Introduction to Biochemistry - End of session 3.4, 3.5, 3.6 assessment

NADH and FADH2 are the electron donors.

Whether or a compound activate, inhibit or does not impact a catalyst is a fact, one need just need to know about it. But these facts make biological sense to maintain homeostasis, therefore we can analyze it from a negative feedback standpoint.

Pyruvate Dehydrogenase (PDH) catalyze the conversion from Pyruvate to Acetyl CoA

CoA  indicate we have abundant reactant, so it should activate PDH.
Acetyl-CoA indicates we have abundant product, so it should inhibit PDH.

Acetyl-CoA starts the citric acid cycle. Therefore it favors energy production.

AMP indicates a low energy state, so it should activate PDH.
ATP indicates a high energy state, so it should inhibit PDH

The citric acid cycle produce NADH from NAD+, therefore

NADH should inhibit PDH
NAD+ should activate PDH

Fatty acid activates PDH, not sure why, but the answer say it does. As we said earlier, it is a fact.

I personally hate this type of problem that requires memorization, and I lose a lot of points from this one :(

I remember Glucagon activates PDK. It make sense for Glucagon to reduce citric acid cycle, so I think it is PDK inactivates PDH, then it must be true that PDP activates PDH.

I checked, while it make sense for calcium (released in muscle for contraction) to activate citric acid cycle, it does so by activating PDP, not by acting on PDH itself.

Therefore the answers are:

PDH is inactivated by the action of PDK (Pyruvate Dehydrogenase Kinase)
PDH is activated by the action of PDP (Pyruvate Dehydrogenase Phosphatase)

Pyruvate is oxidative decarboxylated to form Acetyl-CoA, that release one molecule of carbon dioxide, but this is outside of the citric acid cycle, do not count this one.

Isocitrate is oxidative decarboxylated to form alpha-ketoglutarate, that release one molecule of carbon dioxide.
alpha-ketoglutarate is oxidative decarboxylated to succinyl-CoA, that release one molecule of carbon dioxide.

Therefore a turn of citric acid cycle release 2 molecules of carbon dioxide. An easy way to note is that two carbons enters the cycle, so two carbons must exit, and the only way to exit is through carbon dioxide.

Isocitrate to alpha-ketoglutarate generate 1 molecule of NADH.
alpha-ketoglutarate to succinyl-CoA generate 1 molecule of NADH.
Malate to oxaloacetate  generate 1 molecule of NADH.

Therefore a turn of citric acid cycle generate 3 molecules of NADH.

Succinate to Fumarate generate 1 molecule of FADH2.

Therefore a turn of citric acid cycle generate 1 molecule of FADH2.

Using the OIL RIG mnemonics, we know that NAD+/FAD does not have the extra electrons and therefore is ix the oxidized state. NADH and FADH2 is in the reduced state.

NADH is reduced into NAD+, two electrons are transferred
FADH2 is reduced into FAD, two electrons are transferred.

Therefore each of these carrier can transfer two electrons at once.
In fact, the half reaction in the next problem shows that is the case for NAD+.

The first half reaction is going in reverse direction and the second half reaction go in forward direction, therefore, the standard reduction potential is simply $ -(-0.320) + 0.045 = 0.365 $.

The standard free energy can be computed using the formula $ \Delta G = -nF\Delta E = -2 \times 96.5 \times 0.365 = -70.445 $, make sure we get the sign correct, a positive reduction potential leads to a negative free energy change.

The free energy change is negative and therefore the reaction is spontaneous.

Complex I accepts electrons from NADH, while complex II accepts electrons from FADH2. Complex III accept electrons from Ubiquinol, while complex IV accepts electrons from cytochrome C.

Memonics - Ubiquinol (-ol) is the alcohol and therefore is the reduced form of Ubiquinone (-one), a ketone.

FCCP destroy the proton gradient, so ATP is not synthesized. The electron transport chain will simply function as usual.

Cyanide blocks the electron transport chain, so oxygen is not consumed. We have no proton gradient, so ATP is not synthesized.

Oligomycin blocks ATP synthase, so ATP is not synthesized, eventually, the proton gradient is built up too much so the electron transport chain will also stop, so oxygen is not consumed.

The $ \beta $ subunit of the $ \alpha\beta $ ring.

You need 15 protons to make a turn, a turn can synthesize 3 ATPs, so it need 5 proton flows to synthesize 1 ATP.

They do that through the glyoxylate cycle, it is essentially this:

Modifying their citric acid cycle by skipped two steps that include carbon dioxide production for partially diverting the intermediates to glucose synthesis.

Step I, Acetyl Coenzyme-A is combined with oxaloacetate to form citrate, and
Step IV, Acetyl Coenzyme-A is combined with glyoxylate to form malate.

(I forgot about step IV when I revisit this problem, need to try harder to memorize things)

The first one catalyze step III and is called isocitrate lyase, the second one catalyze step IV and is called malate synthase.

The molecule in the blank box that remains and feeds the cycle is called glyoxylate, and the molecule in the blank box that leaves the cycle is called succinate.

In the glyoxylate cycle, 2 acetyl-Coenzyme A is needed to create 1 oxaloacetate. 1 oxaloacetate is needed to create 1 phosphoenolpyruvate, and 2 phosphoenolpyruvate is needed to synthesize 1 glucose. So we need 4 acetyl-Coenzyme A molecule to synthesize 1 glucose molecule.

Plant has glyoxysome, a specialized organelle for the glyoxylate cycle.

The glyoxysome does the first step, assimilate acetyl coenzyme A to form glyoxylate and excess succinate.

The rest of the citric acid cycle happens in the mitochondria as usual.

The gluconeogenesis pathway happens in the cytoplasm.

When pathogen challenge via infection, the immunoresponsive genes are upregulated, this leads to itaconic acid production, itaconic acid inhibits the isocitrate lyase, and finally inhibits the glyoxylate cycle.

Step I, II, VI, VII, VIII are shared, the only not shared reactions are the step III, IV and V. III and IV are key steps to skip because they lost carbon through oxidative decarboxylation, step V is skipped simply because we don't have Succinate-Coenzyme A.

Obviously, B cannot do the glyoxylate cycle, it just die when there carbohydrate is scarce, both A and C can do glyoxylate cycle.

In addition to that, C can survive itaconic acid, so C can survive immune response.