Guest blah Posted May 26, 2005 Posted May 26, 2005 hello I am new to this, and I don't even know if what i m doing now is correct:) erm could I have some help on this exam questionnn plllss thanks alot Protein A was purified using ammonium sulf. precipitation and cation xchange chromatography. The total protein at each step was 500mg, crude extract; 200mg after precipitation; and 10mg after cation xchange. The total activity of Prtotein A was 25 units, crude extract; 15 units, after precipitation; and 7,5 units after cation xchange. How do I determine the %yield, purification factor, and specific activity of protein A at the final step?
donkey Posted May 26, 2005 Posted May 26, 2005 hopefully someone else will reply to confirm/correct me as I'm doing this from memory and I've never actually had any real life experience with this stuff (i'm just an undergraduate) but I think the following is correct. ok well my understanding is that your % yield is 30% (7.5/25 x100) Basically your total activity is directly proportional to the amount of protein A you have. Therefore if you lose some of your target protein (protein A) you lose activity as has happened. Therefore you yield goes down. It should simply be a process of working out how much your final amount of protein A is compared with the initial amount. This can be worked out using the activity, what % of final total activity do you have compared with your initial amount? I've calculated that above to be 30%. --- I think your final purification factor is 15. I assume this basically means how pure is the final sample compared with the original, crude, sample. So again, your total activity is related to the total amount of protein A you have. So in the crude extract you have 500mg of protein containing protein that has 25 units of activity. In your final one you have 10mg containing protein that has 7.5 units of activity. In other words, in the crude extract (500/25 = 20) 20mg would contain enough protein to give a total activity of 1 unit. In the final extract (10/7.5 = 1.333) 1.333mg would contain enough protein to give a total activity of 1 unit. Therefore it's easy to see that 20mg is a lot more than 1.333mg (i.e. it's less pure - it has other material there). In fact it's (20/1.333333333333333 = 15.0) 15 times more mass for the same activity. Therefore it's 15x less pure. So you could say that the final sample has been purified 15x compared to the crude extract. --- As far as I understand it, specific activity is simply units/mg so that's be 7.5/10 = 0.75 for the final sample and it'd be 25/500 = 0.05 for the crude extract. Again it's easy to see that 0.75 is 15x as large as 0.05 confirming the purification factor. hope that makes sense/helps
Guest blah Posted May 26, 2005 Posted May 26, 2005 thanks so much for replying. I calculated the same values for purification factor and for specific activity. But the thing I am not sure about is %yield. In my opinion yield is the amount of protein purified after a step. In my case we have 10mg at the last step while at first step (crude extract) we had 500mg so i think yield= (10/500)*100% =2% what do u think? I don't know .. your answer sounds more reasonable.
Skye Posted May 26, 2005 Posted May 26, 2005 Yeah, I imagine they would want the yields as a percentage of the original crude extract.
donkey Posted May 27, 2005 Posted May 27, 2005 yeah it's not exactly clear what yeild it wants. good luck!
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now