# problem with spectrophotometer - (Sep/09/2004 )

Hi all;

I'm new here and alsa in laboratory.I have a problem with spectrophotometer.I put into the spectrophotometer kuwette 990 microliter blank and 10 microliter my sample protein. Than read the result at 280nm as 0.053.What does it mean?Is that mean I have 0.053 miligram protein in 10 microliter?Thanks...(sorry for my terrible english )

Burcu

-BURCU:)-

yes and no... - it's never that simple

1) your measured OD is on the low end of a possibly correct measurement. the OD should be between 0.1 and 1 (some say 0.05 - 1.5) for reliable measuring.

2) the OD 280 depends on the number of aromatic aminoacids in your protein, as they absorb the 280nm light. a formula for rough estimation of protein concentration is
od280*dillutionfactor= concentration [mg/ml]
0.053 * 100 = 5,3 mg/ml protein (as you suggested: 53µg/10µl)

But

you should check for contamination with nucleic acids, therfore get the OD260nm, too. If your OD280/OD260 is <0.6 you shoud use the formula

(1.55 * OD280 - 0.76 * OD260) *dillutionfactor = concentration [mg/ml]

3)For extact determination of your protein concentration you have to calculate the molar extinction coefficent, which is individual for every protein. Then there are additional correction factors if your protein is denatured or native.

4)For determination of protein concentration there are loads of commercial kits out there which usually base on the bradford or lowry methods. But they have pro and cons that you have to keep in mind, too.

mike

Thank you very much for your reply Mike.Your answer do my job.I have another question.I am working on protein purification for my thesis and I planned to do acetone precipitation and after ammonium sulphate precipitation.In your opinion is there any drawback of this method?

thank you very much for your time
Burcu:)

-BURCU:)-

aceton precipitation should be ok, worked fine for me....

mike

hello Burcu
as ur OD reading is 0.053
that means ur concentration of protein in mixture is 0.053mg/ml
and
after all
your mixture volume is now 1ml (990 micro litre of blank water+ 10 micro litreof sample)
so ,
10 micro litre of sample contains 0.053milligram of protein(because all the proteins has come from your sample, but not from ur blank water)
again.

10 micro litre of sample contains 0.053milligram of protein
1 micro litre of sample contains 0.053/10 milligram of protein =0.0053milligram
1000 microlitre of sample contains 5.3milligram of protein
that means
1 ml of ur sample contains 5.3milligram of protein
thank you for your nice problem
bye
ashok

-ash_bt143-

hello mike
can you give a reference to us for this calculation formula.
thanks by now...

heval

-heval-

the formula is out of a lecture-script for students around here. I try to find the guy who originally made this script and aks him for the literature, ok?

mike

So, I've been told that the formular is from:

O. Warburg, W. Christian (1941), Biochem. Z. 310, 384 - 421

and should have been refined in:

H. M. Kalckar (1947), J. Biol. Chem. 167, 461
E. Layne (1957), Meth. Enzymol. Vol. 3, 454

I haven't read the original papers, though. I was just told by the guy who had put the script together. So don't blame me if the above papers aren't what they should be!

mike