Question

In a DNA molecule, the base pair adenine and thymine is held together by two hydrogen bonds (see Fig. 16.5). Let’s model one of these hydrogen bonds as four point charges arranged along a straight line. Using the information in the figure, calculate the magnitude of the net electric force exerted by one base on the other.

Answer 1

The electrostatic force of attraction or repulsion between two charges is equal to the product of their charges and inversely proportional to the square of the distance between them.

F = kq1q2/r2

 

Here, k is the Coulomb’s constant, q1 is the charge, q2 is the charge, and r is the distance between the charges.

 

Take right hand directed forces as positive and left hand directed forces as negative. The electrostatic forces of Adenine on Thymine are shown below.

 

The distance between the hydrogen and the carbon is,

rCH = 0.18 nm + 0.12 nm

       = (0.3 nm)(10-9 m/1 nm)

       = 0.3 × 10-9 m

 

The electrostatic force of repulsion due to hydrogen on carbon is given as follows.

FCH = kqCqH/(rCH)2

 

Here, k is the Coulomb’s constant, qC is the charge on the carbon, qH is the charge on the hydrogen, is the distance of the hydrogen and the carbon.

 

Substitute (0.4)(1.6 × 10-19 C) for qC, (0.3)(1.6 × 10-19 C) for qH, 9.0 × 109 N∙m2/C2 for k, and 0.3 × 10-9 m for rCH.

FCH = (9 × 109 N∙m2/C2)(0.3)(1.6 × 10-19 C)(0.4)(1.6 × 10-19 C)/(0.3 × 10-9 m)2

        = 3.072 × 10-10 N

 

Therefore, the force of repulsion between the hydrogen and carbon is 3.072 × 10-10 N.

 

The distance between the hydrogen and the oxygen is,

 

rCH = 0.18 nm

        = (0.18 nm)(10-9 m/1 nm)

        = 0.18 × 10-9 m

 

The electrostatic force of attraction due to hydrogen on oxygen is given as follows.

FOH = kqOqH/(rCH)2

 

Here, k is the Coulomb’s constant, qO is the charge on the oxygen, qH is the charge on the hydrogen, rOH is the distance of the hydrogen and the oxygen.

 

Substitute (0.4)(1.6 × 10-19 C) for qO, (0.2)(1.6 × 10-19 C) for qH, 9.0 × 109 N∙m2/C2 for k, and 0.18 × 10-9 m for rCH.

FOH = [9 × 109 N∙m2/C2)(0.4)(1.6 × 10-19 C)(0.4)(1.6 × 10-19 C)]/(0.18 × 10-9 m)2

        = 8.533 × 10-10 N

 

Therefore, the force of repulsion between the hydrogen and oxygen is8.533 × 10-10 N.

 

The distance between the nitrogen and the carbon is,

rCN = 0.12 nm + 0.18 nm + 0.12 nm

       = (0.42 nm)(10-9 m/1 nm)

       = 0.42 × 10-9 m

 

The electrostatic force of attraction due to nitrogen on carbon is given as follows.

FCN = kqCqN/(rCN)2

 

Here, k is the Coulomb’s constant, qC is the charge on the carbon, qN is the charge on the nitrogen, rCN is the distance of the carbon and the nitrogen.

 

Substitute (0.4)(1.6 × 10-19 C) for qC, (0.3)(1.6 × 10-19 C) for qN, 9.0 × 109 N∙m2/C2 for k, and 0.42 × 10-9 m for rCH.

FCN = [(9 × 109 N∙m2/C2)(0.3)(1.6 × 10-19 C)(0.4)(1.6 × 10-19 C)]/(0.42 × 10-9 m)2

        = 1.567 × 10-10 N

 

Therefore, the force of attraction between the nitrogen and carbon is 1.567 × 10-10 N.

 

The distance between the nitrogen and the oxygen is,

rCN = 0.12 nm + 0.18 nm

       = (0.3 nm)(10-9 m/1 nm)

       = 0.3 × 10-9 m

 

The electrostatic force of attraction due to nitrogen on oxygen is given as follows.

FON = kqOqN/(rON)2

 

Here, k is the Coulomb’s constant, qO is the charge on the oxygen, qN is the charge on the nitrogen, rON is the distance of the nitrogen and the oxygen.

 

Substitute (0.4)(1.6 × 10-19 C) for qO, (0.3)(1.6 × 10-19 C) for qN, 9.0 × 109 N∙m2/C2 for k, and 0.3 × 10-9 m for rCN.

FON = [(9 × 109 N∙m2/C2)(0.3)(1.6 × 10-19 C)(0.4)(1.6 × 10-19 C)]/(0.3 × 10-9 m)2

        = 3.072 × 10-10 N

 

Therefore, the force of repulsion between the nitrogen and oxygen is 3.072 × 10-10 N.

 

The net force acting on the Thymine due to Adenine is the sum of the individual forces of Adenine atoms on the individual atoms of Thymine.

Fnet = FCH – FOH + FON - FCN

 

Substitute 3.072 × 10-10 N for FCH, 3.072 × 10-10 N for FON, 1.567 × 10-10 N for FCN, and 8.533 × 10-10 N for FOH.

Fnet = 3.072 × 10-10 N - 8.533 × 10-10 N + 3.072 × 10-10 N - 1.567 × 10-10 N

         = -4 × 10-10 N

Therefore, the magnitude of electrostatic force of Adenine on Thymine is 4 × 10-10 N.

Therefore, the magnitude of electrostatic force of Adenine on Thymine is 4 × 10-10 N.