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!
Protein labeling calculation
Started by Joana, Aug 04 2009 04:00 AM
4 replies to this topic
#1
Posted 04 August 2009 - 04:00 AM
#2
Posted 04 August 2009 - 08:21 AM
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
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
#3
Posted 04 August 2009 - 09:54 PM
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?
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:
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
Regards
#4
Posted 04 August 2009 - 11:27 PM
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:
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?
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:
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
Regards
#5
Posted 05 August 2009 - 12:10 AM
Joana, on Aug 4 2009, 11:27 PM, said:
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:
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?
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:
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
Regards
Yes, for 20 ul reaction 1.4ul dye is correct.













