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SDF for martini beads
- mferraro
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9 years 6 months ago #4106
by mferraro
SDF for martini beads was created by mferraro
Dear Martini Users,
I'm interested in calculate spatial distribution function for a molecule around a protein. I'm going to consider the SDF for a single representative bead and I have to accurately choose the bin width for generating an appropriate grid and avoiding large errors. Is it correct to set it between 0.47 and 0.52 nm, in order to consider the average Martini beads size? Doing so, every cell should be large enough to correctly take into account the coordinates of the bead (and its size) in that point, right?
Can such a big value influence the outcome of my analysis?
Thank you very much for your precious suggestions!
Maria.
I'm interested in calculate spatial distribution function for a molecule around a protein. I'm going to consider the SDF for a single representative bead and I have to accurately choose the bin width for generating an appropriate grid and avoiding large errors. Is it correct to set it between 0.47 and 0.52 nm, in order to consider the average Martini beads size? Doing so, every cell should be large enough to correctly take into account the coordinates of the bead (and its size) in that point, right?
Can such a big value influence the outcome of my analysis?
Thank you very much for your precious suggestions!
Maria.
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9 years 6 months ago #4107
by mnmelo
Replied by mnmelo on topic SDF for martini beads
Hi Maria,
I would take a finer binning. A lot of the peaks/features in RDFs/SDFs have, understandably, widths roughly equal to the bead size. In order to be able to properly resolve those features you should have a binning that's smaller than their size. I typically go for 0.1 nm.
Now, a (rather long) note on precision-related noise:
If you do a naïve binning and counting, your distributions will be affected by the underlying granularity of the coordinates. And this depends on how you stored your trajectory.
Here's a simple example of the problem:
Let's suppose you stored your Martini trajectory in .xtc format, with a 0.01 nm precision (a typical setup for Martini). You now try using g_density to calculate densities of given particles along a given axis; and you choose a binwidth of 0.1 nm.
You'll have 10 precision points per bin. However, when g_density compares the coordinate values to the bin edges some will overlap. Due to machine precision differences, some bins may end up with 11 precision points (both edges become included), or with 9 (both are excluded). This will show up as +10% or -10% jitter in your distributions, that you can't get rid of by sampling more.
The solution, in this simple 1D case, is to still take bins of 0.1 nm, but make the edges start somewhere between precision points (say, if your box has a 10 nm side, the bins would run from -0.005 to 10.005). This ensures all bins always have 10 precision points within. The point is to have bin edges always fall between precision points. You can also see that having a finer .xtc precision mitigates the issue, but won't solve it entirely -- it's best, and in my opinion more elegant, to adjust the bin edges.
For a multidimensional case things could get more complicated if you're doing an RDF:
The binning is then performed on the distance coordinate r, which has unequally spaced precision points. This means some bin edges could fall on top of precision points. I did some calculations and, for a 5 nm radius, 0.01 precision, and binwidth 0.1 starting from 0.005, you never get a bin edge overlapping a precision point. This is good.
RDF normalization will be another source of precision-related noise, however. Let me know if you are worried about this and I will post another note on how to possibly circumvent it.
Good luck, and let me know if you need more clarification.
Manel
I would take a finer binning. A lot of the peaks/features in RDFs/SDFs have, understandably, widths roughly equal to the bead size. In order to be able to properly resolve those features you should have a binning that's smaller than their size. I typically go for 0.1 nm.
Now, a (rather long) note on precision-related noise:
If you do a naïve binning and counting, your distributions will be affected by the underlying granularity of the coordinates. And this depends on how you stored your trajectory.
Here's a simple example of the problem:
Let's suppose you stored your Martini trajectory in .xtc format, with a 0.01 nm precision (a typical setup for Martini). You now try using g_density to calculate densities of given particles along a given axis; and you choose a binwidth of 0.1 nm.
You'll have 10 precision points per bin. However, when g_density compares the coordinate values to the bin edges some will overlap. Due to machine precision differences, some bins may end up with 11 precision points (both edges become included), or with 9 (both are excluded). This will show up as +10% or -10% jitter in your distributions, that you can't get rid of by sampling more.
