I. TCA Cycle
A. Regulation of PC, Pyruvate Carboxylase
1. PC is the “anaplerotic” (filling up) reaction for the TCA; it is also a G’neo rxn: pyr + ATP
+ CO
2
 oxac + ADP
(Oxidation of 18 of the 20 AA also produces TCA ints.)
2. ACoA activates PC; PC requires that ACoA be bound to it in order to be active a. MR (for G’neo): ACoA is the product of PDH an alternative use of pyr b. ML: If [ACoA] is high, then pyr is not needed for PDH, can be “conserved as CH
2
O” and converted to G ACoA cannot be converted to G: PDH is irreversible AND THERE IS NO
BYPASS c. MR (for TCA): ACoA reacts with the product of PC, d. ML: if [ACoA] is high, oxac is needed to react with it, PC must produce oxac. (same
MRD for G’neo and TCA)
B. Purpose of TCA: oxidize C from food molecules
1. e- from C are transferred to NAD + or FAD, which carry them to ET
2. Compare the oxidation states of C in G, lactic acid, and CO
2
. The average is zero for the
C in G or lactate, but it is +4 in CO
2
.
C. Calculating oxidation states for C atoms, based on atoms/groups bonded to them :
1. Count -2 for =O and for -O -
2. Count -1 for -OX (-OH; -OR; -SR)
3. Count +1 for –H
4. Count nothing for ANY/ALL C
5. sum of above counts, plus that for the C must equal the charge shown
D. Citrate Synthase :
1. oxaloacetate must bind to CS and change its conformation in order for ACoA to bind.
2. Otherwise, if CS had only one, active conformation, ACoA could bind and:
ACoA + H2O

acetate + CASH.
To “reverse” this, 2ATP equivalents are consumed, wasted.
E. Conversion of pyruvate to CO
2
1. The #1 C is converted to CO
2
in PDH.
2. The #2 C of pyruvate is released as CO
2
100% in the “2nd turn” of the TCA (0% in 1st)
3. The third C of pyr is released 50% in the “3rd turn” of TCA and the other 50% will be equally distributed among the 4th C’s of oxac (12.5% X 4). One half of this total (the 2 end C’s) will be released in each round: 25% in the 4th turn; 12.5% in 5th; etc...
F. Pyruvate dehydrogenase (PDH)
In E Coli, a 60 subunit “multi enzyme” complex of 4.6 X 10 6 g/mol.
G. PDH Regulation by Product Inhibition:
1. ACoA can donate its acetyl group to the “acetyl group carrier” of PDH, preventing it from accepting an acetyl group from pyruvate
2. NADH can donate an epair to (one of) the “e pair carrier(s)” of PDH, preventing it from accepting an e-pair from pyruvate a. MR: NADH and ACoA are direct products of PDH b. ML: When [NADH] and/or [ACoA] is/are high, there is not a need to produce more.
Also, pyr isn’t needed as a fuel, can be conserved as CH
2
O:
PDH is irreversible AND has no bypass .
H. PDH Regulation by phosphorylation/dephosphorylation
The enzyme (PDH) is inactivated when Pi is added to one of its residues in a reaction catalyzed by PDH Kinase .
This kinase is activated by the products of PDH, NADH and ACoA (MR, ML above).
The Kinase is inhibited (and so PDH is “ activated ”) by:
1. pyruvate: MR: direct substrate of PDH. ML: when [pyr] is high, there is plenty for use as fuel or for storage as FAs.

