4
$\begingroup$

I am looking to purchase Pyruvate Kinase from the Sigma Website, they state the volume in Kilo Units (KU) i.e. 1, 5 or 25 KU. It also states there are 350-600 units / mg protein.

Does this mean one unit is one protein? so 1KU is 1000 units of protein? What is the relationship between the KU and mg? and how can I use this relationship to calculate my concentration in um or mg.

EDIT

In this paper (look at the methods section) they calculated the concentration of 25,000 units of calmodulin in 0.5 ml to be approximately 75um which is 40,000 units/mg. How did they calculate this?

Improve this question
$\endgroup$
1
  • $\begingroup$ if there are 600 units in 1 mg surely there must be more than 1 mg for 1000 units 1000/600 = 1.67 mg? $\endgroup$
    – user5433
    Jan 15, 2014 at 22:19

2 Answers 2

Reset to default
5
$\begingroup$

From the page linked from your link:

Unit Definition

One unit will convert 1.0 μmole of phospho(enol)pyruvate to pyruvate per min at pH 7.6 at 37 °C.

So, the unit is defined by activity, and there is no way to know how many molecules or milligrams of protein are included

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ What about in the paper I referenced? They calculated the concentration of calmodulin to be approximately 75um. $\endgroup$
    – harpalss
    Jul 11, 2012 at 13:50
  • 2
    $\begingroup$ After some serious thinking it is possible to calculate the mg. On the website it states there is 350-600 units/mg protein. So if I bought the 1KU to work out the mg and assuming I receive 600 units I would divide 600/1000 = 0.6 mg. I completely missed the obvious here. $\endgroup$
    – harpalss
    Jul 11, 2012 at 17:01
  • $\begingroup$ You can obviously calculate the moles or weight by purifying the enzyme and weighing it... $\endgroup$
    – nico
    Jul 11, 2012 at 17:03
  • $\begingroup$ @TomD: sure, but that's the way one would do it! $\endgroup$
    – nico
    Jul 16, 2012 at 5:25
  • $\begingroup$ @TomD: I am not saying HE should do it, I am saying that is one method to do it. The Biuret method would also work, although it only gives you an approximate concentration. If weighting is not an option MS or similar should do. $\endgroup$
    – nico
    Jul 16, 2012 at 11:27
5
$\begingroup$

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 2.7.1.40) 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.
Improve this answer
$\endgroup$
1
  • $\begingroup$ As far as I can see, "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." is correct? $\endgroup$
    – user338907
    Feb 17, 2022 at 21:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .