Question

A volume of 100uL of 0.15 M NaCl is put into wells A2-A5 of a 24 well microplate. A 200uL sample of BSA (10 mg/ml) is then put into well A1 and a serial dilution is made by taking 100 uL out of A1 into A2 etc.

What is the final concentration of BSA in well A4?

Answer 1

For this question we need to concentrate only on BSA.

For well 1,

weight of BSA put into well 1 = (10 mg/ml) x 200ul = 2mg

when 100ul is taken out it contains 1mg BSA which is put into well 2.

Now in well 2, mmoles of Nacl already present = 0.015mmoles = 0.87mg of Nacl

So now in 200ul 0.87mg of Nacl and 1mg of BSA is present in well 2.

again when transferred in well 3, 0.5mg of BSA will be present in well 3 and 0.25mg of BSA will be present in well 4 in 200ul

so final concentration of BSA in well 4 = 1.25g/l