2. ADP: MR: PDH feeds fuel to TCA which pushes ET and ATP production in OP. ML: when [ADP] is high, ATP is being consumed and ATP production is needed.
3. Ca 2+ : MR: Ca 2+ is released in a muscle cell in response to a nerve impulse and is required to trigger protein movement in muscle contraction . This consumes ATP ML: “same” as high [ADP]: need for ATP production.
Ca 2+ also activates PDH phosphatase (PDHPase) so that PDH gets dephosphorylated and activated.
I. The TCA is a Major Intersection of Metabolism
Numerous compounds are produced from or converted to TCA intermediates
1. G’neo consumes TCA intermediates, as does production of some AAs
2. Oxidation of 18 of the 20 AA also produces TCA ints.
II. Electron Transport, ET
ET: the oxidation of NADH and FADH
2
A. NET reaction: 2NADH + O
2
as e
+ 2H +
are passed to O
2
 2NAD + + 2H
2
O.
.
This occurs in three multistep rxns catalysed by large, multisubunit, integral membrane multienzyme complexes of the inner mitochondrial membrane, which surrounds the matrix , where the TCA occurs.
B. Mitochonrial Membranes
1. Protein is a major structural and functional component of membranes.
2. Molecules up to 10-kD permeate the outer membrane
3. The inner membrane is about 75% protein, about 25% phospholipid by weight
C. ET and OP are Tightly Coupled
1. Each of the three multienzyme complexes supplies energy for oxidative phosphorylation to produce one ATP.
2. Each of the 3 complexes acts as a “proton pump” expelling H + from the matrix.
3. The common intermediate which couples ET and oxidative phosphorylation, (OP) is
NOT H + , it is the H + gradient (high [H + ] outside the matrix and low [H + ] inside).
D. Net Reactions of ET Complexes (balanced for one O
2
)
1. NADHDH net reaction : 2NADH + 2H
+
+ 2Q

2NAD
+
+ 2QH
2
2. Cytochrome Reductase net reaction:
2QH
2
+ 4 cyto c (Fe
3+
)

2Q + 4H
+
+ 4 cyto c (Fe
2+
)
3. Cytochrome Oxidase net reaction:
4 cyto c (Fe
2+
) + 4H
+
+ O
2

4 cyto c (Fe
3+
) + 2H
2
O.
E. 1. The actual e- transport is to and from: flavin ring, Fe ions in Fe-S centers and hemes of cytochromes, Coenzyme Q, and Cu ions.
2. In NADHDH (aka NADH-Q oxidoreductase) the major free energy release is in e- transfer to flavin (FMN). So e -
F. E o
from FADH
2 of TCA enter “below” this point and only 2 ATP are produced.
’, the Standard Reduction Potential
1. The reduction potential indicates the tendency to accept e
-
(s); the more

the E o ’, the greater the tendency.
2. For example: succinate + CO
2
+ 2 e
-
 α-kg E o ’ = -0.67V
NAD
+
+ H
+
+ 2 e

NADH E o ’ = -0.32V a. NAD
+
has more

(less negative) E o ’ and a greater tendency to accept e- than does succinate CO
2
.

