BIO Base Pairing in DNA, II. Refer to Exercise 21.23 ....

BIO Base Pairing in DNA, II. Refer to Exercise 21.23 . Figure E 21.24 shows the bonding of the cytosine and guanine molecules. The $\mathrm{O}-\mathrm{H}$ and $\mathrm{H}-\mathrm{N}$ distances are each 0.110 $\mathrm{nm}$ . In this case, assume that the bonding is due only to the forces along the $\mathrm{O}-\mathrm{H}-\mathrm{O}, \mathrm{N}-\mathrm{H}-\mathrm{N},$ and $\mathrm{O}-\mathrm{H}-\mathrm{N}$ combinations, and assume also that these three combinations are parallel to each other. Calculate the net force that cytosine exerts on guanine due to the preceding three combinations. Is this force attractive or repulsive?

BIO Base Pairing in DNA, II. Refer to Exercise 21.23 . Figure E 21.24 shows the bonding of the cytosine and guanine molecules. The $\mathrm{O}-\mathrm{H}$ and $\mathrm{H}-\mathrm{N}$ distances are each 0.110 $\mathrm{nm}$ . In this case, assume that the bonding is due only to the forces along the
$\mathrm{O}-\mathrm{H}-\mathrm{O}, \mathrm{N}-\mathrm{H}-\mathrm{N},$ and $\mathrm{O}-\mathrm{H}-\mathrm{N}$ combinations, and assume also that these three combinations are parallel to each other. Calculate the net force that cytosine exerts on guanine due to the preceding three combinations. Is this force attractive or repulsive?

Key Concepts

DNA Base Pairing

DNA base pairing refers to the specific hydrogen-bonded interactions between complementary nucleotide bases in DNA, such as cytosine binding with guanine. This pairing underlies the double helical structure of DNA and is essential for accurate replication and genetic information transfer.

Hydrogen Bonding

Hydrogen bonding involves an attractive force where a hydrogen atom, covalently bonded to an electronegative atom (like oxygen or nitrogen), interacts with another electronegative atom. In DNA, these bonds, though weaker than covalent bonds, are crucial for the stability and specificity of the base pair interactions.

Molecular Force Calculation and Superposition

Molecular force calculation involves determining the net force on a molecule by vectorially summing the contributions from individual interactions, such as hydrogen bonds. The principle of superposition is used here to account for both the magnitude and direction of each bond's force, allowing a comprehensive analysis of the total interaction between the paired bases.

Chemical Bond Geometry

Chemical bond geometry concerns the spatial arrangement of atoms within a molecule as determined by bond lengths and angles. Understanding this geometry is vital when evaluating how individual hydrogen bonds add up, as the orientation and distance between interacting atoms critically influence the net attractive or repulsive forces observed between molecules.

Transcript

00:01 In this problem, we're going to be covering how to calculate the force between two charged particles.

00:07 In this problem, we have two molecules, cytosine and guanine, and we want to find out what is the force between these two molecules.

00:16 Now, the atoms of these molecules that are perpetuating the force are drawn here in blue and red here.

00:23 So this first row, for example, o -h -o, so the left atoms here are for cytosine, the right are for.

00:31 For guanine.

00:33 The atoms in red are all charged negative e.

00:37 The atoms in blue are all positive e.

00:41 And i also have the distances of these parts.

00:44 So the distance between h and n is 0 .1 ,1 nanometers, which is the same distance between the oxygen and hydrogen up on the left here.

00:55 Okay.

00:57 Now, in order to calculate the forces between these atoms, we will be applying coolum's law.

01:05 So coolum's law is f is equal to k which is the coolum constant times the charge of one particle times the charge of the other particle divided by the distance of those two charged charged particles square.

01:25 So this tells us the force between two charged particles.

01:31 And so to calculate the force between cytosine and guanine, we just need to apply this equation over and over again, add up the forces, and then we'll get the net force out.

01:43 So let's do this for our second row.

01:51 So we'll start off by just calculating what the force is between the two nitrogens.

01:58 Right? so applying this, we have charge e and negative e, or negative e and negative e...

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