1. Use the (modified) NH-diffusion pulse sequence from biopack to measure a series of 1D spectra with different gradient values for the N15, N15/C13, or N15/C13/D labeled protein in 90%/10% H2O/D2O buffer. (For nsp10~12kDa and ubiquitin~8.5kDa, array “gzlvl2 from 2000 to 20000 with interval =3000, and total 7 values are enough here. Note: nt = N*16, N is 1, 2, 3…)
2. Use vnmrj to process the 1D spectra, find a strongest peak and use “peak” command to get the intensity values under each gradient value.
3. Calculate D based on the next equation:
A(g2) = A(0)exp[-γ2δ2g2D(Δ-δ/3)],
Where γ is the gyromagnetic ratio for protons, Δ is the diffusion period time, and δ is the duration time of the gzlvl2 gradient pulse. (Δ and δ values can be gotten from the pulse sequence.)
After the equation transforming:
ln(A(g2)) = –[γ2δ2(Δ – δ/3)*D]*(g2) + ln(A(0)) → y = -a*x + b
Use excel to do the linear fitting, and get the slope “a”, so D = a / [γ2δ2(Δ-δ/3)].
4. Based on Stokes-Einstein equation, D = kT/6πηRs,
Where k is the Boltzman constant, T is the absolute temperature (e.g. 278K here), η is the viscosity of the solvent (for water, η = 0.01gcm-1s-1), Rs is the Stokes hydrodynamic radius of the spherical particles, so
Rs = (2.2/D) x 10-6 nm.
5. Since the proteins have the same approximate specific weight (density), 1.37g/cm3, the simple estimation for the relationship between the molecular weight M(Da)and its spherical radius R(nm) (assume the molecule is spherical shape) is as the following:
M (g)/6.023*1023 = 1.37(g/cm3) *V(cm3)
V = 1/(1.37*6.023)*10-23 (cm3) = 1.212*10-3 * M (nm3)
On the other hand, assuming the protein molecule is spherical shape,
V = (4π/3)*R3,
So, R = [3V/(4π)]1/3 = 0.066*M1/3 (nm)
For a protein (assuming it’s spherical) with its known MW, its radius should be at least has the R value, its Stokes hydrodynamic radius Rs should be more accurate, and bigger than its R.
For example here, for nsp10 (MW~12.5kDa), after measurement, D = 1.30*10-6 cm2/s; so its Rs = 1.69 nm and R = 1.53 nm.
For ubiquitin (MW~8.5kDa), D = 1.57 *10-6 cm2/s; so its Rs = 1.40 nm and R =1.35 nm.
Based on the known data and the measured D, we can estimate if a new protein exists as a monomer or oligomer…
1. JMR 2004, 166, 129-133.
2. Biological Procedures Online, 2009, 11(1), 32-51.
Edited on 12/22/2014