Biology_2112_problemkey-101510

Biology_2112_problemkey-101510 - Biology 102, Fall 2004...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Biology 102, Fall 2004 Problem set 5 For the second exam Entropy & Equilibrium Constant Problem Suppose that a molecule has 10 different configurations. The 10 configurations can be divided into 2 groups, one group has 3 members (group I) each with the energy E 1 , and a second group (group II) with 7 members each with an energy E 2 . Question: At equilibrium, what is the equilibrium constant between group I and group II in terms of energy and entropy? Problem set 5 Answer Key Equilibrium Constant Problem Suppose that a molecule has 10 different configurations. The 10 configurations can be divided into 2 groups, one group has 3 members (group I) each with the energy E 1 , and a second group (group II) with 7 members each with an energy E 2 . Question: At equilibrium, what is the equilibrium constant between group I and group II in terms of energy and entropy? Note: division is indicated by a / . Be careful with super & subscripts Answer: Group I conformations: I(1), I(2), I(3) Group II conformations: I(4), I(5), I(6), I(7), I(8), I(9), I(10). [I] = [I(1)] + [I(2)] + [I(3)] [II] = [I(4)] + [I(5)] +, + ,+, +, + [I(10)] [II] [I] = 7 [ I(4) ] because [I(4)] = [I(5)] =, . [I(10)] 3 [I(1)] and [I(1)] = [I(2)] = [I(3) [II] [I] = - [ {W 4 W 1 } T kln 7/3 ] / k T W 4 W 1 is E(14) & kln 7/3 = S 4- S 1 . [II] [I] = [ ( E T S) / k T ] NOTE the Energy equation utilizes the difference between the base configuration of the 2 groups thus exp {(W4-W1)-T(k*ln7 k* ln 3)}/kT Biology 102 Problem set 5 Sometimes a Thymine is found opposite a Guanine in double stranded DNA. The base pairing pattern is shown in the schematic diagram shown below. Single bonds are single lines. Double bonds are double lines. represents an Hydrogen bond. O Major Groove C Thymine N H O C C O H N Guanine C Minor Groove H N H Put in the water molecules hydrogen bonded to the groups when the bases are paired and when they separated in water. Determine if there is a net change in the number of hydrogen bonds in the two situations. Remember that the water molecules displaced when the bases are paired can hydrogen bond with each other....
View Full Document

Page1 / 11

Biology_2112_problemkey-101510 - Biology 102, Fall 2004...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online