## Tuesday, June 27, 2017

### Introduction to Biochemistry - Quiz 2.3.2

Enzyme catalyze reactions by making
"Weak non-covalent interaction between enzyme and substrate" and
"Transient covalent bond between enzyme and substrate".

Permanent bond would be useless for catalyzing reaction.
Binding energy is the result of these interactions.
Re-solvation is not defined.

D is the binding energy - this is the reduction of free energy in the transient state.

B is the activation energy of the enzyme catalyzed reaction.

I failed this one, twice, I still do not understand, the answer is

Weak non-covalent interaction between substrate and enzyme, and
Energy from desolvation of the substrate.

What I don't understand is if "Weak non-covalent interaction between substrate and enzyme" is an answer, then why "Transient covalent interaction that position the substrate in the active site" not an answer?

Overall - I am just confused by my "type-checker" - an energy is energy, an interaction and an interaction, type mismatch! And interaction can lead to the reduction of energy, sure, but an interaction is not an energy!

Desolvation is the process of replacing environment - substrate bonds with substrate - enzyme bonds.

### Introduction to Biochemistry - Quiz 2.3.1

At any time, the reaction proceeds in both direction - even not during equilibrium.
In an equilibrium that is biased towards product - for example, a strong acid/base reaction, the product concentration could be higher than that of the substrate.
In an equilibrium that is biased towards substrate - for example, a weak acid/base reaction, the product concentration could be lower than that of the substrate, and the substrate will simply never used up.

The only correct answer is "There is no net change in the concentration of the reactants".

A reaction is spontaneous if the product has lower free energy than the substate, a law in thermodynamics.

Both graph depict a spontaneous reaction.
If all other conditions are equal, the reaction in graph B is likely to proceed more rapidly than the reaction in graph A.

This is because graph B has a more stable transition state, a lower activation energy barrier to overcome.

Catalysts decreases the free energy of the transition state in order to reduce the activation energy of the reaction.

Basically make the graph A becomes graph B above.

An enzyme increases the rate of a reaction by stabilizing the transition state.

## Sunday, June 25, 2017

### Introduction to Biochemistry - End of session 2.1, 2.2 assessment

The interaction strength varies from ionic bond - hydrogen bond - Van der Waal's force, so hydrogen bond does not act over a longer or shorter distance, and it is not the weakest. The only difference is that because hydrogen bond is caused by a polar bond, so it is a directional force, unlike the others.

The answer is: "Hydrogen bonds are directional, both other types of noncovalent bond are not directional".

I did a similar problem before and got tricked, so I am not going to be tricked again. In this case, Arginine has a long side chain with a lot of bonds that can be rotated, none of the others can compare to.

It is negatively charged, so the last two options are gone, it does not have a benzene ring, so it is not aromatic, therefore the answer is "Negatively charged, small, polar"

This is simply using the editor to create the structure again :)

There are no consecutive sequence of alpha region and beta region, therefore it is a loop.

The dotted lines are hydrogen bonds between the amide group and the carboxlate group. N is relatively electronegative than hydrogen is, so it pull the electrons to the N side and leave a partial charge on hydrogen, on the other hand, the oxygen to relatively electronegative than carbon so it leaves a partial negative charge on oxygen. So these partial charges attracts each other to form the hydrogen bond.

The hydrogen bond are useful to stabilize the structure of the alpha helix.

cis or trans conformation make sense only when we talk about double bond. The peptide bond participate in a resonance structure so that it is "sometimes " double. The sometimes double characteristics make it planar. And only on such planar structure we say cis or trans.

The bonds are mostly trans (with the exception of proline) is because of steric clashes.

The answer are: "The peptide bond has trigonal geometry" and "To prevent steric clash between adjacent R groups".

We started with the Unfolded protein, because of "Hydrophobic Effect", we dropped 140 KJ/mol. The Enthalpy change bring us another 150 KJ/mol down. However, the "Chain Conformational Entropy" drop causes us to gain 220 KJ/mol, the end state is its "Native Fold".

