Hi,
In Folin Ciocalteu test for measuring total phenolic content , once I run the test with 50 % dilution and another time with 1/10 dilution of sample as well as standard. In another word, in the first test I added 500 microliter of sample/standard to each tube and in the second test I added 100 microliter of sample/standard to each tube because the numbers from 500 microliters sample were too high. Surprisingly, the numbers I read for the standard solution was quite the same even though in the first test I had added 500 microliter of different concentration of standard solution and in the second test I had added 100 microliter of them. Is this normal? Many thanks for your opinion.
Were both samples very dark blue in color? Or was the 100 uL significantly less dark to the eye?
I agree with Dr Guo;
Prepare several dilutions 1/2, 1/5, 1/10 and make your conclusion.
Dear Pery,
I agree with Anque Guo and Zvi Harduf,
Make series of dilutions so that absorbance shown at the selected wavelength is in the range of 0.3 to 1.7 , i. e. in the range of optimism conditions and then go for the calculation.
Hello Pery,
Yes, everybody here is correct. But they did not explain why they were correct. Remember that the absorbance is equal to the negative log of the transmittance. And that transmittance is equal to the radiant flux transmitted by a substance divided by the radiant flux received by a substance. So an absorbance of 1 means that 90% of the light was absorbed (10% transmitted) and an absorbance of 2 means that 99% of the light is absorbed. Now it becomes clear why most spectrophotometers never both to show absorbance much higher than 2. An absorbance >2 means that more than 99% of the light is absorbed by the solution in the cuvette. But the instrument can't really tell how much is being absorbed. It depends upon the design of the spectrophotometer. All the spectrophotometer can really do is measure the intensity of the light going into the sample and HOW MUCH LIGHT GETS TO THE DETECTOR after it passes through the cuvette. Let's take a far-fetched example and assume that "all" of the light is absorbed (this requires either infinite absorbtivity, infinite pathlength or infinite concentration but we will put that aside for the moment.) Now the detector should read "0". But it will not. There is a baseline of electronic noise that generates a signal even when no light is present. Also, there is light leakage--stray light that enters vents from the source or that bounces off dust particles that pass through the initial beam. This means that very dark solutions, solutions that are absorbing far more than 99% of the light at a particular wavelength , will still read an absorbance between 2 and 3 (depending on the manufacturer, the age, the environment, etc.) And even a 5-fold dilution still is absorbing much more than 99%. But the background noise and the light seepage are generating a signal. Since the noise and the light seepage are about the same at both solutions, you get about the same value even though one should be five-fold less.
That is why I asked if it was dark - it proved to me that you were transmitting
Dear Pery
the prepared samples could reach to plateau response of the detector. Yu can apply higher dilution factor to your samples in order ot obtain the linear response. Consider that generally the absorbance may be comprised into 0.1-0.8 range to gave high linearity.
- Badges
- Science method
- Similar topics
- Chemistry
- General Biochemistry
- Biochemical Techniques
- Analytical Chemistry