Calculations for the Purification of a Recombi-nant Protein The purification of a recombinantprotein is carried out starting with 100 liters of aclarified cell lysate (i.e., the cells have been lysed,and the cell debris has been removed to give a clar-ified solution), which has a total protein concentra-tion of 0.36 mg/ml and a recombinant protein con-centration of 2.2 U/ml, where U denotes units ofbiological activity of the recombinant protein. It isknown that the completely pure recombinant pro-tein has a specific activity of 40.0 U/mg. Purifica-tion is continued until a chromatography step thatyields 2.0 liters of a fraction containing the protein,with a total protein concentration of 1.11 mg/ml anda recombinant protein concentration of 43.2 U/ml..For the recombinant protein, calculate the.starting and ending purity, the starting and endingspecific activity, and the percentage yield and foldpurification through the chromatography step.:

In starting volumestarting volume is 100 L i.e. 105 mL.In ending volumeEnding volume is 2 L i.e.…

Answered: Calculations for the Purification of a…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Question

Calculations for the Purification of a Recombi-
nant Protein The purification of a recombinant
protein is carried out starting with 100 liters of a
clarified cell lysate (i.e., the cells have been lysed,
and the cell debris has been removed to give a clar-
ified solution), which has a total protein concentra-
tion of 0.36 mg/ml and a recombinant protein con-
centration of 2.2 U/ml, where U denotes units of
biological activity of the recombinant protein. It is
known that the completely pure recombinant pro-
tein has a specific activity of 40.0 U/mg. Purifica-
tion is continued until a chromatography step that
yields 2.0 liters of a fraction containing the protein,
with a total protein concentration of 1.11 mg/ml and
a recombinant protein concentration of 43.2 U/ml..
For the recombinant protein, calculate the.
starting and ending purity, the starting and ending
specific activity, and the percentage yield and fold
purification through the chromatography step.
:

Expert Solution

Step 1: Calculating the amount and activity of protein in the starting and final volume

In starting volume

starting volume is 100 L i.e. 10⁵ mL.

$A m o u n t space o f space t o t a l space p r o t e i n space equals space 0.36 m g space m L to the power of negative 1 end exponent space cross times space 10 to the power of 5 space m L space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.36 cross times 10 to the power of 5 space m g$

$A c t i v i t y space o f space r e c o m b i n a n t space p r o t e i n equals 2.2 space U space m L to the power of negative 1 end exponent space cross times space 10 to the power of 5 space m L space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2.2 cross times 10 to the power of 5 space U$

In ending volume

Ending volume is 2 L i.e. 2000 mL.

$A m o u n t space o f space t o t a l space p r o t e i n space equals space 1.11 m g space m L to the power of negative 1 end exponent space cross times space 2000 space m L space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2220 space m g$

$A c t i v i t y space o f space r e c o m b i n a n t space p r o t e i n equals 43.2 space U space m L to the power of negative 1 end exponent space cross times space 2000 space m L space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 86400 space U$