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Monday, November 15, 2010
Lecture 34
Announcements:
1.
Assignments for this week:
Reading and problems as in LG.
No more PyMOL assignments until the last week of
class.
2. Office hours and review times, dates, rooms for the last week of classes are the
same as during the rest of the semester.
No office hours or reviews for next week (the
week of the Thanksgiving Break).
3. The Racker Lectures happen this week: Nobel Prize winner Mario Capecchi will
give two talks:
Thursday night at 8 PM in Call Auditorium in Kennedy Hall (a talk to a general
audience), “The making of a Scientist: An unlikely journey”
Friday at 4 PM in G10 Biotech (a sciencefocused talk), “Gene targeting into the 21
st
Century: Mouse models of human disease from cancer to neuropsychiatric disorders”
4. Alternative meets Western medicine: sponsored by the Ithaca Free Clinic, on Wed
11/17 at 5:30PM live demonstrations by a shiatsu massage therapist and an
acupuncturist; tips for stress reduction for college students.
p. 238
The
mitochondrial matrix is often referred to as the "N" side
,
and the
intermembrane space
, which is contiguous with the cytosol because of the
permeability of the mitochondrial outer membrane,
is often referred to as the "P"
side
.
This is because, relative to each other, the matrix is
Negative
and the
intermembrane space (and the cytosol) is
Positive
.
This separation of charge is a
result of:
1. proton pumping from the inside to the outside;
2.
consumption
of H
+
from the matrix at Complex III during the Q cycle to form UQH
2
,
and
consumption
of H
+
from the matrix at Complex IV to form O
2
.
This is the proton and electric potential gradient we have been discussing.
The free energy available (starting from NADH) is as follows:
NADH
+
H
+
+
1/2 O
2
NAD
+
+
H
2
O
NAD
+
/ NADH
E
=
0.28 V
(
in real
mito. matrix, not 1M!
)
1/2 O
2
/ H
2
O
E
=
+0.78 V
(
in real
mito. matrix
)
∆
E
=
+1.06 V
∆
G
=
nF
∆
E
=
205 kJ/mol
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View Full DocumentThe free energy of transport to form the gradient can be written using the pH, since
protons are what are being transported:
∆
G
T
=
RT ln ([H
+
]
cyt
/[H
+
]
matrix
)
+
zF
∆ψ
=
2.3 RT (pH
cyt
 pH
matrix
)
+ F
∆ψ
Typical
values:
T
=
310 K
(pH
cyt
 pH
matrix
)
=
(7  7.5) = 0.5
(note: the matrix always
has higher pH)
∆ψ
= +0.15 V (always negative in the matrix)
∆
G
T
=
3.0 kJ/mol
+
14.5 kJ/mol
=
+17.5 kJ/mol (per mole H
+
transported)
Notice that the difference in electrical potential is much more important that the
difference in H
+
concentrations in typical animal mitochondria!
This is because so
much other metabolism occurs in the mito matrix that its pH must not be allowed to rise
very high.
The mito matrix is
buffered
by the presence of the many proteins therein,
with their many ionizable groups.
Remember that the above 17.5 kJ/mol (per mole H
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 Fall '09
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