You are purifying a new Streptococcal protease. In your first purification, you

Question

You are purifying a new Streptococcal protease. In your first purification, you have a pure solution at 1.2 mg protein/ml.

You obtain an Abs(280 nm) of 1.63. What is the absorptivity (“extinction coefficient”) on a WEIGHT basis? This means use mg/ml and don’t worry about molarity. It is common when working with proteins.

In your second purification, you use the extinction coefficient you obtained earlier to keep track of your concentration. At one point after purification you obtain an Abs > 3.0, which is too high to be measured accurately. So, you dilute the protein 1:10, and get an absorbance of 0.327.

What is the protein concentration in mg/ml after the dilution?

What is the protein concentration in mg/ml before the dilution?

Explanation / Answer

A = ecl

e = A/cl

= 1.63/1.2= 1.35 ml/ mg* cm provided that pathlength is 1 cm.

2)A = 0.327

e = 1.35 ,

c= A/e

= 0.327/1.35 = 0.242 mg/ ml after dilution

Protein was diluted in 1:10 ratio which means 1 ml of intial concentration was diluted to 10 ml. Therefore concentration of in 10 ml = intial concentration of protein /ml= 0.242*10 = 2.42mg/ml

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