Question

BIO Base Pairing in DNA, II. Refer to Exercise 21.21. Figure E21.22 shows the bonding of cytosine and guanine. The O—H and H—N distances are each 0.110 nm. In this case, assume that the bonding is due only to the forces along the O—H—O, N—H—N, and O—H—N combinations, and assume also that these three combinations are parallel to each other. Calculate the net force that cytosine exerts on guanine due to the preceding three combinations. Is this force attractive or repulsive?

Short answer

Total net force by cytosine exerting by guanine is $1.26\times {10}^{-8}N$and is an attractive force.

Step-by-step solution

Step 1:

Force exerting by O-H on O, let the left direction be positive and the right direction be negative.

Force between Oxygen and Hydrogen $F_{O-H}$is

$$\begin{gathered} F_{O-H}=\frac{K{e}^2}{R_{O-H}^2} \\ =\frac{K{e}^2}{{\left(R_{O-O}-R_{O-H}\right)}^2} \\ =\frac{\left(9\times {10}^9\right){\left(1.6\times {10}^{-19}\right)}^2}{{\left[(0.29-0.11)\times {10}^{-9}\right]}^2} \\ =7.11\times {10}^{-9}\mathrm{N} \end{gathered}$$

Therefore, force between Oxygen and Oxygen $F_{O-O}$is

$$\begin{gathered} F_{O-O}=\frac{K{e}^2}{R_{O-O}^2} \\ =\frac{\left(9\times {10}^9\right){\left(1.6\times {10}^{-19}\right)}^2}{{\left[0.29\times {10}^{-9}\right]}^2} \\ =-2.74\times {10}^{-9}\mathrm{N} \end{gathered}$$

Step 2:

Force exerting by H-N on N, let the left direction be positive and the right direction be negative.

Therefore, the force between Hydrogen and Nitrogen is $F_{N-H}$is

$$\begin{gathered} F_{N-H}=\frac{K{e}^2}{R_{N-H}^2} \\ =\frac{K{e}^2}{{\left(R_{N-N}-R_{N-H}\right)}^2} \\ =\frac{\left(9\times {10}^9\right){\left(1.6\times {10}^{-19}\right)}^2}{{\left[(0.3-0.11)\times {10}^{-9}\right]}^2} \\ =6.38\times {10}^{-9}\mathrm{N} \end{gathered}$$

The force between Nitrogen and Nitrogen is

$$\begin{gathered} F_{N-N}=\frac{K{e}^2}{R_{N-N}^2} \\ =\frac{\left(9\times {10}^9\right){\left(1.6\times {10}^{-19}\right)}^2}{{\left[0.3\times {10}^{-9}\right]}^2} \\ =-2.56\times {10}^{-9}\mathrm{N} \end{gathered}$$

As the third part is the same as the first part with the same force$F_{O-H}$

Step 3:

So, the total net force is

$$\begin{gathered} F_{net}=(2\times 7.11-2\times 2.74+6.38-2.56)\times {10}^{-9} \\ =1.26\times {10}^{-8}N \end{gathered}$$

Hence, Total net force by cytosine exerting by guanine is $1.26\times {10}^{-8}N$and is an attractive force.