The two sides of the DNA double helix are connected by pairs of bases (adenine,
ID: 477556 • Letter: T
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. (Figure 1) shows the thymine-adenine bond. Each charge shown is ±e, and the HN distance is 0.110 nm .
Part 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 OHN and the NHN combinations, assuming that these two combinations are parallel to each other. Remember, however, that in the OHN set, the O exerts a force on both the H+ and the N, and likewise along the NHN set.
Part B
Calculate the force on the electron in the hydrogen atom, which is 5.29×102 nm from the proton.
Part C
Fill in the blanks using any of these words: Oxygen, 100, nitrogen, 1000, 10, hydrogen
The bonding force of the electron in the _______ atom is a factor of ______ larger than the bonding force of the adenine-thymine molecules.
Explanation / Answer
F=k|q1q2| /r2=ke^2/r^2
O-H-N:
O--H+:
F=(8.99×10^9N m2/C2)(1.60×10^-19C) ^2 / (0.170×10^-9m)2= 7.96×10^9-9Nattractive
O--N-+ :
F=(8.99×10^9N m2/C2)(1.60×10^-19C) ^2 / (0.280×10^-9m)2= 2.94×10^9-9Nrepulsive
N-H-N:
N--H+:F=(8.99×10^9N m2/C2)(1.60×10^-19C) ^2 / (0.190×10^-9m)2= 6.38×10^9-9Nattractive
N--N-:F=(8.99×109N m2/C2)(1.60×10^-19C) ^2 /(0.300×10^-9m)2= 2.56×10^9-9Nrepulsive
F attractive= 1.43×10^-8N
F repulsive= 5.50×10^-9N
F net= 8.80×10^-9N attractive
Calculate the force on the electron in the hydrogen atom, which is 5.29×102 nm from the proton.
F =ke^2/r^2 = 8.22 x10^8N
The bonding force of the electron in the hydrogen atom is a factor of 10 larger than thebonding force of the adenine-thymine molecules
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