Determine the volume of Protein A resin needed to process a

Determine the volume of Protein A resin needed to process a batch of 1000 L of clarified feed. The feed titer is 2.5 mg/mL and you will load your Protein A resin to a dynamic binding capacity of 35 mg/mL. How much resin would be required if you would like to purify the entire batch in one Protein A experiment (one cycle). How much resin would be needed if you could cycle the resin 20 times?

Solution

a) since protein concentration is 2.5 mg/ml.

means 2.5 gm/ 1000 ml or 1L.

means 2.5 kg/ 1000 L. or 2500000 mg/ 1000 L.

now binding capacity of resin is 35 mg/ml.

means 35 gm/ L. or 350 gm/ 10 L or 3500 g/100 L.

now- for 3.5 kg of protein we need 100 L resin so

for 2.5 kg of protein we need- 100*2.5/3.5 L of resin which is approximate 71.43 L of resin.

b) now if we recycle the resin 20 times, means we need 20 times less resin i.e. 71.43 L / 20= 3.57 L of resin.

in other way (from answer to question) 3.57 L resin will bind to 125 gm of protein (3.57* 35 g).

if we use this resin 20 times then protein loaded to 3.57 L resin will be 125 gm* 20 = 2500 gm or 2.5 kg of protein which is total amount of protein.