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From the definition of crystal and the main differences between crystalline and amorphous material, it is known that crystal formation requires ordered bonding between atoms or molecules.

How can the protein with its irregular structure (when compared to organic and inorganic materials) make ordered bonding, since I often read about the crystal structure of the a protein and its ligand or inhibitor obtained by X-ray?

And is there any difference in crystal structure of an organic and inorganic material?

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    $\begingroup$ Molecular crystal isn't anything unusual... $\endgroup$ – Mithoron Dec 27 '15 at 21:06
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Imagine an infantry unit of soldiers marching in file. Each of them may be quite irregular, but together they form a repeating pattern. And that's exactly what happens with protein molecules in a crystal.

When we say that a protein is irregular, we mean it on a different level. Indeed, one molecule of our protein is irregular if we are talking about the amino acid residues in it, of which there are probably a hundred: here goes Ala, then Gly, then all of a sudden Leu - what will come next? You can never tell. But wait, there is another molecule of the same protein sitting nearby, and it looks exactly the same as this one: Ala, Gly, Leu and so on. And there is yet another next to it, also similar, and many more of them, and together they form an ordered crystal structure which makes it possible for us to diffract X-rays on it and measure the intensities and then do some hardcore math and voila, here is your structure.

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Ivan basically gave a nice and clear example of what is going in. I’m going to offer a more in-depth explanation.

If you were to consider possible crystal structures of just one type of atom, you can boil the possible structures down to a set of similar structures, the Bravais lattices. Only 14 of these exist with different constraints: i.e. the C-centred orthorhombic lattice requires $a \ne b \ne c \ne a$ and $\alpha = \beta = \gamma = 90^\circ$, with atoms placed at the corner of each unit cell and at the centre of the $a,b$ plane. If $a = b$, we could no longer describe this as C-centred orthorhombic; instead it becomes tetragonal ($a = b \ne c$; $\alpha = \beta = \gamma = 90^\circ$) with a smaller unit cell. Because each position is filled by a spherical atom, all symmetry operations allowed by the lattice itself are present in the final crystal.

Now assume, similar to what Ivan mentioned, that you replace the spherical atoms with an organic or inorganic molecule; say ammonia $\ce{NH3}$. Take another unit cell, and replace the spherical atoms with methylamine ($\ce{CH3NH2}$). And take a third and replace the with ethylmethylamine ($\ce{CH3-NH-CH2CH3}$). In the first case we still have a highly symmetric molecule which is nonetheless much less symmetric than a sphere is. Not all of the symmetry elements that are possible in a certain Bravais lattice actually remain when we do the replacement. We will end up with a certain space group — something less symmetric overall. There are 230 different possible space groups.

Note that ammonia has three intrinsic planes of symmetry in its molecular structure while methylamine has one and ethylmethylamine has zero. Therefore, the latter must crystallise in a less symmetric space group than the former two, simply because there is less symmetry present in the molecule.

If we move on to chiral molecules, say butan-2-ol, the symmetry has to be reduced even more. If we have an enantiopure substance, then in no way is it possible that a plane of symmetry transformes one molecule into another; the number of possible space groups reduces itself to the 65 Sohncke space groups — all those that do not contain centres or planes of symmetry.

The transition from a chiral small molecule to a huge chiral protein is then trivial. In the most simple of cases, just put the proteins in a nice orderly fashion next to one another to create a primitive lattice — voilà.

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