The solution, in this simple 1D case, is to still take bins of 0.1 nm, but make the edges start somewhere between precision points (say, if your box has a 10 nm side, the bins would run from -0.005 to 10.005). This ensures all bins always have 10 precision points within. The point is to have bin edges always fall between precision points. You can also see that having a finer .xtc precision mitigates the issue, but won't solve it entirely -- it's best, and in my opinion more elegant, to adjust the bin edges.
For a multidimensional case things could get more complicated if you're doing an RDF:
The binning is then performed on the distance coordinate r, which has unequally spaced precision points. This means some bin edges could fall on top of precision points. I did some calculations and, for a 5 nm radius, 0.01 precision, and binwidth 0.1 starting from 0.005, you never get a bin edge overlapping a precision point. This is good.
RDF normalization will be another source of precision-related noise, however. Let me know if you are worried about this and I will post another note on how to possibly circumvent it.
Good luck, and let me know if you need more clarification.
Manel
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9 years 6 months ago #4108
by mferraro
Replied by mferraro on topic SDF for martini beads
Thanks so much for such precious elucidations!
Maria.
Maria.
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9 years 5 months ago #4170
by mferraro
Replied by mferraro on topic SDF for martini beads
Hello Manel,
I'm back again, but this time I'd like to ask something about the correlation between the reverse Boltzmann relationship and grid counts obtained from SDF.
I mean, even if largely debated, the Free Energy could be directly related to the ratio between atom densities (or counts) in two different region of the sampled space. Some papers report this method with atomistic force fileds.
Anyway, when we use MARTINI beads, we are considering density spots for more than an atom at a time, so if we apply that formula and obtain a Free Energy, how many atomic contributes should I consider as included in that value? If I choose a representative atom, could I directly relate that value to a simple estimate of the binding free energy for the whole molecule?
And finally, can the ligand efficiency (LE) rule be applied also to MARTINI beads?
I'm afraid of making a conceptual mistake using the reverse relationship with MARTINI force field.
Thank you in advance for your willingness.
Maria
I'm back again, but this time I'd like to ask something about the correlation between the reverse Boltzmann relationship and grid counts obtained from SDF.
I mean, even if largely debated, the Free Energy could be directly related to the ratio between atom densities (or counts) in two different region of the sampled space. Some papers report this method with atomistic force fileds.
Anyway, when we use MARTINI beads, we are considering density spots for more than an atom at a time, so if we apply that formula and obtain a Free Energy, how many atomic contributes should I consider as included in that value? If I choose a representative atom, could I directly relate that value to a simple estimate of the binding free energy for the whole molecule?
And finally, can the ligand efficiency (LE) rule be applied also to MARTINI beads?
I'm afraid of making a conceptual mistake using the reverse relationship with MARTINI force field.
Thank you in advance for your willingness.
Maria
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9 years 5 months ago #4195
by mnmelo
Replied by mnmelo on topic SDF for martini beads
Hi Maria,
Sorry it took me so long to get back to you.
For your first question I'd say the point is moot, since you're talking about a ratio and it won't matter if you count 1 or 4 atoms per bead. Densities will be scaled by whichever mapping factor you choose, but that will be cancelled out in the ratio.
For the ligand efficiency you should indeed take into account the number of atomistic heavy atoms, not CG beads. This because MARTINI is geared to reproduce experimental deltaG's, without any sort of scaling. The normalization by the number of atoms should be then the same as for LE values computed with explicit models, otherwise the values won't be comparable.
Cheers,
Manel
Sorry it took me so long to get back to you.
For your first question I'd say the point is moot, since you're talking about a ratio and it won't matter if you count 1 or 4 atoms per bead. Densities will be scaled by whichever mapping factor you choose, but that will be cancelled out in the ratio.
For the ligand efficiency you should indeed take into account the number of atomistic heavy atoms, not CG beads. This because MARTINI is geared to reproduce experimental deltaG's, without any sort of scaling. The normalization by the number of atoms should be then the same as for LE values computed with explicit models, otherwise the values won't be comparable.
Cheers,
Manel
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