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Sample Problem 1 TABLE 3-5 A Purification Table for a Hypothetical Enzyme Fraction volume (ml) Total protein (mg) Activity (u

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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%

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