Materials and Methods
The polypeptide is free swivel chain and has levers which collide to
water molecules. The collision in the protein folding in the cell
follows the Brownian movement. It is much stronger. The calculation of
the translational molecular kinetics is based on the translational
enthalpy data. Thus, the kinetic energy of the water molecule is,
\begin{equation}
\sim 2.0(kcal/mol)=\frac{1}{2}mv^{2}\nonumber \\
\end{equation}And the velocity of water molecule gives,
\begin{equation}
2.0\ (kcal/mol)\ =\frac{1}{2}\times 18(g/mol)\times v^{2}\nonumber \\
\end{equation}\begin{equation}
v^{2}=20\ (kcal/g)\nonumber \\
\end{equation}1 calory is 4.1868 J (1 cal = 4.1868J), so
\begin{equation}
v^{2}=20\times 4.1868\times 10^{3}(J/g)\nonumber \\
\end{equation}\(\ \ \ \ \ \ =20\times 4.1868\times 10^{3}(N\bullet m/g)\)
\begin{equation}
\ \ \ \ \ \ =20\times 4.1868\times 10^{3}(kg\bullet m/s^{2}\bullet m/g)\nonumber \\
\end{equation}\begin{equation}
\ \ \ \ \ \ =20\times 4.1868\times 10^{6}{(m}^{2}/s^{2})\nonumber \\
\end{equation}Hence, velocity of water molecule is
\begin{equation}
v=\sqrt{20\times 4.1868}\times 10^{3}(m/s)\nonumber \\
\end{equation}\begin{equation}
\ \ \ \ =\ \sqrt{83.736}\times 10^{3}(m/s)\nonumber \\
\end{equation}\begin{equation}
\ \ \ \ =9.151\times 10^{3}(m/s)\nonumber \\
\end{equation}This velocity is fast enough to rotate backbone’s residue with
torques(Serway, R. A. and Jewett, Jr. J.W., 2003; F. H. Stillinger,
1975).
Torsional inertia in the swivels are different for each amino acid. In a
backbone, only one residue rotates, the others do not. While the former
has small inertia, the latter has larger one. The difference of
torsional inertia comes from the type and number of atoms and the length
of side chain, and the mass and length of the backbone on both sides of
the residue. The middle residue has a small rotational inertia, because
the rotational inertia is proportional to the square of the distance.
The following(Figure 1) represents only two residues among many in the
backbone. Among these connected residues, only two residues with the
least rotational inertia were shown. Between them, A has smaller
rotational inertia than B. When water molecules collide, A rotates
first. B is a residue that stops while residue A rotates, and then
returns as soon as A stops.
Residue A, has rotational momentum Pa is,
\(P_{A}=\ m_{A\ }{R_{A}}^{2}\omega_{A}\)
\(I_{A}=\ m_{A}{R_{A}}^{2}\)
If B rotates, the rotation momentum Pb of B is,
\begin{equation}
P_{B}=m_{B}{R_{B}}^{2}\omega_{B}-\ P_{A}\nonumber \\
\end{equation}When A does not rotates, the rotation momentum Pb of B
is
\(P_{B}=\ m_{B}{R_{B}}^{2}\omega_{B}\ \)
If A rotates, the rotation momentum Pb of B is
\(P_{B}=\ m_{B}{R_{B}}^{2}\omega_{B}-\ P_{A}\)
If A stops rotation, Pa becomes zero and
Pb is not reduced. So B can rotate more easily. If A
rotates, Pa becomes larger and Pb is
strongly reduced. Therefore, B cannot rotate easily.
In figure 2, the angle between the
axel with A residue and B1 is 104.5°. Assuming that B rotates while A
also rotates, residue A must handle this rotational movement of residue
B. However, if it is 90°, because the radius of rotation(R =
1×R0) is the longest, the rotational inertia of residue
B is affected to residue A. Since, residue A can hardly turn without the
strongest force because rotational inertia(I = R2×m) is
the maximum.
To receive the half of the magnitude of the force mentioned above, it
must be 135 °. The radius of B2 is,
\begin{equation}
R=\ cos45\times R_{0}=\ \frac{1}{\sqrt{2}}R_{0}\nonumber \\
\end{equation}And inertia I is,
\begin{equation}
\ I=mR^{2}=\frac{m{R_{0}}^{2}}{2}=\frac{1}{2}I_{0}\nonumber \\
\end{equation}The bond angle between the two residues A and B is 104.5°. The inertia
of B2 at 104.5° is,
\begin{equation}
I={{(R}_{0}\times\cos(14.5))}^{2}\times m={0.97}^{2}\times I_{0}=0.94\times I_{0}\nonumber \\
\end{equation}This is almost the same as the value of 90° with the most difficulty. In
conclusion, residue B2 cannot rotate with
B1 at the same time.
In short, all of the above are summarized as follows. If there are many
residues of different sizes, they do not rotate at the same time. The
residue with the smallest inertia rotates first. As soon as it stops
rotating, the residue with the next smallest inertia turns. When stops
rotating, the next residue with more inertia than this rotates. In the
case of multiple residues, the rotational inertia is ordered from the
smallest.
The coordinates of short polypeptide with important points above was
set. And it was compared with the structure from solution NMR. The
cotranslational folding and torsional movement of atoms which are
different from others was proved. The most representative model of each
NMR assay was compared with the structure from the new folding
algorithm. Typical structure alignment algorithms including TM-align(Y.
Zhang and J. Skolnick, 2005) were not used in the comparison. These
algorithms moves frame with insertions and deletions even in the case of
the comparison of identical amino acid sequence. The result of the
comparison was represented by logPr and RamRMSD(S. Jung, et al., 2011).
And it was illustrated in the graph of torsion angle along the residue
number(Figure 3). The change of the potential energy during the initial
folding and following optimization is also displayed in the graph(Figure
4).