8.8 (a)e−(E^2 −E^1 )/kBT=e−(5×^10− (^27) J/(1.38× 10 − (^23) JK− (^1) )(298 K)
=e−1.2×^10
− 6
≈0.9999
(b)e−(E^2 −E^1 )/kBT=e−(5×^10
− (^27) J/(1.38× 10 − (^23) JK− (^1) )(4 K)
=e−^9 ×^10
− 5
≈0.999
8.9
= 2.4
8.10This simple equation represents the foundation for his scientific career.
8.11 (a) With only one configuration, ln 1 =0 so S=0; (b) with two
configurations, ln 2 =0.69 and S =0.7kB; (c) with 10 configurations
ln 10 =2.3 and S=2.3kB.8.12 The proteins can fold into intermediate states that lead to the final
folded state and additional proteins can assist in the folding process.8.13 A profile of the different energies for a protein in different configura-
tions. The energy is minimal at the true, folded state although some
proteins may have closely lying states.
8.14 In many cases, proteins immediately form elements of secondary
structures, and the presence of the secondary structural elements
contributes to folding of the overall protein into the proper state.8.15
8.16 Prion-based disease represented a new mechanism that was inde-
pendent of other infectious diseases.8.17 The formation of βsheets extended over multiple copies of prions
provides a mechanism for aggregate formation.8.18 Most treatments are based upon attacking invading agents such
as bacteria. Such treatments are not effective towards stopping the
misfolding of a protein.CHAPTER 9
9.1 (a)(b)(c) 9.
(. )(.
==
×
×
−
−h
meλ66 10
9 1 10 10
34
31Js
kg 001010 73 10^51
)
.
×
− =× −
mms9
.
(. )(.
==
×
×
−
−h
meλ66 10
91 10 10
34
31Js
kg ).
×
− =× −
10
10 73 10^61
mms9
.
(. )(.
==
×
×
−
−h
meλ66 10
91 10 20
34
31Js
kg ).
×
− =× −
10
10 36 10^61
mmsN
N
conformation ee
conformation(^1) ERT
2
==−−Δ/ (()2 10×−^31 J mol−/(.)8 31J/(Kmol().^298 K =e0 81= 045.
q
e
=..
−
− =
1
1
24 110
q
eh
hkTB kTB(. )(
= − / =
×
−1 −
1
6 626 10^34
νν Js 5 500 3 0 10
138 101101
23 1cm cm s
JK−−
−−×
×
)(. )
(. ))( ) 298 K
ANSWERS TO PROBLEMS 453
9781405124362_5_end.qxd 4/29/08 9:17 Page 453