6.8 A fit of the data yields a midpoint potential of 250 mV. The protein
is cytochrome c.6.9
=−51.8 kJ mol−^1.
6.10 The change in the Gibbs energy is the sum of the individual changes:
−61.0 +30.5 =−31.4 kJ mol−^1.
6.11 The change is the sum of the two values, or −64.8 kJ mol−^1.6.12 The midpoint changes from 450 to 210 mV with an apparent pKA
of 5.6.13P865 of the bacterial reaction center has a higher midpoint potential
than the heme and so an electron can be transferred from the
cytochrome to the reaction center.
6.14 Photosystem II has a higher midpoint potential so electron transfer
from the cytochrome to the photosystem II is energetically favorable.6.15
6.16
6.17 Since the motion occurs during the series of electron-transfer steps
that occur in the protein, the change in the distance of the Fe 2 S 2
cluster to the other cofactors changes and subsequently the rate of
electron-transfer changes (Chapter 10).
6.18 The observation of the rotation in three discrete steps would pro-
vide a very consistent model with the biochemical data.CHAPTER 7
7.1 M s−^1
7.2 Nothing can be said about the individual rates, but the rate con-
stant for the forward reaction is 10 times bigger than that of the
reverse reaction.
7.3 There will be no change.
7.4 Four half-lives are required, 600 s.
7.5 t1/2=0.69/k=150 s; four half-lives are required, 600 s.Imz=+=++mz−−(.)(1
2
1
2
1
2
Mg^2222 Mg Cl Cl 05 2))(.)().^221
2
+−=10 1 15
Imz=+=++ mz−− (.)()1
2
1
2
1
2
Na 22 2Na Cl Cl 05 1 ++−=(.)( ).1
2
05 1^205
=− , +
(
30 500
0
J/(Kmol) 8.315 J/(Kmol)(298 K) ln..)(.)
(. )
00025 0 00175
0 00225
ΔΔGGRTln[][]
[]
=°+
ADP P
ATP
iANSWERS TO PROBLEMS 449
9781405124362_5_end.qxd 4/29/08 9:17 Page 449