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I was reading a paper in which a recombinant protein (His-6 tagged) is expressed in E Coli (BL21 DE3). The yield of the enzyme isolated from the culture is reported as 10-30 mg per L of bacterial culture. (After purification by Ni-affinity Chromatography)

First, is this a reasonable concentration for an enzyme? I thought it was rather high.

Second, when they measure the concentration of the product of the enzymatic reaction that was found to be merely 1-3 mg/Litre of bacterial culture.

Is this typical or unusual for the product concentration to be an order of magnitude smaller than the enzyme concentration? I assumed that since enzymes were in catalytic amounts they ought to be at lower concentrations?

PS. They use SDS-PAGE followed by the Bradford method for determining the protein titres. How accurate is the Bradford method?

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Remember that the concentrations you report are in mass units rather than mole units. If the protein has a mass of 40,000 Da, reasonable for many proteins, then 20 mg/liter means $0.5$ x $10^{-6}$ mol/liter. A typical small-mass product of an enzyme reaction might have a mass about 200, so its apparently lower mass concentration of 2 mg/liter is $10$ x $10^{-6}$ mol/liter, 20 times higher than the enzyme concentration in terms of numbers of molecules per liter. The mole or molecule ratios are what you should think about in terms of catalytic function.

Furthermore, from what you provide we don't know whether the enzyme was produced under circumstances that allowed it to express its activity; there may have been limitations on substrate availability, presence of inhibitors that would be removed in later purification, and so forth. In this scenario, evidently to obtain enzyme for downstream use, the emphasis is typically on getting large amounts of the specific enzyme regardless of whether some reaction product is formed. I don't find the enzyme concentration particularly high for something that the investigators are trying to over-express as in this case.

The Bradford reaction is a mainstay of standard biochemistry, as it's simple and straightforward. Different proteins can bind the reagent to somewhat different extents, so there is a second-order issue of whether the protein used for the standard curve and that being measured are similar enough in that respect. In practice, what's important is to provide enough information in the methods that are described so that someone could repeat the work to determine the relation between the Bradford measurement and some other more accurate method for this particular protein.

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  • $\begingroup$ Ah! Yes. I blundered. Totally forgot about converting to molar concentrations. My bad! $\endgroup$
    – curious_cat
    Apr 9, 2016 at 15:46
  • $\begingroup$ Also, the enzyme mass is approx. 60,000 Da & the reaction-product is MW=200. So your guesses were bang right on target! :) $\endgroup$
    – curious_cat
    Apr 9, 2016 at 15:52
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    $\begingroup$ @curious_cat, mass versus mole confusions are easy to make. They become less frequent with practice, but can still occur. For the Bradford method, I would generally assume that bovine serum albumin was used as the standard if it wasn't specified. If a particular manufacturer's version of the Bradford test kit was specified, the kit itself might have a standard protein that the manufacturer's information would disclose. $\endgroup$
    – EdM
    Apr 9, 2016 at 17:12
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    $\begingroup$ With respect to issue (1), the gold standard is to purify the protein in large amounts and weigh it, then use that batch of the protein as the standard for other assays whether colorimetric, mass spectrometric, or based on binding to an antibody (e.g.,ELISA). Stable-isotope labels can also be used along with combinations of liquid chromatography and tandem mass spectrometry for absolute quantitation of a protein that has been digested into smaller known peptides. See this Wikipedia page as an introduction. $\endgroup$
    – EdM
    Apr 9, 2016 at 18:19
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    $\begingroup$ For issue (2), it is in principle possible to express an enzyme in a cell-free system that contains no substrates for the enzyme. That's not as efficient as bacterial over-expression, but the principle is the same: you don't necessarily have to produce product from an enzyme just to get the enzyme itself expressed. In fact, a non-hydrolyzable mimic of an enzyme substrate, IPTG, has been used for decades to drive expression from the lac operon that naturally controls $\beta$-galactosidase production. $\endgroup$
    – EdM
    Apr 9, 2016 at 18:27

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