Question

If a protein can be induced to crystallize, its molecular structure can be determined by X-ray crystallography. Protein crystals, though solid, contain a large amount of water molecules along with the protein. The protein chicken egg-white lysozyme, for instance, crystallizes with a unit cell having angles of \(90^{\circ}\) and with edge lengths of \(7.9 \times 10^{3} \mathrm{pm}, 7.9 \times 10^{3} \mathrm{pm}\), and \(3.8 \times 10^{3} \mathrm{pm}\). There are eight molecules in the unit cell. If the lysozyme molecule has a molecular weight of \(1.44 \times 10^{4}\) and a density of \(1.35 \mathrm{~g} / \mathrm{cm}^{3}\), what percent of the unit cell is occupied by the protein?

Short answer

The protein occupies approximately 5.95% of the unit cell.

Step-by-step solution

Calculate the Volume of the Unit Cell

The unit cell is described by three edge lengths: \( a = 7.9 \times 10^3 \text{ pm} \), \( b = 7.9 \times 10^3 \text{ pm} \), and \( c = 3.8 \times 10^3 \text{ pm} \). Convert these units into centimeters (since 1 pm = \( 10^{-10} \) cm): \( a = 7.9 \times 10^{-7} \text{ cm} \), \( b = 7.9 \times 10^{-7} \text{ cm} \), \( c = 3.8 \times 10^{-7} \text{ cm} \). The volume \( V \) of the unit cell is given by \( V = a \times b \times c \). Substitute the converted values and calculate:\[ V = (7.9 \times 10^{-7} \text{ cm}) \times (7.9 \times 10^{-7} \text{ cm}) \times (3.8 \times 10^{-7} \text{ cm}) \approx 2.37 \times 10^{-18} \text{ cm}^3 \]

Calculate Mass of Protein in the Unit Cell

The molecular weight (molar mass) of the protein is given as \( 1.44 \times 10^4 \text{ g/mol} \). We know there are 8 molecules per unit cell. The molar mass indicates the mass per mole, and using Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ molecules/mol} \), the mass of one molecule is:\[ \text{Mass of one lysozyme molecule} = \frac{1.44 \times 10^4 \text{ g/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \approx 2.39 \times 10^{-20} \text{ g/molecule} \]Thus, the mass of 8 molecules in the unit cell is:\[ \text{Mass of protein in unit cell} = 8 \times 2.39 \times 10^{-20} \text{ g} \approx 1.91 \times 10^{-19} \text{ g} \]

Calculate Volume of Protein in the Unit Cell

Using the calculated mass of protein in the unit cell (\(1.91 \times 10^{-19} \text{ g}\)) and the given density of the protein (\(1.35 \text{ g/cm}^3\)), we can find its volume, since density = mass/volume. Therefore, the volume \( V_p \) of the protein in the unit cell is:\[ V_p = \frac{1.91 \times 10^{-19} \text{ g}}{1.35 \text{ g/cm}^3} \approx 1.41 \times 10^{-19} \text{ cm}^3 \]

Calculate Percent Occupied by Protein

The percent of the unit cell occupied by the protein is the volume occupied by the protein divided by the total volume of the unit cell, multiplied by 100%:\[ \text{Percent Occupied} = \left( \frac{1.41 \times 10^{-19} \text{ cm}^3}{2.37 \times 10^{-18} \text{ cm}^3} \right) \times 100\% \approx 5.95\% \]

Key concepts

Protein Crystallization

Protein crystallization is a vital step in the structural analysis of proteins using x-ray crystallography. This method involves the formation of solid crystals of proteins, in which the molecules are precisely arranged in a repeating pattern. Achieving successful protein crystallization requires ideal conditions, such as optimal pH, temperature, and the correct concentration of protein and precipitants.
This process allows scientists to lock the protein in a solid structure, making it possible to study its 3D configuration in detail.

• Crystallization captures the protein in its active conformation.
• Solvent water molecules are often included in crystals, maintaining a natural-like state.

Understanding the precise molecular arrangement aids in determining how proteins perform biological functions. With this knowledge, researchers can glean insights into the workings of enzymes, receptors, and complex multi-protein structures.
While crucial, not all proteins easily crystallize, and some may require substantial trial-and-error to achieve the correct conditions. Nonetheless, the ability to crystallize a protein is a key step toward determining its structural and functional properties.

Unit Cell Volume Calculation

The unit cell is the smallest repeating unit in a crystal lattice, and its volume is crucial for understanding how much space various molecules occupy within the crystal. To calculate the volume of a unit cell, one must first obtain the dimensions of its edges.
In our problem, these edges are given as:

• 7.9 × 10⁻⁷ cm
• 7.9 × 10⁻⁷ cm
• 3.8 × 10⁻⁷ cm

Each measurement corresponds to a different dimension of the cell.
The volume (V) of this cell is calculated by multiplying these three lengths together: \[V = 7.9 \times 10^{-7} \text{ cm} \times 7.9 \times 10^{-7} \text{ cm} \times 3.8 \times 10^{-7} \text{ cm} \approx 2.37 \times 10^{-18} \text{ cm}^3\].

By converting the initial measurements (in picometers) to centimeters, we maintain consistent units for calculation. Understanding the unit cell volume helps in determining what percentage of the cell is occupied by the crystallized protein, and how much is filled by the solvent or empty spaces.

Molecular Structure Determination

Molecular structure determination through x-ray crystallography enables scientists to visualize the 3D arrangement of atoms within a protein. Once a protein crystal is acquired, it is exposed to x-rays, producing diffraction patterns based on the scattering of x-rays by the electrons in the atoms.
These diffraction patterns can be interpreted mathematically to unveil the atomic positions, providing a detailed view of the protein's architecture. Such data is invaluable for identifying the functional regions of a protein, including:

• Active sites, where biochemical reactions occur.
• Binding sites for other molecules, such as substrates or inhibitors.

The precise structural information obtained not only supports biological research but also aids in drug design, as researchers can simulate how a drug might interact with a target protein.
Being able to determine the molecular structure is crucial for understanding the protein's role within the cell and how it can be manipulated for therapeutic purposes. Despite the complexity of x-ray crystallography, the rewards in advancing scientific knowledge and medical applications make it an indispensable tool.