# Help determining Absorption Coefficient & Concentration from Uv-Vis reading [closed]

hope you're doing well. I'm having some trouble determining the absorption coefficient and concentration values of a protein from its UV absorption results.

The experiment in question involves an ion exchange chromatogrpahy column to elute hemoglobin from an HB/BSA mixture, then quantitating the amount and concentration of hemoglobin in the samples and fractions. With the calculations that I'm currently using, I obtain an absurdly low yield, less than 2%, and I'm not sure where I'm going wrong.

My results are comparable to the sample results we were provided for the experiment, so I'm assumming that the error is a calculation error on my part. If someone could check over my approach, I would greatly appreciate it. The values below are completely made up, this is just to check my process.

Suppose I was working with a 2% Hemoglobin 0.1% BSA mixture.

Then, the (w/v) concentration of the haemoglobin should be 2 g/100 mL. If 0.3 mL of the mixture were added to the column, then a mass of $$\frac{2\text{ g}}{100 \text{ mL}} \times 0.3 \text{ mL} = 0.006 \text{ g}$$ of haemoglobin was added to the column.

Okay, say you collected a number of fractions from your column, pipetted them into a microtiter plate with buffer blanks and a 0.05% Hb standard solution as references, and measured the absorbance of the samples at a particular wavelength.

You want to determine the absorption coefficient for your 0.05 % (w/v) standard solution, so you do the following:

First, you average the buffer blank absorbance values, say its something like 0.035 and subtract this average from all absorbance values to obtain adjusted values.

Convert the 0.05% (w/v) concentration into mg/mL: 0.05 % (w/v) $$\rightarrow$$ 0.05 g/100 mL = 0.5 mg/mL.

Sample calculation:

Suppose this well has 50 uL of the 0.05% standard solution.

Then, the mass of Hb present in the well is equal to m = (0.5 mg / mL) (50 uL x 1 mL/1000 uL) = 0.025 mg.

If the volume of solution in the well is 50 uL Hb standard + 150 uL buffer = 200 uL, then the concentration of Hb in the well is equal to the mass of HB present divided by the volume of solution in the well. That is, [Hb] = $$\frac{0.025 \text{ mg}}{200 \text{ mL}} \times \frac{1000 \mu\text{L}}{1 \text{ mL}}= 0.125 \text{ mg/mL}$$

And similarly for the other standard wells.

Then, by Beer’s law, the absorption of the absorbing species is the product of the absorption coefficient, the path length, and the concentration of the sample solution. Hence, $$A=elc \implies e =A/lc.$$ If your plate has a path length of 0.7cm, and your sample has an adjusted absorbance value of 0.360, then the absorption coefficient e is given by \begin{align*} e & = \frac{A}{lc} \\&= \frac{0.360}{(0.7 \text{ cm})(0.125 \text{ mg/mL})} \\ &= 4.114 \,(\text{mg/mL})^{-1} \text{ cm}^{-1}\tag{1} \end{align*} Yes? Then, suppose you perform the same calculation for two other standard solution wells, and get values of 4.13 and 4.09 for e respectively. You take the average of the three values for an absorption coefficient of 4.11 (mg/mL)^-1 cm^-1.

You then try to use this value to calculate the concentration of Hb present in the other wells:

For simplicity, say you ran a teeny tiny plate with only four wells, whose (adjusted) absorbances were Well 3A: 0.203 (25 uL of Fraction 1, 175 uL buffer) Well 3B: 0.651 (25 uL of Fraction 2, 175 uL buffer), and Well 4A: 0.417 (50 uL of Fraction 1, 150 uL of buffer), and Well 4B: 1.315 (50 uL of fraction 2, 150 uL of buffer).

You calculate the concentrations of Hb in each well as follows

3A: By Beer’s Law, the concentration of Hb in solution is given by \begin{align*} c &= \frac{A}{el}\\ &= \frac{0.203}{((4.114(\text{mg/mL})^-1 \text{cm}^-1)(0.7 \text{ cm})}\\ &= 0.0705 \text{ mg/mL} \tag{2} \end{align*} This is the concentration of Hb in the well however, so to find the concentration of Hb in the fraction, you find the amount of Hb present in the well: \begin{align*} m &= c \times \text{ Volume of solution in well}\\ & = \frac{0.0705 \text{ mg}}{\text{mL}} \times \frac{0.2}{ \text{ mL}}\\ & = 0.014 \text{ mg } \tag{3} \end{align*} Then divide it by the volume of the fraction solution in the well: \begin{align*} \text{Fraction concentration } & = \frac{0.014 \text{ mg}}{ 0.025 \text{ mL}}\\ & = 0.564 \text{mg/mL} \tag{4} \end{align*} The same calculation applies to the other wells. To calculate the total mass of haemoglobin recovered, you take the sum of the value determined in (3) for each well.

Although these values are made up, is this the correct approach? I can’t find any examples showing how to do this, and we weren’t taught how to do so in the lab or in class (and it’s been a while since I’ve done any sort of stoichiometric calculations) so I’m hoping, but not sure whether this is actually correct.

• Please explain your sentence (I copy) : Okay, say you collected a number of fractions from your column, pipetted them into a microtiter plate with buffer blanks and a 0.05% Hb standard solution as references, and measured the absorbance of the samples at a particular wavelength. ... Which volume do you pipette ? And how much standard do you pipette ? Oct 22, 2021 at 15:44
• Hi @Maurice, thank you for replying. The hypothetical arrangement here involves collecting fractions from the column in volumes of 4 mL. Each well of the microtitre plate has a total volume of 200 uL. The "blank" wells contain only 200 uL of your buffer, the Hemoglobin standard wells contain varying volumes (25, 50, 100 uL etc) of 0.05% Hb Standard solution with the remaining volume made up by buffer, and the sample wells each contain a particular volume from a given fraction (20 uL, 40 uL) with the remaining volume made up by buffer. Oct 22, 2021 at 16:15
• There are too many variable parameters. I am lost. Sorry ! Oct 22, 2021 at 19:10
• It would be nice if you could choose a meaningful title. Oct 22, 2021 at 22:48
• A bit late, but this is my guess: the error is likely in the way the total yield was computed from the amount of protein determined for the individual wells. You should not just add up the amount of protein in each well, since this the sample volumes in each well represent only an aliquot of each fraction collected. You should instead scale those amounts by the ratio of the aliquot volume to that of the fraction. You may have done this after the fact, but that is not clear from your exposition. Jul 20, 2022 at 7:12

## 1 Answer

Wait: 1)You're using several standards with the same concentration? 2)The samples (well, all but one) have higher absorbances than your standard? 3)Half of the samples have absorbances above 1?

I think I see at least part of the problem.

Higher absorbance means less light is getting through to the detector. Due to limitations of the instrumentation (stray light, noise in electronics) this means that the instrument response will deviate from the expected linear behavior at higher concentrations.

Suggestions:

1. Prepare multiple standards with different concentrations, and plot absorbance vs concentration. Use linear regression to check how well a straight line fits your data. The first time around, you'll want a lot of data points over a wide range of concentrations.
2. If your correlation coefficient is too low (I generally use r^2>=0.998 for UV-vis), try dropping data points one at a time from the high concentration end of the graph until you get a good linear fit to your data.
3. Dilute all your samples so they fall within the range of standards you decided to keep (preferably close to the middle, but it doesn't have to be perfect).