BIO Base Pairing in DNA, I. 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 E21.21 shows the bonding of thymine and adenine. Each charge shown is±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, theO-exerts a force on both theH+and theN-, 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.

Short Answer

Expert verified
  1. Net force by thymine exerting on adenine is 8.89×10-9N, and it is an attractive force.
  2. The force between electron and proton in the hydrogen atom is 8.23×10-8N, and its ratio is 9.3

Step by step solution

01

Step 1:

Given data:

RNH=0.11×109mRNO=0.28×109mRNN=0.3×109mqN=qH=qO=e

02

Step 2:

Force exerting by O on H-N, let the left direction be positive and the right direction be negative.

Therefore, the force between oxygen and hydrogen.

FOH=Ke2ROH2=Ke2RONRNH2=9×1091.6×10192(0.280.11)×1092=7.97×109N

Forces Between oxygen and nitrogen

FON=Ke2RON2=9×1091.6×101920.28×1092=2.9×109N

03

Step 3:

Force exerting by H-N on N, let the left direction be positive and the right direction be negative.

Therefore, force between hydrogen and nitrogen is

FNH=Ke2RNH2=Ke2RNNRNH2=9×1091.6×10192(0.30.11)×1092=6.38×109N

Force between nitrogen and nitrogen is

FNN=Ke2RNN2=9×1091.6×101920.3×1092=2.56×109N

04

Net force

So, the total net force is

Fnet=(7.972.9+6.382.56)×109=8.89×109N

And force is directed towards the left which is attractive.

Hence, the Net force by thymine exerting on adenine is role="math" localid="1668165900263" 8.89×10-9N, and it is an attractive force.

The force between electron and proton in a hydrogen atom

FH=Ke2r2=9×109×1.6×101920.529×10192=8.23×108N

The ratio is

R=8.23×1088.89×1099.3

Therefore, The force between electron and proton in the hydrogen atom is 8.23×10-8N, and its ratio is 9.3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer’s heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

An electrical conductor designed to carry large currents has a circular cross section 2.50 mm in diameter and is 14.0 m long. The resistance between its ends is 0.104Ω. (a) What is the resistivity of the material? (b) If the electric-field magnitude in the conductor is 1.28 V/m, what is the total current? (c) If the material has 8.5×1028free electrons per cubic meter, find the average drift speed under the conditions of part (b).

A 5.00-A current runs through a 12-gauge copper wire (diameter

2.05 mm) and through a light bulb. Copper has8.5×108free electrons per

cubic meter. (a) How many electrons pass through the light bulb each

second? (b) What is the current density in the wire? (c) At what speed does

a typical electron pass by any given point in the wire? (d) If you were to use

wire of twice the diameter, which of the above answers would change?

Would they increase or decrease?

We have seen that a coulomb is an enormous amount of charge; it is virtually impossible to place a charge of 1 C on an object. Yet, a current of 10A,10C/sis quite reasonable. Explain this apparent discrepancy.

Consider the circuit of Fig. E25.30. (a)What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors? (b) What is the power output of the 16.0-V battery? (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery? (d) Show that the power output of the 16.0-V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.

Fig. E25.30.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free