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$\bullet$$\bullet$ Base pairing in DNA, I. The two sides of the DNA dou-ble helix are connected by pairs of bases (adenine, thymine,cytosine, and guanine). Because of the geometric shape ofthese molecules, adenine bonds with thymine and cytosinebonds with guanine. Figure 17.43 shows the thymine-adeninebond. Each charge shown is $\pm e,$ and the $\mathrm{H}-\mathrm{N}$ distance is0.110 $\mathrm{nm}$ . (a) Calculate the net force that thymine exerts onadenine. Is it attractive or repulsive? To keep the calculationsfairly simple, yet reasonable, consider only the forces due tothe $\mathrm{O}-\mathrm{H}-\mathrm{N}$ and the $\mathrm{N}-\mathrm{H}-\mathrm{N}$ combinations, assumingthat these two combinations are parallel to each other.Remember, however, that in the $\mathrm{O}-\mathrm{H}-\mathrm{N}$ set, the $\mathrm{O}^{-}$exerts a force on both the $\mathrm{H}^{+}$ and the $\mathrm{N}^{-},$ and likewise alongthe $\mathrm{N}-\mathrm{H}-\mathrm{N}$ set. (b) Calculate the force on the electron inthe hydrogen atom, which is 0.0529 $\mathrm{nm}$ from the proton.Then compare the strength of the bonding force of the elec-tron in hydrogen with the bonding force of the adenine-thymine molecules.

a) $F_{n e t}=8.89 \times 10^{-9} \mathrm{N}$ , Attractive.b) $F_{H}=8.23 \times 10^{-8} \mathrm{N}, R=9.3$

Physics 102 Electricity and Magnetism

Chapter 17

Electric Charge and Electric Field

Gauss's Law

Electric Potential

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Shown here are models of t…

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(See Exercise 21.21.) (a) …

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Draw the structures of ade…

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In the double-helical stru…

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Draw the structures of cyt…

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In DNA the nucleic acid ba…

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Adenine and guanine are me…

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Draw guanine and cytosine …

03:07

DNA attains a secondary st…

00:50

Uracil is one of the bases…

08:52

Indicate whether each of t…

00:54

in this problem will be applying this force law to all the different combinations of interactions. But throughout this problem Q. One times cute, too, and they have some value around. It will always be equal t squared, since the charges of all the interactions will evolve, e chart and access charge be. And so first we're going to look at the O. H. Interaction. And as part of this interaction, I'm first going to look at the O. H. And so the force between these will be equal to K, which is 8.99 times 10 to the ninth Times Square, which is 1.6 times 10 to the negative 19. In that square over r squared in the R value is given to the problem. It's just the distance between the so in age, and it's 0.17 times 10 to the minus nine and that square. And so when you do this out, you get 7.96 times 10 to the minus nine Newton's and because they are opposite sign, this is attractive. This is an attractive horse. Now we're going to do the oh minus with the n minus, so first we did this interaction, and now we're going to do this interaction. So I'm not going to repeat all the number country here because it's very similar. The only thing that really changes is the artist. And since the distance between the an end is different than the distance between the O and H, what you'll get is 2.94 times 10 to the minus nine newness. And this is repulsive, since they have, like, charges. And so let's move on to the other action, the N H in interaction. So first we're going to do that. And then we're going to do that so that in a church interaction and just to be technical, it's in minus eight. Plus, the force is going to be 638 times. Tend to the negative onions if you cross it in again. The only difference is the distance is changing. The distance between in H is different than the rest of the distances. That's the only thing that separated these numbers from each other, and this force is attractive because there are also signs, and then the in minus and minus gives a force value that is 2.56 times 10 to the negative nine mutants, which is repulsive. So now we have these four forces to our tracks into repulsive, and we need the total force. And so the the total attractive force. If you summed the attractive forces, I would give you one point for three times 10 to the negative eight and the toll repulsive force that's the attractive forced. The toll repulsive force is five point I, zero times 10 to the negative nine. And so the net attractive force, by taking this minus this is equal to 8.80 times 10 too negative nine noon. And because the attractive forces larger than a repulsive force, this net force is attractive. So it's trapped in there. And so this is how strong this whole molecule was being held to the other molecules. And that's answer to party in part B. We just need to figure out the force between electron and 100 Adam. So again we have a physical to K E squared over our square. But then playing in these values 8.99 times, 10 to the nine. So that's K and then he is 1.6 year old times tent into your 19 and that's squared. And then our for the hydrogen atom is 0.529 times 10 to the minus night and then squared. And so this gives 8.22 times 10 to the negative eight noons, which is about a factor of 10 larger than this value here. Because this is to the negative, eh? And this is to the negative nine. So it's 10 times larger, and that completes.

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