Question

At 280 nm the chemokine CCL19 has an extinction coefficient of 8730 M-1cm-1. A solution of CCL19 is diluted 50 fold. After dilution (in a cuvette with a 1 cm pathlength) the absorbance at 280 nm is 0.112. What is the concentration of the undiluted CCL19 solution?

978 M

1.28 X 10^-5 M

641 µM

1.00 x 10^6 mM

Answer 1

Absorbance and extinction coefficient is directly proporsional to concentration if the solution is diluted to 50 fold the exctinction coefficient also reducted to 50 times

therefore extinction coefficient for 50 fold diluted solution

8730 M^-1 cm^-1 / 50 = 174.6 M^-1 cm^-1

A = Cl

A = absorbance; = extinction coefficient; C = concentration; l = path length

A = 0.112

= 174.6 M^-1 cm^-1

C = ?

l = 1 cm

C = A/l

C = 0.112 / 174.6 M^-1 cm^-1 x 1 cm = 0.000641 M = 641 M

therefore answer is option C = 641 M