Tables for
Volume F
Crystallography of biological macromolecules
Edited by E. Arnold, D. M. Himmel and M. G. Rossmann

International Tables for Crystallography (2012). Vol. F, ch. 5.2, pp. 152-153   | 1 | 2 |

Section 5.2.4. Algebraic concepts

E. M. Westbrooka*

aMolecular Biology Consortium, Argonne, Illinois 60439, USA
Correspondence e-mail:

5.2.4. Algebraic concepts

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Let V be the volume of one unit cell of the crystal. Let [m_{c}] be the total mass within one unit cell, and [m_{m}], [m_{bs}] and [m_{fs}] be the masses, within one unit cell, of the macromolecule, bound solvent and free solvent, respectively. Let [\rho_{c}], [\rho_{m}], [\rho_{bs}] and [\rho_{fs}], respectively, be the densities of a complete macromolecular crystal, its unsolvated macromolecule, its bound-solvent compartment and its free-solvent compartment. Let [\varphi_{m}], [\varphi_{bs}] and [\varphi_{fs}], respectively, be the fractions of the crystal volume occupied by the unsolvated macromolecule, the bound solvent and the free solvent. By conservation of mass, [m_{c} = m_{m} + m_{bs} + m_{fs}. \eqno(]The volume fractions must all add to unity: [\varphi_{m} + \varphi_{bs} + \varphi_{fs} = 1. \eqno(]The density of the crystal is the total mass divided by the unit-cell volume: [\rho_{c} = {{m_{c}} \over V} = {{m_{m}} \over V} + {{m_{bs}} \over V} + {{m_{fs}} \over V}. \eqno(]The mass in each solvent compartment is the product of its density and the volume it occupies: [m_{bs} = \rho_{bs} V\varphi _{bs}, \qquad m_{fs} = \rho _{fs} V\varphi_{fs}. \eqno(]The mass of the macromolecule in the cell can be defined either from its partial specific volume, [\overline{\upsilon}_{m}], the unit-cell volume, V, and the molecules' volume fraction, [\varphi_{m}], or from the molar weight, M, the number of molecular copies in the unit cell, n, and Avogadro's number, [N_{o}]: [m_{m} = V\varphi_{m}/\overline{\upsilon}_{m} = nM/N_{o}. \eqno(]Now ([link]) may be rewritten as [\rho_{c} = \varphi_{m}/\overline{\upsilon}_{m} + \rho_{bs} \varphi _{bs} + \rho_{fs} \varphi_{fs}. \eqno(]Define a mean solvent density, [\rho_{s}]: [\rho_{s} = {\rho_{bs} \varphi_{bs} + \rho_{fs} \varphi_{fs} \over \varphi_{bs} + \varphi_{fs}}. \eqno(]This allows ([link]) to be rewritten as [\rho_{c} = \varphi_{m}/\overline{\upsilon}_{m} + (1 - \varphi_{m})\rho_{s}. \eqno( ]Upon rearrangement, this gives expressions for the volume fraction of a macromolecule and for the molecular-packing number: [\eqalignno{\varphi_{m} &= {\rho_{c} - \rho_{s} \over (\overline{\upsilon}_{m})^{-1} - \rho_{s}} = {n\overline{\upsilon}_{m} M \over VN_{o}}, &\cr n &= {VN_{o} \over \overline{\upsilon}_{m} M} {\rho_{c} - \rho_{s} \over (\overline{\upsilon}_{m})^{-1} - \rho_{s}}. &(\cr}]In ([link]), all terms can be measured directly, except [\rho_{s}]. The treatment of [\rho_{s}] will be discussed in Section 5.2.7[link]. ([link]) defines the total macromolecular mass in the unit cell, [m_{m} = nM/N_{o}], from a measurement of the crystal density [\rho_{c}]. If M were known from the primary sequence of the molecule, this measurement determines the molecular-packing number, n, with considerable certainty. If the molar weight were not accurately known, it could be determined by measuring the crystal density.

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