Question

The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the geometric shape of these molecules, adenine bonds with thymine and cytosine bonds with guanine. The figure (Figure 1) shows the thymine-adenine bond. Each charge shown is ±e, and the H−N distance is 0.110 nm.

A) Calculate the net force that thymine exerts on adenine. To keep the calculations fairly simple, yet reasonable, consider only the forces due to the O−H−N and the N−H−N combinations, assuming that these two combinations are parallel to each other. Remember, however, that in the O−H−N set, the O− exerts a force on both the H+ and the N−, and likewise along the N−H−N set.

B) Is the net force attractive or repulsive?

C) Calculate the force on the electron in the hydrogen atom, which is 5.29×10−2 nm from the proton.

D) Compare the strength of the bonding force of the electron in hydrogen with the bonding force of the adenine-thymine molecules.

Answer 1

Answer:

Explanation:

Attached is an image of the bond from search.

Take the force that diected towards thymine as negative

For the combination O-H-N:

The net force for this combination is,

$F=-F_{OH}+F_{ON}\\\\=\frac{Ke^2}{r^2}+\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[(0.280-0.110)\times 10^{-9}m]^2}+\frac{1}{0.280\times 10^{-9}m)^2}]\\\\=-5.03354\times 10^{-9}N$

For the combination N-H-N:

The net force for this combination is,

$F'=F_{NN}-F_{HN}\\\\=\frac{Ke^2}{r^2}-\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[0.300\times 10^{-9}m]^2}-\frac{1}{((0300-0..110)\times 10^{-9})m)^2}]\\\\=-3.822\times 10^{-9}N$

The net force that exerted by the thymine on the Adenine is,

$F_{net}=F+F'\\\\=-5.03354\times 10^{-9}N-3.822\times 10^{-9}N\\\\=-8.8558\times 10^{-9}N$

In three significant figures, the net force is $8.86\times 10^{-9}N$

b)

The negative sign indicates that the force is attractive since it acts towards the thyme.

c)

The force o electron due to the proton is,

$F=\frac{Ke^2}{r^2}\\\\=\frac{(9.0\times 10^9Nm^2/C^2)(1.6\times 10^{-9}C)^2}{(5.29\times 10^{-11}m)^2}\\\\=8.233\times 10^{-8}N$

Since the electron and proton are oppositely charged, the force acting on the electron is directed towards the proton.

d)

Take the ratio of the above forces:

$\frac{F}{F_{net}}=\frac{8.233\times 10^{-8}N}{8.233\times 10^{-8}N}\\\\=9.3$

Thus, the bonding strength of the electron in hydroen atom is 9.3 times to the bonding force of the adenine thymine molecules.