Problem 16-37: The two strands Tymine (T) of the helix-shaped DNA molecule are h
ID: 716997 • Letter: P
Question
Problem 16-37: The two strands Tymine (T) of the helix-shaped DNA molecule are held together by electrostatic forces as shown in H- Figure 16-44 in the textbook. Assume that the net average charge (due to electron sharing) indicated on H and N atoms is 0.2e and on the indicated C and O atoms is 0.4e. Assume also that atoms on each molecule are separated by 1.0x 1010m. 280mAdenine (A) - 0.300 nm - I.I I nm (a) Estimate the net force between a thymine and an adenine. For each bond (red dots) consider only the three atoms in a line (two atoms on one molecule, one atom on the other) (b) Estimate the net force between a cytosine and a guanine (c) Estimate the total force for a H-C DNA molecule containing 103 pairs of such molecules. Cytosine (C) - -0.290 nm- Guanine (G) 0.300 nm S 0.290 nm 1.08 nmExplanation / Answer
The force between 2 charges is attractive if they are of opposite nature else it is repulsive. Let us start solving by considering part (a). To calculate the net force between thymine and adenine, we need to essentially compute the force that one of them is applying on the other one.
Part (a)
Looking at the O----H--N bond, there is an attractive force between Oxygen and hydrogen whereas there exists a repulsive force between Oxygen and Nitrogen.
The magnitude of both these forces is given by the formula: F = k q1q2/r2 ------(1), where k is a constant with a value of 9 x 109 N, q1 and q2 are the charges involved and r is the distance between the charges. Let us 1st convert the charge on each atom to SI units.
qO = 0.4e = 0.4 x (1.6*10-19) C = 0.64 x 10-19 C, where qO is charge on oxygen.
qH = qN = 0.2e = 0.2 x (1.6*10-19) = 0.32 x 10-19 C, qH is charge on hydrogen and qN is charge on Nitrogen.
dO-H = 0.28 x 10-9 - 0.1 x 10-9 = 0.18 x 10-9 m, dO-H is the distance between oxygen and hydrogen.
dN-O = 0.28 x 10-9 m(given in diagram), dN-H is the distance between oxygen and nitrogen.
Using equation 1 labelled above,
FO-H = (9 x 109)*(0.64 x 10-19)*(0.32 x 10-19)/(0.18 x 0.18 x 10-9 x 10-9) N = 56.9 x 10-11 N
FO-N = (9 x 109)*(0.64 x 10-19)*(0.32 x 10-19)/(0.28 x 0.28 x 10-9 x 10-9) N = 23.5 x 10-11 N.
These both forces are opposite in direction therefore Fnet1 = (56.9 - 23.5) x 10-11 N = 33.4 x 10-11 N.
The direction of this force is towards the oxygen atom on Thymine.
Similarily, lets compute the force in the N--H-------N bond,
Fnet2 = FNH - FNN = 9 x 109 x 0.32 x 10-19 x0.32 x 10-19 x (1/(0.2 x 10-9)2 - 1/(0.3x10-9)2) = 12.8 x 10-11
Ftotal = Fnet1 + Fnet2 = 46.2 x 10-11 N------ It's an attractive force( since its positive)
Part (b)
There are 3 bonds involved here,
Ftotal = F1 + F2 + F3 , where F1, F2, F3 are the net forces in the individual bonds.
F1 = FOH - FON = [0.08x 9 x 10-9 x (1.6x10-19)2/(0.19x10-9)2] - [0.08 x 9 x 10-9x(1.6x10-19)2/(0.29x10-9)2
F2 = FNH - FNN = [0.04x9x10-9 x (1.6x10-19)2/(0.2x10-9)2] - [0.04x9x10-9x(1.6x10-19)2/(0.3x10-9)2]
F3 = FOH - FON = [0.08x9x10-9x(1.6x10-19)2/(0.19x10-9)2] - [0.08x9x10-9x(1.6x10-19)2/(0.29x10-9)2]
Therefore, FNET = F1 + F2 + F3 = 71.1 x 10-11 N
Part (C)
The total net force on a DNA molecule cannot be determined unless we know how many out of the 105 pairs are A-T and how many are G-C pairs.