Integral (4)

Problem:

$ \int\frac{d\theta}{A + B \cos\theta} $

Solution:

Let $ t = \tan\frac{\theta}{2} $, $ \cos \theta = \frac{1 - t^2}{1 + t^2} $, $ d\theta = \frac{2}{1 + t^2}dt $

$ \begin{eqnarray*} & & \int\frac{d\theta}{A + B \cos\theta} \\ &=& \int\frac{\frac{2}{1 + t^2}dt}{A + B \frac{1 - t^2}{1 + t^2}} \\ &=& \int\frac{2dt}{A(1 + t^2) + B(1 - t^2)} \\ &=& \int\frac{2dt}{(A + B) + (A - B)t^2} \\ &=& 2\sqrt{\frac{1}{(A + B)(A - B)}}\tan^{-1}\left(\sqrt{\frac{A - B}{A + B}}t\right) \\ &=& 2\sqrt{\frac{1}{(A + B)(A - B)}}\tan^{-1}\left(\sqrt{\frac{A - B}{A + B}}\tan\frac{x}{2}\right) \\ \end{eqnarray*} $

The second last step comes from a previously derived result which can be found here.

Integral (3)

Problem:

$ \int\frac{dt}{P + Qt^2} $


Solution:

Let $ \sqrt{\frac{P}{Q}}y = t $, so $ \sqrt{\frac{P}{Q}}dy = dt $.


$ \begin{eqnarray*} & & \int\frac{dt}{P + Qt^2} \\ &=& \int\frac{\sqrt{\frac{P}{Q}}dy}{P + Q\left(\sqrt{\frac{P}{Q}}y\right)^2} \\ &=& \int\frac{\sqrt{\frac{P}{Q}}dy}{P + Q\frac{P}{Q}y^2} \\ &=& \int\frac{\sqrt{\frac{P}{Q}}dy}{P + Py^2} \\ &=& \sqrt{\frac{P}{Q}}\int\frac{dy}{P + Py^2} \\ &=& \sqrt{\frac{1}{PQ}}\int\frac{dy}{1 + y^2} \\ &=& \sqrt{\frac{1}{PQ}}\tan^{-1} y \\ &=& \sqrt{\frac{1}{PQ}}\tan^{-1} \left(\sqrt{\frac{Q}{P}}t\right) \\ \end{eqnarray*} $

Introduction to Biochemistry - Quiz 3.7.4

Enzyme limited step is usually valves:

PFK-1 catalyze the reaction to convert fructo-6-phosphate into fructo-1,6-bisphosphate, it is irreversible and is enzyme limited.

Hexokinase catalyzes the reaction to convert glucose into glucose-6-phosphate, it basically traps the glucose in the cell and it is enzyme limited

Pyruvate kinase catalyze the reaction to convert phosphoenolpyruvate to pyruvate, the reaction is exergonic and it is enzyme limited.

Therefore the only substrate limited choice above is aldolase.

Pyruvate kinase is allosteric activated by fructo-1,6-bisphosphate. I googled on this one.
It doesn't make sense for ATP or citrate to activate pyruvate because they are products. I am just unsure if it is fructo-1,6-bisphosphate or fructo-2,6-bisphosphate.

It is allosterically inhibited by ATP and alanine. It makes sense for ATP because it is the end product, alanine I just googled it.

When blood glucose is high, fructo-6-phosphate accumulates in the muscle due to downstream inhibition of PFK-1. This leads to an inhibition of the upstream enzyme hexokinase, and ultimately glucose remains in the blood and is taken up in liver cells. Liver glucokinase is not inhibited by its product, so liver cells continue to take up glucose.

Glucose is shunted towards fatty acid synthesis, and
Glucose is used to synthesize glycogen.

Both are used for energy storage, as excess energy must be stored.

Introduction to Biochemistry - Quiz 3.7.3

AMP is at low abundance in cell, so a small decrease in [ATP] leads to large changes in [AMP].

Whatever added is an inhibitor, citrate make sense to inhibit glycolysis because that indicate high energy state. ATP also indicate a high energy state and that's why it also inhibit glycolysis.

It is an allosteric activator (I cheated - I forgot and Googled the result), and
It decreases the sensitivity of PFK-1 to ATP. (The make sense because we wanted to activate glycolysis)

A is Fructose
B is Fructose-1,6-bisphosphate
C is Fructose-6-phosphate
D is Fructose-2,6-bisphosphate

Therefore

C is the substrate for PFK-1 and PFK-2.
B is an allosteric regulator of PFK-1
D is the product of PFK-1

Introduction to Biochemistry - Quiz 3.7.2

Cells are in dynamic equilibrium. The answer is:

Cells alter metabolic fluxes in response to perturbation to restore metabolic reactions to an steady state.

Some reactions are actually maintained far away from equilibrium state.

Enzyme-limited reactions are often far from equilibrium, while substrate-limited reactions are near equilibrium.

Thermodynamically, A and D would have gone more towards the right hand side but it doesn't. That way they are most likely enzyme-limited.

Allosteric regulation and covalent modification cannot last that long, it has to have something to do with gene expression. Genes are transcribed into RNA and then RNA are translated into protein. The translation process is rather rigid, therefore the things that change is transcription. That's why the answer is

Altered transcriptional regulation of metabolic enzymes.

The official answer is allosteric regulation. Think about the Phospho-Fructo-Kinase that catalyze the conversion of Fruto-6-Phosphate to Fruto-1,6-Bisphosphate, the enzyme is activated by allosteric regulation by various energy state factors.

Introduction to Biochemistry - Quiz 3.7.1

CTP inhibits ATCase activity
When dissociated into subunits, one of them did not have catalytic activity, and bound CTP.

Homotropic refers to using the same substrate to regulate.
Allosteric refers to using different sites to regulate.

So we have

The binding of a substrate to one subunit promotes the transition to the R state in the other subunits of the enzyme.

We want to heterotropic, so it is not the second choice.
It is allosteric, so it is not the active site, not the first choice
We want activation, so we want R state.

Therefore the answer should be the 4th choice.

A small molecule binds the enzyme at a site other than the active site, stablizing the R state of the enzyme.

CTP is the ultimate product of the metabolic pathway that begins with ATCase. When CTP is in sufficient quantity, it blocks its own production at a very early step.

T is tight, R is relax, therefore the answer is:

The R state has a high affinity for the enzyme substrates; the T state does not.

The lecture spent some time talking about these MWC and KNF models, but I am not sure how much I understood it because we never applied these models to do prediction in the lecture or exercises. Here are just the memorized answer for the two coming questions.

The conformation of each monomer of the enzyme is connected. If one monomer switches to the R state, they all do. The binding of a substrate to a monomer favors the R state of that monomer (and thus the whole protein).

The conformation of each monomer of the enzyme is independent, but the conformation of a monomer is totally dependent of substrate binding. When unbound, the monomer is in the T state, and switches to the R state upon substrate binding. The R state of one monomer favors a switch to binding and the R state in the other monomers.

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.