molar extinction coefficient calculation

8 posts / 0 new
Last post
biochemsitry12
biochemsitry12's picture
molar extinction coefficient calculation

Hi, im having loads of difficulties trying to calculate the molar extinction coffient for phosphate at 670nm.

I currently have a stand curve, with the x-axis in micro grams of phosphate and the y-axis with absorbance at 670nm.

could someone explain how to do it i know i have to use the formula e=A/CxL (L is 1cm) but im not sure how to convert the micrograms of phosphate and use it to get the coefficient :-S

if you need any additional information please let me know, i would really appericate if someone could suggest a method to calculate it based on micrograms of phosphate. thanks

Jon Moulton
Jon Moulton's picture
OK, here's the first piece -

OK, here's the first piece - from here you can probably finish it up.

Your problem is that you have a mass of phosphate instead of a concentration.  What volume is your phosphate dissolved into?  Convert grams phosphate to moles phosphate, then divide by the volume -- that will give you a molarity.  For a molar extincition coefficient, you need molarity (the units of the C factor in your extinction equation is moles/liter in order to give the extinction coefficient its usual units).

biochemsitry12
biochemsitry12's picture
hey,

hey,
oh yes i forgot to state the volume.
the phosphate is dissolved in 0.1ml, so if i convert the x axis into Moles and re draw the standard curve with the axis now in molarity (M) could i take the gradient of the line and would that give me the e value for the Abs at 670nm?

Jon Moulton
Jon Moulton's picture
You're headed in the right

You're headed in the right direction, but you need to convert not to moles but to molarity (moles/liter).  Moles is a step along the way, but then you need to take the volume into account to reach the molarity.

biochemsitry12
biochemsitry12's picture
at the moment i am slightly

at the moment i am slightly confused.
the reaction is the between AMP + water -----> adenosine + pi
AMP is 0.1ml, 10mM
we are provided with the absorbances of pi at various masses (ug)
so for example, 20ug pi has an absorbance of 0.15 measure at 670nm.
so for my calculations, i am assuming that there is 20ug/0.1ml and i convert that to g/l (0.2g/l), then divding it by 94.97 (mr for pi) to get moles (2.1x10^-3), from here i divided the absorbance value 0.15 by 2.1x10^-3 and gained a value of 71.4. i am assuming that this is my coffient.
is there something i am doing wrong? i am wrong for assuming that because there was 0.1ml AMP, that means that the phosphate in ug was dissolved in 0.1 ml?

Thanks you very much

biomedical
biomedical's picture
Hi, i'm working on the same

Hi, i'm working on the same calculation, and have come against similar problems.
I got a value of 601.04 as the coefficient. the answer seems wrong since its such a small number, but I was wondering if this is because the mass of Pi is small?

Jon Moulton
Jon Moulton's picture
@biochemsitry12:  Your

@biochemsitry12:  Your calculations look fine, but I wonder if you are not telling me about a dilution.  After the reaction, was the 0.1 ml of product solution diluted when placed into the spectrophotometer cell?  Many spec. cells require a larger volume of fluid to immerse the beam path.  If there is a dilution, you must take that dilution factor into account.  You know the concentration in the reaction solution, you know the absorbance measured in the spectrophotometer, but often the concentration in the reaction solution is diluted into the spec. cell and therefore the concentration measured in the cell is lower than the actual concentration of the reaction solution.  For instance, if you diluted the 0.1 ml to 1 ml you have a tenfold dilution and the measured absorbance would be multiplied by ten to describe the absorbance of the undiluted reaction solution (and yes, that is likely out of the linear response range where Beer's law is true, but you are not making a measurement at the very high absorbance; it works for the calculation of an extinction coefficient).  You do the calculation of the extinction coefficient based on the mass of phosphorous in the reaction solution and the calculated absorbance that the reaction solution would have if Beer's law was true for that concentration (that is, corrected from the diluted absorbance measured in the spectrophotometer).

@biomedical: it is not as simple as a size or mass correlating with absorbance.  Absorbance is due to the allowed electron energy state transitions in the molecules.  Small molecules can have big extinction coefficiants, big molecules can be transparent at some wavelengths.  It depends on what quanta can be absorbed, what are the allowed amounts of energy that the molecule can take up by moving electrons from lower energy states to more excited states. 

In some cases, size of the molecule can correlate with extinction coefficient.  In DNA for example, each base acts as an antenna for absorbing UV photons of particular energies.  Because DNA contains many repeated subunits, each bearing a base, a bigger DNA molecule does have a higher UV extinction coefficient, roughly the sum of the UV extinction coefficients of its constituant bases (we should be talking about specific wavelengths, but I am not going to start looking up absorbance curves for the bases today). So, a 1000-nucleotide DNA molecule contains a lot of bases (1000), each contributing to the overall molecule's extinction coefficient.

Janet Gonzalez
Janet Gonzalez's picture
molar extinction

I have an old spectrophotometer in the lab and my enzyme Ab max = 486 nm but the machine has only a 450 nm filter.  Is there a calculation that I can use to convert from one wavelength to another?