# Protein labeling calculation - (Aug/04/2009 )

How to calculate this?

I need to label my protein. The dye should be 10 molar excess in reaction.

The labelled molecule need to be in 100 uM concentration.

Labelled molecule:

MW= 2000 (1mg/ml)

need to be diluted to 100 uM solution

M= 1 mg/ 2000 mg/mmol =5 x 10E-4 mM

V1=500ul = 500x 10E-6l

C1= 100 x 10E-6l

V2=c1 x V1/c2

V2= 100 ul

dilution: 100 ul molecule + 400 ul buffer

Dye:

MW= 720.66

C= 10mg (1mg/100ul DMSO)

M= 10mg/720.66 mg/mmol

0.013876 mM

10 molarity excess of dye to labelled molecule

M (of molecule)= 100 uM = 100 x 10E6 l

total volume of reaction= 20 ul = 20 x 10E6 l

Dye amount in reaction?

V2(dye)=c1 x V1/c2

V2(dye x 10 excess) =1.44 ul

is this right at all??

help highly appreciated!

I don’t think so.

According to my calculations you have 13,876mM (not 0.013876 mM) dye and for 500ul reaction you should put 100ul molecule ( to have 100uM solution) and 36ul dye ( to have 1mM solution) and 346ul buffer.

Regards

The total volume what I wanted for my labeling was 20 ul; I just wanted do a stock of 100 uM molecule first (ok unnnecessary I know .

So since you calculated that the tot. vol would be 500 (25 times more), so then my answer 1.4 is correct (for 20ul tot. volume). Right?

gotmog on Aug 4 2009, 09:21 AM said:

Regards

And yes, it is 0.013876 M not mM (just wrote it wrong but calculated as M)!

Joana on Aug 4 2009, 09:54 PM said:

So since you calculated that the tot. vol would be 500 (25 times more), so then my answer 1.4 is correct (for 20ul tot. volume). Right?

gotmog on Aug 4 2009, 09:21 AM said:

Regards

Joana on Aug 4 2009, 11:27 PM said:

Joana on Aug 4 2009, 09:54 PM said:

So since you calculated that the tot. vol would be 500 (25 times more), so then my answer 1.4 is correct (for 20ul tot. volume). Right?

gotmog on Aug 4 2009, 09:21 AM said:

Regards

Yes, for 20 ul reaction 1.4ul dye is correct.