Suppose that I have an HPLC with a variable wavelength UV detector, as well as two pure protein samples that differ by a point mutation (otherwise same concentration, extinction coefficient, buffer conditions, etc.). If I run them through the same column sequentially and integrate the A280 peaks after baseline subtraction, should I expect the areas under the curves to be the same? I guess another way to ask the question is whether the absorbance unit is something that can be compared from run to run, or is the elution volume/peak shape the only physically meaningful quantity for comparison?


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There is nothing absolute in HPLC related measurements. Take the easiest example first. Think of a separation of a racemate using a chiral column on a UV detector. After injection, you will see two peaks, with different retention times, and equal peak areas. Their absorbance at the peak maximum will be different because both peaks will have different widths. Change the flow rate and now your peak areas change but their peak ratio will still be 50:50.

The same ideas apply in your case. You state that everything is same (conc., molar extinction coefficient etc.). In HPLC, your peak areas will be affected by flow rates only given the rest of the conditions remain identical.

  • $\begingroup$ so if I understand you correctly, if the flow rate is the same in two runs then the area should be identical? $\endgroup$
    – Drecate
    Sep 13, 2021 at 22:21
  • $\begingroup$ Yes, if the concentration and molar extinction coefficients are identical. $\endgroup$
    – AChem
    Sep 13, 2021 at 22:37
  • $\begingroup$ Why does flow rate change peak area? I was under impression that same amount of analyte will result in the same peak area. Is it because of the solvent which absorbs at the same wavelength (and its amount changed)? But in that case peak ratios will be different too.. $\endgroup$ Sep 14, 2021 at 15:53
  • $\begingroup$ @StanislavBashkyrtsev, Yes, this is the tricky part of HPLC-UV. It is a concentration sensitive detector (another type is mass sensitive detector). So at low flow rate, you would expect that peak areas are larger, because the analyte is spending more time in the detector (recall peak area is integration with respect to time). $\endgroup$
    – AChem
    Sep 14, 2021 at 16:00
  • $\begingroup$ But the height of the signal is lower at low flow rate, which should compensate for the longer exposure. Or do you mean that the same portion of analyte is being measured again and again with the low flow rate and therefore each individual molecule contributes to the area more? $\endgroup$ Sep 14, 2021 at 17:27

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