b. If all reactants and products were present at 1M (except H
+
: pH = 7), α-ketoglutarate would give up 2 e
-
to form succinate + CO
2
and NAD
+
would accept the 2 e
-
. (This occurs by way of 2 reactions in the TCA) c. To obtain the reaction that would occur spontaneously, one of the half-reactions must be reversed, the sign of its E o ’ must be changed, and the sum of the E o ’ values must be positive. So the first reaction is reversed, its E o ’ becomes + 0.67 v, which added to –0.32 v for the 2 nd
rxn gives Δ E o ’ = +0.35 v.
3. Free Energy and Reduction Potential
ΔG°’ = -nF(ΔE o ’)
= -2mole(96.5kJl/v mol)(+0.35 v) = -67.3kJ.
This is enough free energy to couple to GTP formation (+30kJ) and have -37 kJ for the reaction.
(F = 96.5kJ/v mol; n = # mol e- cancelled in balanced rxn)
G. The Enzyme Complexes of ET
1. NADH –Q oxidoreductase (NADHDH): a. The 2e from NADH are passed to flavin mono nucleotide (FMN) a tightly bound prosthetic group of this 25 subunit complex. deltaG°’ is a large negative for this step. b. The 2e are then passed to a series of iron – sulfur clusters, and then to coenzyme Q.
Q has a long hydrophobic chain (C
50
) which keeps it “dissolved” in the inner membrane in which it moves freely. c. The 2e of FADH
2
from succinate dehydrogenase are passed through FeS proteins to
Q, “below” the energy releasing step of NADHDH.
2. Cytochrome Reductase: a. Cytochrome reductase has 3 subunits that are the major actors: cytochrome b, cytochrome c1, and an iron-sulfur protein (ISP) b. It also has 2 binding active sites for Q binding: i. Qo, which is toward the o uter side of the lipid bilayer (where stigmatellin is bound) between the Fe-S cluster and heme b
L
; and ii. Qi which is toward the i nner (matrix) side. c. Both Qo and Qi are in the portion of the protein that spans the inner membrane and is surrounded by the long -CH
2
- chains of the lipid bilayer in which Q moves. d. The reactions catalyzed by cytochrome reductase are called the “ Q cycle ”. They involve passing e
-
from QH2 to cyto c by way of the ISP and cyto c1, and from one one QH
.
-
to another by way of hemes bL and bH, producing Q:
2-
.
The Q Cycle: Cycle 1
1. At Qo, QH2 donates 1e
-
to the ISP, which donates it to cyto c1, which passes it to cyto c;
QH2 also donates its 2H
+
to the aqueous medium outside, producing Q
.-
( notice “dot” and
“negative charge”)
2. Q
.-
donates 1e
-
to heme bL which passes it to heme bH; Q is released from Qo
3. Q binds at Qi, accepts the 1e
-
from bH, and reverts to Q
.-
.
Net: QH2 + cyto c (Fe
3+
)  Q
.-
+ cyto c (Fe
2+
) + 2H
+
(outside)
The Q Cycle: Cycle 2

1. Another QH2 goes through steps 1 and 2 of cycle 1, but in step 3 the electron from bH is donated to Q
.-
at Qi. When this Q:
2 -
binds 2H
+
on the matrix side, one of the 2QH2’s that was consumed is regenerated.
Net for cycle 2:
Q
.-
+ cyto c (Fe
3+
) + 2H
+
(matrix)

Q + cyto c (Fe
2+
) + 2H
+
(outside)
Q Cycle Net Reaction
Sum for cycle 1 + cycle 2:
1: QH
2
+ cyto c (Fe
3+
)

Q
.-
+ cyto c (Fe
2+
) + 2H
+
(outside)
2: Q
.-
+ cyto c (Fe
3+
) + 2H
+
(matrix)

Q + cyto c (Fe
2+
) + 2H
+
(outside)
Net:
QH
2
+ 2 cyto c (Fe
3+
) + 2H
+
(matrix)

Q + 2 cyto c (Fe
2+
) + 4H
+
(outside)
When QH
2
binds at Qo, it is in proximity to the ISP, donates e
-
to it. Then the conformation changes, placing Q
.-
in proximity to heme b
L
.
3. Cytochrome Oxidase a. Four e
-
from 4 cyto c (Fe
2+
) are passed to O
2
(along with 4H
+
, producing 2H
2
O) by cytochrome oxidase (8 subunits). b. An important point here is that O
2
binds between Fe
2+
and Cu
+
and 4 e
are passed to O2 almost at once. Three of the e
-
come from the changes in the oxidation states of Fe (to 4+) and Cu to (2+). The fourth comes from a tyrosine side chain hydroxyl (forming an oxide radical).
The rest involves entry of e
-‘s and H+’s to form H
2
O and return Fe and Cu to Fe
2+
and Cu
+
. c. Steps in the cytochrome oxidase reaction: i. O
2
binds between Fe
2+
and Cu
+
and 4 e
are passed to O2 almost at once. Three of the e
-
come from the changes in the oxidation states of Fe (to 4+) and Cu to (2+). The fourth comes from a tyr side chain hydroxyl (forming an oxide radical), which also donates its H
+
. ii. An e
-
from cyto c and an H
+
from the matrix return tyr (Y) to the –OH form.
Another H
+
converts the OH
-
on the Cu
2+
to water. iii. Another e
-
converts Fe from 4+ to 3+ and an H + converts O
2 -
to OH
-
. iiii. Then another H
+
produces water and two more e
-
return Fe to 2+ and Cu to 1+.
Ready to go again!
H. Proton Pumping a. It is obvious how the Q cycle can contribute to an H+ gradient. Another mechanism for ET enzymes is the proton pump:
1: nH+ bind to enzyme side chains on the matrix side of the inner membrane.
2. E accepts e-, providing energy to cause a change in conformational that puts the side chains outside.
3. H+ is released outside the matrix.
4. E releases e- and reverts to original conformation

