A 50 mL crude extract of skeletal muscle contains 32 mg of protein per mL. Ten m

Question

A 50 mL crude extract of skeletal muscle contains 32 mg of protein per mL. Ten microliter of the extract catalyzes a reaction at a rate of 0.14 micromol product formed per minute. The extract was frationated by ammonium sulfate precipitation, and the fraction precipitating between 20% and 40% saturation was redissolved in 10 mL. This soluction contains 50 mg/mL protein. Ten microliters of this purified faction catalyzes the reation at a rate of 0.65 micromol/min.
a) what is the degree of purification (fold purification)?
b) what is the percent yield of the enzyme recovered in the purified fraction?

Explanation / Answer

To find fold purification: The first sample will always have a fold purif. of 1, since it is the entire sample before purifying it; everything has been recovered. The remaining sample folds can be found by dividing each specific activity by the specific activity of the crude sample (that's a way to remember the crude sample's fold will be 1). To find percent yield: The first sample will always have a percent yield of 100%, since again, nothing has been lost. The remaining percent yields of the enzyme are found by dividing the total activity of each purification step by the crude extract's total activity and multiplying by 100 (this is how the crude extract will always be 100%).

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