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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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