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BIO Base Pairing in DNA, II. Refer to Exercise 21.21. Figure E21.22 shows the bonding of cytosine and guanine. The O—H and H—N distances are each 0.110 nm. In this case, assume that the bonding is due only to the forces along the O—H—O, N—H—N, and O—H—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?

Short Answer

Expert verified

Total net force by cytosine exerting by guanine is 1.26×10-8Nand is an attractive force.

Step by step solution

01

Step 1:

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

Force between Oxygen and Hydrogen FO-His

FOH=Ke2ROH2=Ke2ROOROH2=9×1091.6×10192(0.290.11)×1092=7.11×109N

Therefore, force between Oxygen and Oxygen FO-Ois

FOO=Ke2ROO2=9×1091.6×101920.29×1092=2.74×109N

02

Step 2:

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

Therefore, the force between Hydrogen and Nitrogen is FN-His

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

The force between Nitrogen and Nitrogen is

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

As the third part is the same as the first part with the same forceFO-H

03

Step 3:

So, the total net force is

Fnet=(2×7.112×2.74+6.382.56)×109=1.26×108N

Hence, Total net force by cytosine exerting by guanine is 1.26×10-8Nand is an attractive force.

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Most popular questions from this chapter

(See Discussion Question Q25.14.) An ideal ammeter A is placed in a circuit with a battery and a light bulb as shown in Fig. Q25.15a, and the ammeter reading is noted. The circuit is then reconnected as in Fig. Q25.15b, so that the positions of the ammeter and light bulb are reversed. (a) How does the ammeter reading in the situation shown in Fig. Q25.15a compare to the reading in the situation shown in Fig. Q25.15b? Explain your reasoning. (b) In which situation does the light bulb glow more brightly? Explain your reasoning.

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.

(a) What is the potential difference Vadin the circuit of Fig. P25.62? (b) What is the terminal voltage of the 4.00-Vbattery? (c) A battery with emf and internal resistance 0.50Ωis inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-Vbattery. What is the difference of potential Vbcbetween the terminals of the 4.00-Vbattery now?

An electron experiences a magnetic force of magnitude4.60×10-15Nwhen moving at an angle of 60.0° with respect toa magnetic field of magnitude3.50×10-3T. Find the speed ofthe electron.

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