Introduction to Biochemistry - Quiz 3.6.3

By stabilizing ATP in an ATP-ATP synthase complex with a free energy similar to the ADP-ATP synthase complex.

The T (tight) conformation.

The O (Open) conformation.

3

The transport of ATP out of the mitochondria [How could the proton motive force help with this? Proton wants to get in the matrix cannot drive anything to get out]

The transfer of protons into the intermembrane space by the electron transport chain. [This is obvious because it works against the proton motive force]

Introduction to Biochemistry - Quiz 3.6.2

It holds one component of the ATP synthase still so that it cannot rotate with respect to the stalk.

It rotates as protons pass through the a chain and transfers rotation to the asymmetric stalk.

It forms the asymmetric 'stalk' that changes the conformation of the ATP synthase.

The c-ring and the a chain.

Introduction to Biochemistry - Quiz 3.6.1

The given reaction is the sum of the first half reaction reversed and the second half reaction, therefore, the $ \Delta E^{t_0} $ is $ -(-0.320) + 0.82 = 1.14V.

Overall, there is two electron transferred, therefore n = 2. Using the formula $ \Delta G = -nFE $, we get the answer $ 2 \times 96.5 \times 1.14 = -220.02 $

A negative change of free energy means the reaction is spontaneous.

The system preserved 200/220 ~= 90% of energy that is derived from the redox reactions. Compare to a typical car engine which is only 10% - 50% efficient, I'd say the electron transport chain is amazingly efficient.

The exergonic transport of electrons is directly coupled to the endergonic transfer of protons from the mitochondrial matrix to the intermembane space.

The exergonic movement of protons from the intermembane space to the mitochondrial matrix is coupled to the endergonic process of ATP synthesis.

FCCP destroyed the proton gradient and therefore we cannot ATP synthesis is not supported.

The pH 7 solution created the proton gradient and valinomycin, by being able to take potassium [but not chloride] ion out of the mitochondrial matrix, also created a charge gradient, this two things together support ATP synthesis.

We do not have a proton gradient here so it does not support ATP synthesis.

The cyanide block the electron transport chain, but since we have an artificial proton gradient build up, we do not need the electron transport chain, so it still support ATP synthesis.

Oligomycin blocks ATP synthase, so regardless of the artificial proton gradient the setting will not support ATP synthesis.

Introduction to Biochemistry - Quiz 3.5.3

The correct answer is D.

Complex I and Complex II takes electrons from NADH and FADH2 respectively and transfer them to Coenzyme Q. Complex III take electrons from reduced Coenzyme Q to cytochrome C. Finally, Complex IV takes electron reduced cytochrome C to oxygen.

NADPH - this molecule is not involved in the electron transport chain.

$ O_2 $

Protons are transmitted across the mitochondrial membrane by the ETC, building up a gradient which is used to drive ATP synthesis.

To transfer a single electron to cytochrome c from the double electron carrier QH2.

To transfer electrons from universal electron acceptors to coenzyme Q.

Here are the equations for the electron transfer steps:

NADH + 5H(m) + Q -> NAD+ + QH(2) + 4H(i)

QH(2) + 2CYTC(ox) + 2H(m) -> Q + 2CYTC(red) + 4H(i)

4CYTC(red) + 8H(m) + O2 -> 4CYTC(ox) + 2H2O + 4H(i)

Taking the double of the first two equations, we get: 

2NADH + 10H(m) + 2Q -> 2NAD+ + 2QH(2) + 8H(i)

2QH(2) + 4CYTC(ox) + 4H(m) -> 2Q + 4CYTC(red) + 8H(i)

4CYTC(red) + 8H(m) + O2 -> 4CYTC(ox) + 2H2O + 4H(i)

And then sum them up, we get

2NADH + 10H(m) + 2Q + 2QH(2) + 4CYTC(ox) + 4H(m) + 4CYTC(red) + 8H(m) + O2

-> 

2NAD+ + 2QH(2) + 8H(i) +  2Q + 4CYTC(red) + 8H(i) + 4CYTC(ox) + 2H2O + 4H(i)

Therefore the answers are:

20 protons

2 molecules of water, and

1 molecule of oxygen

The number of protons worth a little more discussion here, obviously, we consumed 22 free protons in the matrix and produced 20 free protons in the intermembrane space. We count that as pumped 20 protons, we could as well say we pumped 22. The two protons, together with the two disassociated from the NADH, are used to form water.

Introduction to Biochemistry - Quiz 3.5.2

It means the reaction has a negative free energy - which means the reaction is spontaneous.

The reaction under question is the first reaction goes in reverse and the second reaction goes in forward, therefore, the $ \Delta E^{t_0} = 0.220 - 0.077 = 0.143 $.

The free energy can be computed using $ \Delta G = -nFE = -1 \times 96.5 \times 0.143 = -13.7995 $

A negative free energy indicates the reaction is spontaneous.

Oxidation is Loss - Reduction is Gain (OIL RIG)

During the first electron transfer, the first electron donor is oxidized and the second electron donor is reduced, therefore, the reaction has a reduction potential given by $ \Delta E = E_2 - E_1 $, where $ E_1 $ is the reduction potential of the first electron donor and $ E_2 $ is the reduction potential of the second electron donor.