In step "A", Anfinsen added 8M urea ...
In step "E", Anfinsen added a small amount ...
In step "B", Anfinsen re-oxidized

In step "C" and "D", Anfinsen dialyze the protein.

The hydrophobic interaction in the protein core and the hydrogen bonding network in secondary structures are the main cooperative interactions.

The nucleation condensation model states that the hydrophobic core forms first while the diffusion collapse model states that the secondary structures forms first.

HSP70 transiently bind to nascently translated protein to avoid it binding to others.
HSP70 bind to it through interactions with hydrophobic peptides.

## Saturday, June 24, 2017

### Introduction to Biochemistry - Quiz 2.2.4

The cytosol is the fluid area inside the cell membrane. Proteins are built there, in particular, with the ribosome near the rough endoplasmic reticulum. When they are built sequentially, the protein isn't well folded yet. This is particularly true if the folding requires interaction with the tail end of the protein, which is not synthesized yet. Multiple protein heads will likely have hydrophobic groups, and if not properly regulated, they would cluster and form aggregates.

The answer that is the closest to the description above is:

"The cytosol is a much more crowded environment than is typical for an in vitro experiment".

This is a repeating theme - an answer a lot of times - it is the hydrophobic interaction

Before two protein molecules forms their own hydrophobic cores, they could merge together so that the hydrophobic part is more deeply buried and therefore more thermodynamically favorable.

The last two answers are plain wrong. Chaperones do not digest protein, and they do not use energy to force protein to fold. Since hydrophobic interaction is the major driving force to get protein to fold, chaperones works by preventing intermolecular hydrophobic interaction, because once that is prevented, the intramolecular hydrophobic interaction will drive the folding of the protein folding by itself.

The answer is: "They prevent intermolecular interaction while the protein folds"

This is just memorizing, and I hate memorization, so I just looked it up.

In its ATP bound state, HSP70 does not bind to its substrate, while it its ADP bound state, HSP70 bound to a substrate.

The idea is that HSP70 has two states, the binding can be make transient,

At high temperature, entropy wins, the protein will denature, and potentially interact with each other to form aggregates.

The answer is "Protein are denatured, and at risk of forming intermolecular aggregates"

This is again a problem on memorization. HSP70 goes through transient cycles of substrate binding, and HSP60 forms a large cages.

### Introduction to Biochemistry - Quiz 2.2.3

The Levinthal's paradox is about the combinatorial complexity of conformation. If there are 100 bonds that can be rotated, and even if we simplify the possible rotation to just 3 choices, there will be $3^{100}$ rotations, that's a huge number. But protein can fold into an unique native fold. If the folding process goes through all possible conformation, that will takes forever.

Therefore protein folding must be a guided process, it cannot possibly be sampling all possible conformations.

The answer is "A protein would essentially never find its native fold by sampling all possible conformations".

It is fun to repeat the math here. Assuming 100 bonds, 3 configuration per bond, and 0.1 ps for sampling a configuration.

The time required is $\frac{(3^{100}) \times 0.1 \times 10^{-12}}{60 \times 60 \times 365}$ years. One can easily compute the value using logarithm and it is $10^{28}$ years, similar but not exactly equal to the value given in the lecture.

=100 * log(3) + log(0.1) - 12 * log(10) - log(60) - log(60) - log(365.25)

As discussed above, the protein folding process is guided. By what? By far, the dominating folding driver is the hydrophobic effect. Once the hydrophobic effect kicks in and bring the hydrophobic groups together, now the molecule is getting close and other interaction can happen. This is cooperation.

The answer is "There is a high degree of cooperativity in protein folding".

The first answer is wrong because we know free energy difference does to explain the speed of the process. Burning tree can release a lot of energy, but until someone set a fire a tree will practically never burns.

