The two sides of the DNA double helix are connected by pairs

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 figures below show the thymine-adenine and cytosine-guanine bonds, respectively. Each charge shown is te, and the H-N and O-H distances are each 0.112 nm. 5. 0.290 H) (H Guanine k 0.280 nm Cytosine C-C Thymine Adenine 0.300--> je k-0.300 nm ,(C)--C nm NC 0.290 nm CN Calculate the net force that adenine exerts on thymine. Is it attractive or repulsive? For simplicity, 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 0 exerts a force on both the H+ and the N, and likewise along the N-H-N set. a. b. Calculate the force on the electron in the hydrogen atom, which is 0.0529 nm from the proton. c. Compare the strength of the bonding force of the electron in hydrogen with the bonding force of the d. Assuming that the bonding is due only to the forces along the O-H-O, N-H-N, and O-H-N, and that these adenine-thymine molecules three combinations are parallel to each other, calculate the net force that cytosine exerts on guanine due to the preceding three combinations. Is the force attractive or repulsive?

Solution

a)-|F|=k|q₁q₂|/r² =ke²/r²

O-H-N:

O^--H⁺:F =(8.99×10⁹N m²/C²)(1.60×10-¹⁹C)²/(0.170×10^-9m)²

= 7.96×109-9Nattractive

O^--N^-+ :F= (8.99×10⁹N m²/C²)(1.60×10-19C)²/(0.280×10^-9m)²

= 2.94×109^-9Nrepulsive

N-H-N:

N^--H⁺:F=(8.99×10⁹N m²/C²)(1.60×10^-19C)²/(0.190×10^-9m)²

= 6.38×109^-9Nattractive

N^--N^-:F=(8.99×10⁹N m²/C²)(1.60×10^-19C)²/(0.300×10^-9m)²=

2.56×109^-9N repulsive

Fattractive= 1.43×10-8N

Frepulsive= 5.50×10-9N

Fnet= 8.80×10-9Nattractive

b)-The bonding force of the electron in the hydrogen atom is a factor of 10 larger than thebonding force of the adenine-thymine molecules.5. If two electrons are each 1.50×10¹⁰m from a proton, as shown below, nd the magnitude and direction of the net electric force they will exert on the proton.The net force on the proton is the vector sum of the forces due to the electrons. Since the distances are the same, the magnitudes of the two forces are the same.F1=F2=ke2/r2= 1.023×10-8.Doing it component by component, we get

F1x= 1.023×10^-8N

F1y= 0F2x=F2cos65= 4.32×10^-9N

F2y=F2sin65= 9.27×10^-9N

Fx=F1x+F2x= 1.46×10^-8N

Fy=F1y+F2y= 9.27×10^-9N

F=qF2x+F2y= 1.73×10-8N= tan-1

FyFx= 32.4

The net force is 1.73×10^-8N

c)then we have to compare the force with the force between the electron and proton in hydrogen atom

d)-F=1.26 *10^-8 N

It is an attractive force.