In order for these reaction to be spontaneous, these reduction potential must be all positive, in other words, $ E_1 < E_2 $.

Therefore, the correct answer is:

Sort them according to their physiological reduction potential (E), lowest to highest.

We have already outlined the theoretical approach in the previous problem, basically, order them according to the reduction potential (the standard condition here is irrelevant, what is relevant is the condition in the cell).

An experimental approach would be to disrupt some known steps in the electron transport chain. If the chain is disrupted, then there wouldn't be consumption of oxygen or production of ATP, but definitely we should still see changes in oxidation states.

Therefore the correct answers are:

Rank the electron carriers in terms of their reduction potential ($ \Delta E $), and
Treat the cells with drugs to disrupt known steps in the electron transport chain and observe the oxidation state of each electron acceptor/donor.

This is fully oxidized Coenzyme-Q (also known as Q or ubiquinone).
The -one prefix reminds me about the ketone, with the two double bond O pointing outward, in reduced state, these ketone group are reduced into hydroxyl group.

It is obviously not a coenzyme Q.
It is not an iron-sulfur center as there is no sulfur.
The only sensible answer is that is a a heme.
Therefore the answer is

A heme functional group, the redox component of the cytochromes.

Introduction to Biochemistry - Quiz 3.5.1

The mitochondrial inner membrane - the electrons are transferred from the mitochondrial matrix to the intermembrane space.

The exergonic transfer of electrons powers the transfer of protons into the intermembrane space. The protons flow down their concentration gradient to power the endergonic synthesis of ATP.

The accept electrons from intermediates of the citric acid cycle, and
They donate electrons to the electron transport chain.

NADPH is different from NADH, they do not donate electrons to the electron transport chain, they donate electrons to power anabolic reactions in the cytosol.

Introduction to Biochemistry - Quiz 3.4.5

I failed this one - it is obvious that when sugar resources are low. It is not so obvious that we can still use the citric acid cycle to do biosynthesis because the citric acid cycle itself create biosynthesis pre-cursors and there is anapleroic reactions to replenish them. However, the glyoxylate cycle is important when microorganism is invading the host. In that case, the host is depleted with the sugar and the microorganism will need the glyoxylate cycle.

To summarize, the answer is

When sugar resources are low, and
During invasion of a host

Plants make use of the glyoxylate cycle during seed germination. and cease it when they are full grown therefore plants will not be affected by inhibitors.

By upregulating isocitrate lyase enzyme to overcome the normal levels of itaconic acid, and
By degrading the itaconic acid.

They become succinate and glyoxylate, glyoxylate can combine with another acetyl group to form malate with malate synthase, the succinate will simply go through the citric acid cycle as usual.

Note that we need to give a reason that is "NOT" true. The answer is:

The inhibition of ICL also inhibits the citric acid cycle of the microorganism and helps further to eradicate the pathogen.

This is simply not true, the ICL is not used in the citric acid cycle at all.

Introduction to Biochemistry - Quiz 3.4.4

The major goal of citric acid cycle is to produce the electron donors for the oxidative phosphorylation step, which means

To produce NADH
To produce FADH2

That's what I answered for the first time and I get partial credits, so I make another trial. I was thinking about it could be the production of GTP, after all it is for the energy.

But I learn a lesson here, citric acid cycle is also used to make precursors for bio synthesis of fatty acids and amino acids.

It is Fatty acids, Amino acids and pyruvate.

The problem is that pyruvate to acetyl coenzyme A released a lot of energy and is essentially irreversible, so there is no hope there. But once it enters the cycle, we cannot escape from the fate that two carbons will be decarboxylated away through the decarboxylation. So while it is possible to get glucose out of the oxaloacetate, the glucose is not made from the acetyl coenzyme A because there is no net conversion of acetate to oxaloacetate.

Iso-citrate to alpha-ketoglutarate and alpha-ketoglutarate to succinyl-coA.
This two oxidative decarboxylation tooks the carbons away.

A single turn in the glyoxylate cycle consume one and produce two oxaloacetate molecules, so it is a net production one one oxaloacetate molecule. To produce two oxaloacetate molecules and therefore 2 PEP molecules, we need two full turns.

Bacteria upregulate the glyoxylate cycle and down regulate the citric acid cycle.

I got this all wrong :(

I thought if the bacteria is feeding on pyruvate, there is no glucose around and it must build glucose, but that was false. Pyruvate itself can be an entry point to gluconeogenesis, therefore the cell can do citric acid cycle.

Once we decided it is the citric acid cycle, the rest follows:

Down regulate glyoxylate cycle
Down regulate gluconeogenesis
Up regulate citric acid cycle
Down regulate isocitrate lyase
Down regulate malate synthase
Unregulated citrate synthase
Upregulate phosphatase of AceK

Note: AceK can be used to phosphatase (i.e. remote the phosphate group) of the isocitrate dehydrogenase, that allows the citric acid cycle to move forward.

Introduction to Biochemistry - Quiz 3.4.3

This is obvious, $ O_2 $ and FAD are not product of the citric acid cycle.

To regenerate citric acid cycle intermediates, they are often used for biosynthesis of other molecules and the citric acid cycle would stop without them replenished.