Question

A commercial preparation of an enzyme is marketed as containing 25.5 U (units of activity) in a 100.0 uL solution. For this enzyme 1 U is defined as the amount that will convert 1 mol of substrate to product in 1 min at 25 C. If you conduct an experiment using 15.0 uL of enzyme at 25 C for 4 min, what quantity of product do you expect to make?

If the commercial enzyme sample in the previous question had a concentration of 2.7 mg/ml, what was the specific activity of the enzyme?

Answer 1

You should aim to do these calculations in your head or on a scrap of paper (i.e. without a calculator). The following logic may be helpful:

• Enzyme units are expressed as µmol substrate converted per min.
• If the question gives enzyme activity in nmol per min, divide by 1000 to convert to µmol.
• Then multiply by the volume to get the total number of units.
• To summarize, the total number of enzyme units is: volume x substrate conversion rate/(1000 if nmol, or 1 if µmol

From the question you can observe for that the

Enzyme activity could be calculated using the following equation:

activity (U/L) = ((absorbance variation/time)/molar extintion coefficent * path length) * 1 micromol * (total reaction volume/total enzyme volume)

= (4 min/1*1) * 1*0.015

= 4*0.015

= 0.06/L

2. Dilution equation = C1V1= C2V2

2.7*1000= x * 0.015

X = 1.8 mg /L