b. The 4 steps in pumping H
+
above must occur in order: can’t have conformational change in step 2 without H
+
bound and e
accepted; can’t revert conformation in step 4 without releasing H
+
in step 3 and e- in step 4. c. Why is the energy release of redox rxns needed at step 2 rather than for conformational change in step 4? Step 2 is where H
+
moves from low [H
+
] to high [H
+
]: energy consumed d. How can H
+ bind where [H
+
] is low and be released where [H
+
] is high ?
The basic group of the side chain is positioned on the inside where its bound H
+
can form an Hbond or ionic bond with another side chain. Then, when the conformation changes, these groups move away from each other.
It is bonded to 2 groups on the inside and 1 on the outside; part of the energy input is to break this bond.
III. Oxidative Phosphorylation, OP
OP is the process of ATP synthesis by ATP synthase , which catalyzes: ADP + Pi

ATP.
It is oxidative in that it is tightly coupled to ET and so to the TCA and other processes that oxidize C from food.
A. ATP Synthase
1. ATP synthase has one major portion (Fo) that is embedded in the inner membrane and another (F1) that projects from it into the matrix like a lollipop.
2. When mitochondria are hit with ultrasonic waves, they are broken up into
“submitochondrial particles” that are inverted vesicles: the lollipops point out. These vesicles can carry out ATP synthesis if supplied with NADH. a. When treated with urea, the lollipops are dissolved into the aqueous medium, the vesicles are smooth, and they cannot synthesize ATP. The soluble protein can hydrolyse ATP but cannot synthesize it; this is the “F
1-ATPase”. b. When F1 is added back to the smooth vesicles without urea, the F1 again projects from the membrane and the vesicles can synthesize ATP.
3. The F1 portion has 9 subunits: α3β3γδε. a. The α
3β3 hexamer has α and β subunits alternating like the sections of an orange. b. The gamma ( γ) subunit has two long α helical segments projecting like an “axle” up between the “wheel” of α and β subunits.
4. The Fo part has a subunit composition ab2cn; where n= 10 in yeast and ~ 10 in E coli. a. The c subunits form a “ring” or “barrel” that is about the same thickness as the inner membrane and is embedded in it. b. The γ and, in E coli, ε subunits are firmly attached to the c barrel and rotate with it. (γ and δ in mitochondria)
5. ATP Synthase is a Molecular Rotary Motor a. The a, b
2
, δ and α
3
β
3
hexamer subunits are stationary, but the c barrel and the γ subunit are caused to rotate by the entry of H+ into the matrix. b.
The gamma subunit is not symmetric. A different “face” of it is in contact with each of the 3 αβ pairs, forcing the 3 βs to always be in 3 different conformations. c. As gamma spins, each β goes through a sequence of conformational changes: O 
L

T

O

L

T

6. Binding Change Mechanism of ATP Synthase a. Properties of the 3 conformations of β:
O – “open”; low affinity for ATP/ADP: releases ATP

L – “loose”; moderately affinity for binding ADP and Pi
T – “tight”; tight binding of ATP: can’t release ATP b. ATP is synthesized on the β subunit that is in the T form. But ATP production requires that T subunit to be converted to the O form for release of ATP. If T-ATP reverts to the L form, ADP + Pi will be released. c. It is the directional sequence of conformational changes: O

L

T

O

L

T

etc, that causes ATP production.
7. Evidence for the Binding Change Mechanism a. When dissolved F
1
“particles” (α
3

3
γδε) were incubated with ADP and Pi in H
2
18
O,
(water in which the O is
18O) it was found that this “ATPase” did make ATP , but didn’t release it: i. No ATP product was measured in the solution but it was observed that the
18
O from the H
2
18
O was incorporated into Pi ii. HOW? F
1
1. ADP + Pi

ATP catalyzed these reactions:
2. ATP + H
2
18
O

ADP +
18
O-Pi
F
1
was required for this to occur
ADP was also required
8. Rotational Motion from H + Flow: rotation of c barrel turns gamma subunit a. An H
+
binds to the asp (-CH3CO2
-
) side chain of a c subunit in the cytosolic halfchannel between the a and c subunits. (This is a water-filled channel on the intermembrane-space side). b. Upon accepting the H
+
, this c subunit changes conformation so as to push the c barrel to rota te this subunit “out” from contact with the a subunit and into contact with the lipid bilayer. c. A c subunit in the –CH
3CO2H form moves into contact with the matrix half-channel, where it can release H
+
into the low [H
+
] environment there. d. Repeating the steps on the previous slide for each of the 10 c subunits: i. each of the c subunits delivers a proton (H
+
) from the outside to the matrix (10H
+ in) ii. rotates the c barrel one complete turn iii. causes the synthesis of 3 ATP by the
 ’s
9. Factors in c Barrel Rotation a. In addition to the change in conformation of the c subunit, protonation of the asp side chain also breaks asp’s ionic bond with an arg side chain of the a subunit. b. This asp-arg bond holds the c subunit in contact with the cytosolic half-channel; but when it breaks and c “pushes off” the barrel can rotate.
10.
“Complete” net reactions for ET and OP (per O
2
, counting 3ATP per NADH):
ET: 20H
+
(in) + 2NADH + 2H
+
+ O
2

2NAD
+
+ 2H
2
O + 20H
+
(out)
OP: 20H
+
(out) +6ADP + 6Pi

6ATP + 20H
+
(in)
11.Experimental Results that Support the above Mechanism of ATP Synthase a. ATP synthesis in synthetic vescicles by ATP synthase can also be “driven” by bacteriorhodopsin, a well known H
+
pump. This protein under goes a conformational change when

it absorbs a photon and transfers an H+ across the membrane as a result (just as the ET enzymes do using energy from redox). The ATP synthesis is light dependent in this system. b. When the α
3

3
hexamer is immobilized on a slide and a fluorescent actin filament is attached to either the end of γ that attaches to the c-barrel or to the distal end of the c-barrel, there is ATP-dependent rotation that “swings the filament” around. It moves in slow 120 
steps @ low
[ATP], spins faster at high [ATP]. This emphasizes that rotation and ATP hydrolysis ( or synthesis ) are coupled.
12. Evidence for the Chemiosmotic Hypothesis for Coupling of EP and OP
The chemiosmotic hypothesis: ET and OP are coupled by the H
+
gradient. ET produces it by pumping H
+
out and OP consumes it by using the entry of H
+
to produce ATP.
1. an intact membrane that does not allow diffusion of H
+
or ions is required for coupling
2. adding acid to the medium of a suspension of mitochondria will increase synthesis of
ATP
3. ET does cause H
+
to exit the matrix
4. ET and OP are uncoupled by substances that transport H+ or charge across the inner membrane
13. Uncouplers of ET and OP a. Dinitrophenol (DNP) is an uncoupler: in the protonated (neutral) form, it can cross the inner membrane, where it can release H
+
: it carries H + into the matrix and it decreases ATP production. (It is also toxic.) b. Valinomycin carries K
+
across the membrane (into matrix) and also decreases ATP synthesis. How/why? The K
+
decreases the charge difference that is produced by the proton pumps of ET pumping H
+
out: K
+
carries that + charge back in. c. ET and OP are also uncoupled for heat production in brown fat cells in infant mammals in nonshivering thermogenesis
IV. Regulation of ET ( and OP and TCA)
A. 1. The rate of ET is controlled by [ADP]; ET rate increases when [ADP] inceases. But no enzyme of ET binds ADP or is regulated by ADP.
2. The [ADP] controls the OP rate by its effect on the rate at which ADP binds to

