Biochemistry - Protein Purification - Discussion
Discussion Forum : Protein Purification - Section 1 (Q.No. 6)
6.
A purified protein sample contains 10 μg of protein and has an enzyme activity of 1 m mole of ATP synthesized/sec (1 unit). What is the specific activity of the final purified sample?
Discussion:
4 comments Page 1 of 1.
Ambiha said:
9 years ago
Specific activity = enzyme in sec/ total protein; units micro moles per sec/mg.
Hence conversion of unit = 1000 micro mole/10^(-2).
= 100,000.
Hence conversion of unit = 1000 micro mole/10^(-2).
= 100,000.
Sree said:
7 years ago
Specific activity = 1/10.
= 0.1 to mg.
= 0.1/1000.
= 0.0001 mg.
This should be working right?
= 0.1 to mg.
= 0.1/1000.
= 0.0001 mg.
This should be working right?
R L Deopurkar said:
7 years ago
1 unit has been defined in the problem as 1 millimole synthesised per second.
It clearly says 10 microgram protein has one unit activity. So Specific activity is I unit/ 10 microgram protein and therefore it is 100units/mg.
This alternative is not given. Why should we change the given definition of the unit of enzyme in the problem?
Agreed that international definition of a unit is different.
It clearly says 10 microgram protein has one unit activity. So Specific activity is I unit/ 10 microgram protein and therefore it is 100units/mg.
This alternative is not given. Why should we change the given definition of the unit of enzyme in the problem?
Agreed that international definition of a unit is different.
Bhushan K said:
5 years ago
I agree @Deopurkar According to definition of unit given in the problem, Specific activity should be 100 units/mg.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers