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The purification of a recombinant protein is carried out starting with 100 liter

ID: 177336 • Letter: T

Question

The purification of a recombinant protein is carried out starting with 100 liters of a clarified cell lysate, which has a total protein concentration of 0.36 mg/ml and a recombinant prootein concentration of 2.2 U/ml, where U denotes units of biological activity of the recombinant protein. It is known that the completely pure recombinant protein has a specific activity of 40.0 U/mg. Purification is continued until a chromatography step that yields 2.0 liters of a fraction containing the protein, with a total protein concention of 1.11 mg/ml and a recombinant protein concentation of 43.2 U/ml.

For the recombinant protein, calculate the starting and ending specific activity, and the percentage yield and fold purification through the chromatography step.

Explanation / Answer

Total volume of a clarified cell lysate = 100 L or 100000 ml

Total protein concentration ( mg/ml) = 0.36 mg/ml

Total protein (mg) = Protein concentration * total volume

= 0.36 * 100000

= 36000 mg

Total recombinant protein concentration (Activity, units/ml) = 2.2 U/ml

Starting total activity (units) = activity * total volume

= 2.2 * 100000

= 220000 units

Starting specific activity (units/mg) = Total activity / total protein

= 220000 / 36000

= 6.1 units/mg

After purification

Total volume = 2 L or 2000 ml

Total protein concentration ( mg/ml) = 1.11 mg/ml

Total protein (mg) = Protein concentration * total volume

= 1.11 * 2000

= 2220 mg

Total recombinant protein concentration (Activity, units/ml) = 43.2 U/ml

Total activity (units) = activity * total volume

= 43.2 * 2000

= 86400 units

Ending specific activity (units/mg) = Total activity / total protein

= 86400/ 2220

= 39 units/mg

Percentage yield = (Total activity / Starting total activity) * 100

= (86400 /220000 ) * 100

= 40 %

Fold purification = Ending specific activity / starting specific activity

= 39 / 6.1

= 6.4