centrifugation. Any fat floating on the surface can be removed by coarse filtration
through glass wool or cheesecloth. However, the solution will still be cloudy with
organelles and membrane fragments that are too small to be conveniently removed by
filtration or low speed centrifugation. These may not be much of a problem as they
will often be lost in the preliminary stages of protein purification, for example during
salt fractionation. However, if necessary they can be removed first by precipitation
using materials such as Celite (a diatomaceous earth that provides a large surface area
to trap the particles), Cell Debris Remover (CDR, a cellulose-based absorber), or any
number of flocculants such as starch, gums, tannins or polyamines, the resultant
precipitate being removed by centrifugation or filtration.
It is tempting to assume that the cell extract contains only protein, but of course a
range of other molecules is present such as DNA, RNA, carbohydrate and lipid as well
as any number of small molecular weight metabolites. Small molecules tend to be
removed later on during dialysis steps or steps that involve fractionation based on size
(e.g. gel filtration) and therefore are of little concern. However, specific attention has to
be paid at this stage to macromolecules such as nucleic acids and polysaccharides. This
is particularly true for bacterial extracts, which are particularly viscous owing to the
presence of chromosomal DNA. Indeed microbial extracts can be extremely difficult to
centrifuge to produce a supernatant extract. Some workers include DNase I in the
extraction buffer to reduce viscosity, the small DNA fragments generated being
removed at later dialysis/gel filtration steps. Likewise RNA can be removed by treat-
ment with RNase. DNA and RNA can also be removed by precipitation with protamineExample 2(cont.)
Therefore in 100 mm^3 of solution A there is 0.8 enzyme unit.
Therefore in 1 cm^3 of solution A there are 8.0 enzyme units.
Since we have 2000 cm^3 of solution A there is a total of 20008.0¼16 000
enzyme units in solution A.
(iii) Using the same approach as in Example 2(i), the protein concentration of
solution B is 3.005/0.0050.89¼ 601 0.89¼535 mg cm^3.
Therefore the total protein present in solution B¼ 120 535 ¼64 200 mg.
(iv) Using the same approach as in Example 2(ii), there are 0.21/0.1¼2.1 units of
enzyme activity in the cuvette. These units came from the 20 mm^3 that was
added to the cell.
Therefore, 20 mm^3 (0.020 cm^3 ) of solution B contains 2.1 enzyme units.
Thus, 1 cm^3 of solution B contains 1.0/0.022.1¼105 units. Therefore,
solution B has 105 units cm^3.
Since there are 120 cm^3 of solution B, total units in solution B¼ 120 105
¼12 600.
(v) For solution A, specific activity¼16 000/208 000¼0.077 units mg^1.
For solution B, specific activity¼12 600/64 200¼0.197 units mg^1.
(vi) Fold purification¼0.197/0.077¼2.6 (approx.).
(vii) Yield¼(12 600/16 000)100%¼79%.319 8.3 Protein purification