# How do I use this molar extinction coefficient?

Asked by Mr_Saturn512 (466) July 10th, 2014

Long story short, we want to be able to use the molar extinction coefficient of a protein to calculate protein concentration instead of constantly having to create a standard curve and calculating from that.

The formula to get the protein concentration is as follows: (A280)(dilution factor) / 51,300 (that’s the extinction coefficient) = protein concentration.

Since it’s the molar extinction coefficient, it’s in these units I’m unfamiliar with in the sense that everything I’m looking up makes me unsure why they rearrange it. Someone said it’s simply “mol/Liter” but I also see “mol/L-1 cm-1” So I’m just confused as to what the units are in the actual equation then.

We then want to convert the answer into mg/mL. That’s the goal here: we want it in mg/mL.

Apparently I have to know the Molecular Weight of this protein, which is 34,620. But I don’t know how to incorporate that into all of this in order to figure this out.

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Molar extinction coefficients are typically in M^-1* cm^-1 (aka 1/(M*cm)). If using a cuvette with a 1 cm path length, using the equation you give above the cm cancels and it gives you concentration in Molar, or Moles/Liter. If you multiply this by molecular weight (g/mol) moles cancel and you get g/L, which is equivalent to mg/mL.

The units are because of the Beer-Lambert law, which you rearranged in the above, A = ebc, where A is the absorbance (A280 in this case), e is the extinction coefficient, b is the path length, and c is the concentration. A has no units, thus the right side must cancel. If c is Molar and b is cm, e must be M^-1*cm^-1. Likewise, when it’s rearranged as above, there’s actually a b in the equation that’s dropped because it’s just 1 usually. So it’s really c = A/(eb). If A has no units, e = M^-1*cm^-1, and b = cm, cm cancels, and M^-1 is in the denominator and thus c is left to simply be M.

BhacSsylan (9520)

Okay awesome, now what if the pathlength isn’t 1 cm? Say .1 cm or lower? I technically am not using a cuvette since the spectrophotometer I’m using allows you to just pipette the amount on a pedastel without the need of a cuvette. But I have the known path length somewhere, just not on me at the moment. I just know it’s something like .1 cm.

Then just multiply the resulting absorbance by the required factor. So if it’s .1 cm (for a nanodrop machine which sounds like you described, this is likely), you’d multiply the absorbence by 10 (or divide by the b of .1, same thing). Sometimes those machines will do that for you since the math is slightly easier at 1 cm, though, so check for sure what it’s giving you.

BhacSsylan (9520)

Either this is just how it works out to be or something is still off, although I feel like I’m really close to figuring this out (thanks for bearing with me).

My resulting concentrations do make sense. . .and yet some are pretty off from the calculations I got using the standard curve equations.

For example, for a 1.0 mg/mL solution, my result was 0.99 mg/mL using the standard curve. However, when I use the equation with the extinction coefficient, I get 1.6 mg/mL. Kind of a big difference on the smalle scale.

Others look okay. For 2.0 mg/mL I got 2.02, then .83 for .8 mg/mL. But then a couple others are just off again, like getting .34 for a .2 mg/mL solution and .2 for a .1 mg/mL solution.

The standard curve doesn’t look wrong either. I graphed it out and the data points I graphed are spot-on in the linearity.

So ultimately, I don’t know what this means – having the two methods split like that.

Hmm, interesting. There could be a few issues, like the fact that the extinction coefficient is just wrong. It relies on a number of assumptions if it’s calculated from the protein sequence (for instance: that all aromatic groups are on the surface, which is probably not true). The fact that 1 is wrong while 2 and .8 are not is strange, I’d measure that again to be sure. But it’s not odd to have it off on either end of a curve, usually linearity in these things only exists within a certain range. And if you made your curve with that spectrometer, it’s already somewhat accounting for that while the coefficient assumes it’s constant. So I wouldn’t be too worried.

In general, it tells you just what you probably already knew: the curve is more onerous to make but more accurate. It seems accurate enough to work for most applications, but depending on what you’re trying to accomplish it may not be exact enough. Another thing you can try is finding another spectrometer and doing the same experiments. Whichever is more accurate should be consistent across the machines.

BhacSsylan (9520)

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