Question

Sample Problem 1 TABLE 3-5 A Purification Table for a Hypothetical Enzyme Fraction volume (ml) Total protein (mg) Activity (units) Procedure or step Specific activity (units/mg) 1,400 1. Crude cellular extract 2. Precipitation with ammonium sulfate 3. Ion-exchange chromatography 4. Size-exclusion chromatography 5. Affinity chromatography 10,000 3,000 400 10 32 100,000 96,000 80,000 60,000 45,000 200 600 15,000 • If affinity chromatography is the final step of purification, what is the yield and fold purification. Protein X has an absorptivity of 0.4 ml/mg/cm at 280 nm. What is the absorbance at 280 nm of a 2.0 mg/mL solution of protein X in a 1 cm cuvette? (Arecl, Beer lambert's law)

Answer 1

1)

Yield is given by:

%Yield = Enzyme Activity after purification Enzyme Activity before purification rification X 100

In this example,

%Yield = 45,000 units 45 100,000 units - x 100 = 1 x 100 = 45%

The Yield is 45%.

Fold purification is given by:

Fold Purification = Specific Activity After Purification Specific Activity After Purification 15,000 10

= 1500.

Thus, the fold enzyme was purified 1500 fold.

2)

Absorbance = $\epsilon$ cl = 0.4 mL/mg/cm * 2 mg/mL * 1 cm = 0.8

So, the absorbance of the protein is 0.8 or 80%