subunits of ATP synthase: a [S] effect on this E.
3. The rates of ET and the TCA are controlled by the OP rate: all three processes go at proportional rates.
B. How OP Controls the Rates of ET and TCA
1. OP consumes the H+ gradient, which is the product of ET. By doing this, it removes the product inhibition effects of the H+ gradient on ET.
2. ET in turn consumes NADH, the product of the TCA, and produces NAD+, the substrate of the TCA. This removes the effects of NADH in product inhibition, and increases [S], increasing the rates of TCA enzymes.
C. At low [ADP]
1. When the rate of ATP synthase is low, the H
+
gradient is consumed slowly and the gradient approaches a maximum.
2. When the gradient is at max (at rest, when ATP consumption is low and [ADP] is low) then net H
+
release by ET on the outside is slowed by the high [H
+
] and H
+
binding on the inside is slowed by the low [H+]: ET enzymes must “pump” H+ in order to catalyze e- transfer.

3. The gradient is the product of ET; when it is high ET is slowed (and when it is low, ET goes).
4. When ET is slow, the rate of NADH consumption slows, and [NADH] increases. This causes product inhibition effects that slow TCA. And ET produces NAD
+
slowly, [NAD
+
] decreases, and enzymes of TCA have a lower [S] supply, slowing them.
D. When [ADP] is high
1. When OP is rapid in response to high [ADP], the rate of H
+
entry into matrix increases and [H
+
] outside is decreased while [H
+
] inside is increased: gradient is less.
2. When the gradient decreases, the rate of H
+
binding on matrix side is faster (higher H
+ inside) and net release outside is faster (lower H
+
outside) so ET redox reactions are faster: rate of
ET is controlled by the level of the H
+
gradient.
3. When ET is fast, the rate of NADH consumption increases, and [NADH] decreases. This decreases the product inhibition effects, so TCA can go. And ET produces NAD
+
rapidly, [NAD
+
] increases, and enzymes of TCA have a greater [S] supply, increasing their rates.
E. LeChatelier’s Principle effects caused by [ADP]:
TCA: ACoA + GDP + Pi + 3NAD
+
+FAD

2CO
2
+ GTP + 3NADH + FADH
2
ET: 20H
+
(in) + 2NADH + 2H
+
+ O
2

2NAD
+
+ 2H
2
O + 20H
+
(out)
OP: 20H
+
(out) + 6ADP + 6Pi

6ATP + 20H
+
(in)
V. Electron Pair Shuttles
The NADH produced in the cytosol (cytoplasm) in G’lys cannot be transported into the matrix for
ET.
A. But its e- pair is transported by the glycerophosphate shuttle in skeletal muscle and brain:
1. The NADH is used to reduce DHAP
2. The glycerol-3-P produced in this reaction is oxidized back to DHAP by an enzyme on the outer surface of the inner membrane.
3. The FADH2 produced in this reaction is consumed by ET.
B. The Malate-Aspartate Shuttle Produces NADH in the Matrix
1. The NADH in the cytosol reduces oxac to malate. The malate enters the matrix and is converted to oxac, producing NADH.
2. The oxac is converted to asp for exit to cytosol because oxac doesn’t exit the matrix.
3. The amino group of asp is transferred to α kg producing glu, which is transported back to the matrix to bring the amino group back.
VI. ATP Produced in G’lys + PDH + TCA + ET + OP (per G)
G’lys, PDH, and TCA (See 62405):
Converted G to 6 CO
2
Produced 2 ATP + 2 GTP
Produced 2 FADH
2
+ 10 NADH
When ET consumes the 2 FADH
2
, OP produces 4 ATP
When ET consumes the 10 NADH, OP produces 30 ATP, for a net of 38 ATP / G
When the glyerol-3-P shuttle is used:
4 FADH2 go into ET, and 8 ATP from OP, and
8 NADH go into ET, and 24 ATP from OP. This nets 36 ATP / G