The third answer is not true either, we know pure protein sample folds itself  in aqueous solution.

The fourth answer is close, but not quite, many of the amino acid do not have a charged side chain, the only charge that always exists is at the terminals, that is just too far away - even if the salt bridge interaction "force" is large, the "chance" for them to be in reasonable proximity is low.

This is my original words when I replied on the website and it is accepted.

Diffusion collision forms secondary structures first, and then rearrange to get to the final fold. Nucleation condensation forms the hydrophobic core first, and then form the secondary structures.

## Thursday, June 22, 2017

### Introduction to Biochemistry - Quiz 2.2.2

If protein has many equally thermodynamically stable conformation, there wouldn't be a unique native fold.

The system of atoms is so huge that one can only simulate the fold of a few residues only, to predict the structure of a large protein is hard.

The native fold of the protein does not seems to have much correlation with temperature except in high temperature it denatures. So kinetic parameters doesn't seems to matter much.

The answer is the second choice - the most thermodynamically stable conformation is its native fold - this is essentially the thermodynamic hypothesis.

All of these are facts, but not all of them support the thermodynamic hypothesis, this two do:

RNase A regains activity when urea is removed - this means the protein can fold back to what it is by itself without help from anything else, and

We can predict the structure of some proteins based on sequence and knowledge of inter-molecular forces - this means protein molecules are just like any physical substances follow physical properties, there isn't any miracle about its fold - it just fold because of the law of Physics.

In high concentration urea - protein denatures. Beta-Mercaptoethanol is used to cleave the disulphide bond. (Disulphide bonds are bond formed between thiol group of two cysteine residues on the side chain). After all these bonds are cleaved, if we remove the urea by dialyzing it, it will become 90% active, however if we re-oxide the protein in the urea so that it forms the disulphide bonds randomly, the protein cannot go back to its native fold even after dialyzing. That is the Anfinsen experiment.

We see that first, protein spontaneously adopt their native fold after the urea is removed by dialyzing. Second, the disulphide bond should be formed after the folding complete, if not, the reoxided protein should also be able to get to the native fold, so we know the native fold specifies the location of the disulfphide bond, not otherwise.

B represents the native fold of the protein because it is the absolute minimum.
C represents a semi-stable folding immediate because it is a local minimum.

## Wednesday, June 21, 2017

### Introduction to Biochemistry - Quiz 2.2.1

When water is exposed to a hydrophobic molecule, it has a more restricted conformation because they must be all aligned in order to form hydrogen bond. Therefore, hiding the hydrophobic surfaces away can increase water's entropy. Although the folding of protein reduce energy, it also significantly reduce its conformational entropy, that's why the increase of water's entropy is important here.

The effect of increasing water's entropy by folding is called the hydrophobic effect and is one of the major force for the folding. Folding does decrease chain conformational entropy, but that is an inhibitor, not a driving force, folding does not increase chain conformational entropy. Water's entropy got increased by the folding, not decrease. Finally, the word "enthalpy" alone is vague, what it really means in this context should be the overall reduction of potential energy by pulling the opposite charges closer, that is indeed a driving force.

Therefore the answers should be "The hydrophobic effect" and "Enthalpy".

Enthalpy and The hydrophobic effect drives protein folding as we discussed above. Increased entropy of water is basically the same as hydrophobic effect, therefore that is not an answer either. The only remaining choice is "Chain conformational entropy". Indeed, the folding decrease the chain's flexibility to change conformation, therefore it reduces entropy and disfavor protein folding.

Hydrogen bond is the key player in stabilizing the secondary structure for both alpha helix and beta sheets. Salt bridges are strong, but we don't have formal charges around in the protein chain. The other forces are weaker than hydrogen bonds.

Protein folding does have an effect on all of the given factors, but in terms of scale, a folded protein is pretty much stuck, but the unfolded one have a lot of freedom to move around, so if I have to guess, i would guess "Chain conformational entropy".

And I am right :)