\frac{0.08k{e}^2}{{\left(2.00\times {10}^{-10}\text{m}\right)}^2}+\frac{0.08k{e}^2}{{\left(3.00\times {10}^{-10}\text{m}\right)}^2}\right)\times \left(\frac{1}{1.0\times {10}^{-10}\text{m}}\right)\frac{k{e}^2}{d^2}\\ =\left(0.03085\right)\times \left(\frac{\left(8.988\times {10}^9\text{N}{\text{.m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\times {\left(1.6\times {10}^{-19}\text{C}\right)}^2}{{\left(1.0\times {10}^{-10}\text{m}\right)}^2}\right)=7.1\times {10}^{-10}\text{N}\end{array}$

Conclusion:

The net force is $7.1\times {10}^{-10}\text{N}$

(c)

To determine

To calculate: The total force for the DNA molecule.

Answer to Problem 51P

The net force is ${\text{6×10}}^{\text{-5}}\text{N}$ .

Explanation of Solution

Given:

The number of pairs of molecules is ${\text{10}}^5$ .

Formula used:

The net force is given as,

  $F_{net}=\frac{1}{2}\left({10}^5\right)\left(F_{C.G}+F_{A.T}\right)$

Calculation:

The total force is calculated by using the given value is,

  $\begin{array}{l}{F}_{net}=\frac{1}{2}\left({10}^5\right)\left(F_{C.G}+F_{A.T}\right)\\ =\left(0.5\right)\left(\text{4}{\text{.6×10}}^{\text{-10}}\text{N}+\text{7}{\text{.1×10}}^{\text{-10}}\text{N}\right)\\ =6\text{×}{10}^{-5}N\end{array}$

Conclusion:

The net force is $6\text{×}{10}^{-5}N$ .