## Wednesday 20 October 2010

### Why can't I extrapolate the Bradford Assay graph if the Beer-Lambert Law applies?

If you would like to test your skills in working with the Beer-Lambert Law, you might like to take the Spectrophotometry tests at Maths4Biosciences.

One question I have been asked several times after the feedback on a practical is if the Beer-Lambert Law, A = ε . c . l, gives a straight line, then why can't I extrapolate the data to find the answer for one of my unknown proteins?

The Bradford Assay is an indirect (or colorimetric) method for estimating protein concentrations. The assay is based on a shift in the absorbance of a dye, Coomassie Brilliant Blue G-250, from 465 nm to 595 nm when it binds a protein under acidic conditions. The Bradford assay measures the amount of bound dye and NOT the amount of protein present (although the two are connected until the dye is all used). As the dye is limiting (i.e. you only put in a set amount), you can reach a point when all the dye is bound, and no matter how much more protein you put in, you will not see a change in absorbance. You can increase the protein level to a point where available dye becomes exhausted so the graph would plateau.

Standard Curve for a Bradford Protein Assay

Any value that falls in the yellow area of the graph (i.e. a concentration of more than 15 µg/ml or an absorbance greater than 1.02) cannot be used as the graph is no longer linear and is starting to plateau. Any absorbance that gives a protein concentration below 15 µg/ml can be used as this is in the linear portion of the graph (i.e. blue area), and the Beer-Lambert Law applies as the dye is not limiting.

In the experiment, you are working in the linear portion of the graph above (the blue area), so the line is straight, i.e., it hasn't plateaued. Therefore, we can say that between 0 and x µg/ml (or 15 µg/ml above), we have a certain absorption range, say 0 to y (where y is about 1.01 in the above figure). However, once we get above x µg/ml, or above the maximum absorbance y, we cannot say whether or not the graph has plateaued, so all we can say is the value is greater than x µg/ml. This may seem somewhat limiting (pardon the pun), but the Bradford assay can, in fact, detect protein concentrations over a 10-fold concentration range.

By the way, if we were measuring the protein directly, say at 280 nm, then we could apply the Beer-Lambert Law. In fact, if we knew the extinction coefficient, we wouldn't even have to do a standard curve.

## Wednesday 6 October 2010

### The Beer-Lambert Law, a straight line and the units of the extinction coefficient

If you would like to test your skills working with the Beer-Lambert Law, you might like to look at the Spectrophotometry tests at Maths4Biosciences.

Some students struggle to understand the relationship between the Beer-Lambert Law and a straight line and work out the units of the extinction coefficient (ε).

The Beer-Lambert Law states:

A = ε . c . l

Where:

A = absorbance
ε = extinction coefficient
c = concentration
l = path length (i.e. the distance the light travels through the sample)

So, what is the connection between this and a straight line, and what are the units of the extinction coefficient?

The units of the extinction coefficient

In my opinion, the extinction coefficient has some of the craziest units out there.
Absorbance (A) has no units, so the units of the extinction coefficient (ε) are determined by how the concentration (c) and path length (l) are being measured. That is, the units of the extinction coefficient must cancel out the concentration and path length units so that the absorbance can have no units!

A worked example.

A = ε . c . l

A = absorbance, units - so put in 1
ε = extinction coefficient, units - unknown
c = concentration, units - milli-Molar, mM
l = path length (i.e. the distance the light travels through the sample), units - cm

So...

A = ε . c . l

And, with units:

[1] = ε . [mM] . [cm]
Rearranging...
[1] / ε = [mM] . [cm]

ε = 1 / ([mM] . [cm])

or

ε = [mM]-1 . [cm]-1

So, the units are mM-1 . cm-1

This can be checked by putting it all back together:

A = ε . c . l

A = ([mM]-1 . [cm]-1) . [mM] . [cm]

This gives:

A = [mM]/[mM] . [cm]/[cm]

So, the mM and the cm cancel each other out, leaving no units for absorbance A.

A straight line

The Beer-Lambert Law:

A = ε . c . l

Where:

A = absorbance
ε = extinction coefficient
c = concentration, units
l = path length

The equation for a straight line is:

y = mx + c

Where:

c = the y-intercept

If you plot concentration against absorbance, then x = concentration and y = absorbance. Plus, from the Beer-Lambert Law, we know that if the concentration is zero, then absorbance must be zero.

A = ε . c . l
A = ε . 0 . l
A = 0
So...

y = mx + c
absorbance = m . concentration + c

From above, if concentration = 0, then absorbance = 0, hence c must be zero
y = mx + c
absorbance = m . concentration + c*
0 = m . 0 + c
c = 0

(* note, this c is the y-intercept and not the concentration)

Therefore...
y = mx + c
absorbance = m . concentration + 0
or
y = mx

Comparing:

y = mx
absorbance = m . concentration

With (and rearranging):

A = ε . c . l

A = (ε . l) . c

y = m . x
If y = absorbance, and x = concentration, then m (the gradient) must equal extinction coefficient (ε) multiplied by the path length, l, or ε . l. As l is typically 1 cm, the gradient, m, must equal the extinction coefficient (ε).

You may also find the following two videos helpful.