MD Mutation485
Contents
check the trajectory
We checked the trajectory with following command:
gmxcheck -f mut436_md.xtc
With the command we got following results:
Reading frame 0 time 0.000 # Atoms 96545 Precision 0.001 (nm) Last frame 2000 time 10000.000
Furthermore, we got some detailed results about the different items during the simulation.
Item | #frames | Timestep (ps) |
Step | 2001 | 5 |
Time | 2001 | 5 |
Lambda | 0 | - |
Coords | 2001 | 5 |
Velocities | 0 | - |
Forces | 0 | - |
Box | 2001 | 5 |
The simulation finished on node 0 Thu Sep 15 19:12:47 2011.
Time | ||
Node (s) | Real (s) | % |
22336.000 | 22336.000 | 100% |
6h12:00 |
The complete simulation needs 6 hours and 12 minutes to finishing.
Performance | |||
Mnbf/s | GFlops | ns/day | hour/ns |
1277.617 | 93.808 | 38.682 | 0.620 |
As you can see in the table above, it takes about 2/3 hour to simulate 1 ns of the system. So therefore, it would be possible to simulate about __ns in one complete day calculation time.
Visualize in pymol
First of all, we visualized the simulation with with ngmx, because it draws bonds based on the topology file. ngmx gave the user the possibility to choose different parameters. Therefore, we decided to visualize the system with following parameters:
Group 1 | Group 2 |
System | Water |
Protein | Ion |
Backbone | NA |
MainChain+H | CL |
SideChain |
igure 1 shows the visualization with ngmx:
create a movie
Next, we want to visualize the protein with pymol. Therefore, we extracted 1000 frames from the trajectory, leaving out the water and jump over the boundaries to make continuse trajectories. Therefore, we used following command:
trjconv -s fole.tpr -f file.xtc -o output_file.pdb -pbc nojump -dt 10
The program asks for the a group as output. We only want to see the protein, therefore we decided to use group 1.
Todo: film und filtered
energy calculations for pressure, temperature, potential and total energy
Temperature
Average (in K) | 297.936 |
Error Estimation | 0.0045 |
RMSD | 0.940566 |
Tot-Drift | 0.00654126 |
The plot with the temperature distribution of the system can be seen here:
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As you can see on Figure 2, most of the time the system has a temperature about 298K. The maximal difference between this average temperature and the minimum/maxmimum temperature is only about 4 K, which is not that high. But we have to keep in mind, that only some degree difference can destroy the function of a protein. 298 K is about 25°C, which is relativly cold for a protein to work, because the temperature in our bodies is about 36°C.
Potential
Average (in kJ/mol) | -1.28176e+06 |
Error Estimation | 85 |
RMSD | 1068.67 |
Tot-Drift | -536.314 |
The plot with the potential energy distribution of the system can be seen here:
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As can be seen on Figure 3, the potential engery of the system is between -1.282e+06 and -1.281e+06, which is a relativly low energy. Therefore this means that the protein is stable. So we can suggest, that the protein with such a low energy is able to function and is stable and therefore, our simulation could be true. Otherwise, if the energy of the simulated system is too high, we can not trust the results, because the protein is too instable to work.
Total energy
Average (in kJ/mol) | -1.05203e+06 |
Error Estimation | 83 |
RMSD | 1308.04 |
Tot-Drift | -531.275 |
The plot with the total energy distribution of the system can be seen here:
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As we can see on Figure 4 above, the total energy of the protein is a little bit higher than the potential energy of the protein. In this case, the energy is between -1.05e+06 and -1.051e+06. But these values are already in a range, where we can suggest that the energy of the protein is low enough so that this one can work.
Pressure
Average (in bar) | 0.998385 |
Error Estimation | 0.0058 |
RMSD | 71.0317 |
Tot-Drift | -0.0436306 |
The plot with the pressure distribution of the system can be seen here:
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As you can see on Figure 5, the pressure in the system is most of the time about 1, but there a big outlier with 250 and -250 bar. So therefore we are not sure, if a protein can work with such a pressure.
minimum distance between periodic boundary cells
Next we try to calculate the minimum distance between periodic boundary cells. As before, the program asks for one group to use for the calculation and we decided to use only the protein, because the calculation needs a lot of time and the whole system is significant bigger than only the protein. So therefore, we used group 1.
Here you can see the result of this analysis:
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As you can see on Figure 6, there is a huge difference between the different time steps and distances. The highest distance is up to 4 nm, whereas the smallest distance is only about 1nm. Therefore, we can see that the protein is very flexible over the time.