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. $\textbf{Figure E21.21}$ shows the bonding of thymine and adenine. Each charge shown is $\pm e$, and the H$-$N distance is 0.110 nm. (a) Calculate the $net$ force that thymine exerts on adenine. Is it attractive or repulsive? 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) Calculate the force on the electron in the hydrogen atom, which is 0.0529 nm from the proton. Then compare the strength of the bonding force of the electron in hydrogen with the bonding force of the adenine-thymine molecules.

Key Concepts

Coulomb’s Law

Coulomb’s law is the fundamental principle describing the force between two point charges. It states that the magnitude of the electrostatic force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This law underpins the calculation of forces in systems ranging from atomic (electron–proton interactions) to molecular (electrostatic interactions in biomolecules).

Hydrogen Bonding

Hydrogen bonding is a type of dipole–dipole attraction that occurs when a hydrogen atom covalently bonded to a highly electronegative atom, such as oxygen or nitrogen, is attracted to another electronegative atom. This interaction, though weaker than covalent bonds, plays a crucial role in stabilizing the three-dimensional structures of biological macromolecules like DNA through base pairing.

Electrostatic Interactions in Molecular Systems

Electrostatic interactions refer to the attractive or repulsive forces between charged entities. In molecular systems, these interactions are critical for determining the stability and structure of complexes. Calculating the net force in such systems involves summing contributions from individual charged components, based on their magnitudes, distances, and spatial arrangements.

Comparative Force Scales in Chemical Bonding

Chemical bonding involves forces acting over different scales; for example, the forces that lead to the formation of a hydrogen bond in DNA operate over nanometer distances, whereas the forces binding electrons to nuclei in an atom operate over much shorter, sub-nanometer distances. Understanding and comparing these scales helps in appreciating the relative strengths of intermolecular versus intramolecular forces, which is essential in fields ranging from biochemistry to quantum chemistry.

Transcript

00:01 Hey everyone, this is question number 21 from chapter 21.

00:04 So in this problem, we are talking about dna and the bonds between, or the attractive forces between thiamine and aden.

00:15 So we are given that the charges are plus or minus e and that the distance between h and 0 .11 nanometers.

00:22 The rest of the given information is in a picture in your book.

00:25 So then we're asked to find part a, the net force that thiamine exerts on adenine.

00:30 Or repulsive, and then b, the force on the electrons and the hydrogen atoms, we're given a radius, and then from the proton.

00:38 And then we're supposed to compare the strength of the bonds.

00:41 So i wrote over here the two attractive forces that we're going to be working with, an o from thymine, and then the hn on adenine, and then the nh on thymine, and the n on aden.

01:00 So just as a real quick, so we can get our heads thinking about directionality because we are asked the direction.

01:07 This o and h, that's going to be attractive.

01:10 The o and the n is going to be repulsive because they're opposite.

01:13 The n and the n are going to be repulsive, and the n and the h are going to be attractive.

01:18 So total, we're going to denote left as positive and right as negative.

01:24 Okay, so let's start by using our kool -o -m equation to find the attractive force between oxygen and hydrogen.

01:40 So we have f -o -h equals k, 9 times 10 to the 9th, newton meter squared per cool -alm squared.

02:08 And then our q -1 and q2 our charges are going to be plus and minus the 3.

02:13 Charge of an electron, which is 1 .6 times 10 to the minus 19th.

02:26 And that is going to be squared because we have two of them.

02:35 And the radius from our picture is 0 .2 minus 0 .11 nanometers.

02:45 So that's going to be 0 .17 times 10 to the minus 9 meters, and that's squared.

02:53 And when we multiply that out, we get a force of 7 .97 times 10 to the minus 9th newton's, and that is positive going to the left.

03:15 Okay, so now we work with the other bond and the other attractive force in this part, so the o in the n, which is going to be repulsive.

03:28 So the net force between oxygen and nitrogen, same equation f -o -n.

03:39 9 times 10 to the 9th.

03:49 Same charges, 1 .6 times 10 to the minus 19 squared.

04:06 And then our radius for this one is 0 .28 times 10 to the minus 9 meters.

04:17 Because we're nanometers and that's squared.

04:20 And that's going to give us a repulsory...

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