Dear researcher,
I did the molecular dynamic simulation of native protein. I am working on protein folding. I want to add the urea molecules in Gromacs. How to calculate the number of urea molecules for adding 6 M urea concentration having the box size (74.3×74.3×74.3 Å).
Molar concentration is the number of molecules per unit of volume; the identity of the molecule does not matter: 1 liter containing 1 mole of molecule A contains the same number of molecules as 1 liter containing 1 mole of molecule B . . .
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The ratio of 6/55.5 (urea/water) for 6 M urea holds when the water and urea accessible volumes are (about) same. Water molecules are smaller than urea molecules: it may happen that some crevices on the protein surface will be wide enough for a water molecule, but too narrow for a urea molecule. Hence, the water-accessible volume is somewhat larger than the urea-accessible volume. Does this difference (and the fluctuation in this difference due to the conformational changes in the protein) need to be factored in when the total number of water molecules is on the order of 10**4? No, it does not. It is safe to follow the simple 6/55.5 ratio for 6 M urea:
1. Add water molecules into the imaginary box containing the protein to obtain the number of water molecules that fit into the water-accessible space of the box.
2. Calculate the number of urea molecules for the intended concentration.
3. Place the urea molecules randomly around the protein (within the imaginary box).
4. Add water molecules to fill the space that is not occupied by the protein or urea molecules.
5. Simulate . . .
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Assuming the MD simulation program can add the water molecules itself, there is no need for the user to explicitly calculate the water-accessible volume of the box. And as mentioned in my first comment, there is absolutely no need to calculate the total volume of the box; the total volume of the box is totally irrelevant (Archimedes, 3rd century B. C.).
The only math the user must do is to calculate the number of urea molecules (N) as follows:
N = (M / 55.5) * W
where M is the intended molar concentration of the urea; W is the number of water molecules in the box.
This will of course work also for other small molecules and counter-ions.
6M solution would have ~3.6 urea molecules per cubic nanometer. Your specified volume is about 410 cubic nanometers. So, about 1477 ought to be close enough (assuming I did the math right).
Start with the volume of 1 liter = 1 cubic dL. Each time you go down to 0.1 of the dimension, you go .001 in volume. Once you get to a nanometer your volume is 1 yoctoliter or 1e-24. Multiply by 6 and Avogadro's #.
The size of the cube does not matter because a part of it is filled with the protein. What matters is the water-filled/accessible volume, i.e. the space in which the urea molecule can actually be located. Assuming that water molecules have access to the entire water-accessible volume of the box, one can calculate the number of urea molecules as follows:
The molar concentration of water: 55.5 M.
6M urea in water: 1 urea molecule per 55.5/6 water molecules.
Dear sir,
Can you give any reference because both the answers are contradictory to each other
Does mixing water with a protein change the number of water molecules in water? No, it does not. The protein pushes water molecules away (outside the simulation box), creating a "cage" for itself (Archimedes, 3rd century B.C.). Does the presence of the protein affect the water/urea molecular ratio? No it does not. The protein does not cause any water or urea molecule to vanish.
I think the point that Martin is getting at is that you should have a specific ratio of water to urea and that this ratio should be independent of the volume and the specific number of molecules. With this in mind, you have to know the displacement volume of the protein to calculate the remaining volume that will be filled with the molecules of water and urea. This doesn't take into consideration that the displacement volume of the protein can change as it folds or unfolds. The ratio w/w .495/1 urea/water, or w/v 60.06g/mol*6 + balance to 1L with water, which turns out to be roughly v/v of 9/16 (urea/water). This still doesn't give you a number though, you still need to know the density of the solution to calculate the number of molecules. Also, without knowing the protein, there is no way to calculate the number of urea atoms that will fit in the specified volume at the specified molarity, but you can have a ratio of water to urea that will stay the same no matter the size of the protein.
Molar concentration is the number of molecules per unit of volume; the identity of the molecule does not matter: 1 liter containing 1 mole of molecule A contains the same number of molecules as 1 liter containing 1 mole of molecule B . . .
---
The ratio of 6/55.5 (urea/water) for 6 M urea holds when the water and urea accessible volumes are (about) same. Water molecules are smaller than urea molecules: it may happen that some crevices on the protein surface will be wide enough for a water molecule, but too narrow for a urea molecule. Hence, the water-accessible volume is somewhat larger than the urea-accessible volume. Does this difference (and the fluctuation in this difference due to the conformational changes in the protein) need to be factored in when the total number of water molecules is on the order of 10**4? No, it does not. It is safe to follow the simple 6/55.5 ratio for 6 M urea:
1. Add water molecules into the imaginary box containing the protein to obtain the number of water molecules that fit into the water-accessible space of the box.
2. Calculate the number of urea molecules for the intended concentration.
3. Place the urea molecules randomly around the protein (within the imaginary box).
4. Add water molecules to fill the space that is not occupied by the protein or urea molecules.
5. Simulate . . .
---
Assuming the MD simulation program can add the water molecules itself, there is no need for the user to explicitly calculate the water-accessible volume of the box. And as mentioned in my first comment, there is absolutely no need to calculate the total volume of the box; the total volume of the box is totally irrelevant (Archimedes, 3rd century B. C.).
The only math the user must do is to calculate the number of urea molecules (N) as follows:
N = (M / 55.5) * W
where M is the intended molar concentration of the urea; W is the number of water molecules in the box.
This will of course work also for other small molecules and counter-ions.
1000g (1L) water divided by the molecular weight of water (18.01528) equals 55.508435061791989910786843168688, or something like that.
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