The figure of 350 - 600 Units per mg refers to the specific activity of the enzyme.
The Unit is International Unit or IU and is usually defined as that amount of enzyme that will catalyze the transformation of 1 micromole of substrate (or product) per min, under defined assay conditions (such as pH, temperature, substrate concentration, presence of Mg++, etc). It is thus a measure of activity.
When the enzyme is pure (no other extraeneous proteins present), the specific activity provides important information about the catalytic capacity of the enzyme.
It is usually calculated by measuring
- the activity of the enzyme preparation under defined assay conditions
- the protein concentration of the same enzyme preparation (using, say, the Lowry or Biuret method for protein estimation).
Alternatively, if the E(1%, 280) is known (see below) and the enzyme is pure, measurement of the absorbance at 280 nm gives a very good estimate of protein content (and the enzyme may be recovered 'unharmed' at the end of the measurement).
Thus, taking a figure of 450 Units/mg for the specific activity of pyruvate kinase,
25 KU (25 Kilo-Units, I presume) contains 500/9 mg (~55 mg) protein.
I notice that the Sigma product sheet provides a figure for E(0.1%, 280) = 0.54.
This means that a 1 mg/ml solution of the protein will have an absorbance at 280 nm of 0.54
E(0.1%, 280) can be used as a very convenient measure of the protein content provided that the enzyme preparation supplied by Sigma is pure.
A 'rule of thumb', useful when the E(0.1%, 280) is unknown, is that a 1mg/ml protein solution has an A280 of 1.
Thus if, say, the A280 (absorbance at 280 nm) of the resuspended lyophilized powder is 1.08 and you have 5 ml of this, the protein concentration is 2mg/ml and you have 10 mg of protein in total. You may wish to assay the enzyme yourself to determine an accurate specific activity.
The EC (Enzyme Commission) Number may also be of interest. For pyruvate kinase (EC 220.127.116.11) see here.
For a great ref on PK (pdf may be downloaded) see here (Ainsworth et al.)
For your second question, I do not have access to that paper from home.
However, if calmodulin has a specific activity of 40 000 Units/mg,
- 25 000 Units is equivalent to 0.625 mg; this is in a volume of
0.5 ml. Therefore, the calmodulin concentration is 1.25mg/ml.
- Taking the molecular weight of calmodulin to be 16 000,
then 16 000 mg /ml (theoretical) would be a 1 Molar solution.
Thus a 1.25 mg/ml solution is about 78 micromolar.