Is it still tetrametric after the mutations?

Hi Norman,

Yes it is. In the meantime I simulated the tetrameric structure using swisslab servers and did my calculations on this pdb. However, I am not sure whether this is the right way to do so. Do you have an idea?

So depending on wether the mutation took place near to the areas at which polymerisation occur, you would need to access the free energy difference between the polymerisation with and without the mutation.

I know that the mutation does not occur within a polymerization domain - my question is: is there a mathematical model to calculate the change that happens to a tetramer structure if I know the free energy difference between a wildtype and a mutant monomer of said protein?

I personally don't think so, as the free energy gain by polymerisation occurs due to effects like shielding hydrophobic surface areas. So the question would be, why do your proteins build tetramers and would the mutation have any influence on that. You could compare solvation free energies between monomers and complexes of wild type and mutation. One cheap option would be MM/GBSA, and the expensive option would be FEP.

The so-called "entropy " doesn't exist at all.

During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

The so-called entropy was such a concept that was derived by mistake in history.

It is well known that calculus has a definition,

any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.

.

.

.

Why did the wrong "entropy" appear ?

In summary , this was due to the following two reasons:

1) Physically, people didn't know Q=f(P, V, T).

2) Mathematically, people didn't know AΔB couldn‘t become AdB directely .

If people knew any one of them, the mistake of entropy would not happen in history.

Please read my paper and those answers of the questions related to my paper in my Projects.

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity

1. Logic of the Second Law of Thermodynamics: Subjectivism, Logical Jump, Interdisciplinary Argumentation.

2. New thermodynamics pursues universality, two theoretical cornerstones:

2.1 Boltzmann formula: ro=A*exp(-Mgh/RT) - Isotope centrifugal separation experiments show that it is suitable for gases and liquids.

2.2. Hydrostatic equilibrium: applicable to gases and liquids.

3. The second and third sonic virial coefficients of R143a derived from the new thermodynamics are in agreement with the experimental results.

3.1. The third velocity Virial coefficient derived is in agreement with the experimental data, which shows that the theory is still correct when the critical density is reached.

4. See Appendix Pictures and Documents for details.