Hi all,
I having the data related to the asparaginase enzymes. In order to calculate the enzyme activity I did few experiments.. As the result I performed standard ammonium sulphate curve and I plotted my test values in the standard graph and I determined the equivalent amount (mM) of ammonium sulphate.. The problem starts here, I couldn't proceed further with the calculations?
Hi there,
For your samples you have an ammonium concentration (be careful as 1mM ammonium sulphate is actually 2mM in ammonium) and the time of the kinetics at which you have analyzed the sample. If you have run the reaction in the right condition (enzyme concentration far less than the substrate concentration and product of reaction formed linearly toward the time), the ratio ammonium formed / time gives the initial rate of the reaction. If you know the concentration of enzyme then you might calculate the activity of the enzyme by dividing the initial rate by the enzyme concentration.
Thanks for your answer @ Dominique Liger.. Assume that I am setting up a reaction with 0.5 ml of Tris-HCl (pH 8.6) along with 0.5 ml of crude enzyme + 0.5 ml of substrate. Incubated for 30 min at 37 oC. the reaction was stopped using TCA. By following Imada et al., asparaginase assay. I got an absorbance value of 0.997 at 450 nm. Here starts my confusion, as you said 1 mole of ammonium sulphate gives 2 moles of ammonia. If i plot the absorbance value to the standard ammonium sulphate curve I receiving a value of 700 mM ammonium sulphate. Now what is the equivalent amount of ammonia liberated as per the fact it should be 1400 mM but what is the amount of ammonia in terms of moles? This part is so confusing that I couldnt solve as I am not a chemist
Hi again,
There must be something wrong with your standard curve or the calculation you make with it. How do you raise the conclusion that 1 OD unit corresponds to 1400mM ammonia? In my memories, ammonia determination with Nessler reagent gives much smaller concentration...
0.25mM - 0.032; 0.50mM - 0.1; 0.75mM - 0.17; 1mM - 0.25; 1.25mM - 0.324; 1.50mM - 0.383; 1.75mM - 0.505; 2.00mM - 0.57. With a value of y = 0.0029X - 0.0353, R2 - 0.9901 if i plot the values for my sample of the absorbance value - 0.997, its showing as 355.96 mM of ammonium sulphate. Since, 1 mole of ammonium sulphate liberates 2 moles of ammonia i calculated as 712 mM of ammonia was liberated.. Am I right with my calculation
Hi again,
First issue : OD of 0.997 is not within the range of your standard curve. Second issue, if OD 0.57 is giving 2mM how do you get that 0.997 is giving 356mM? It should give something around 4mM if linearity is respected... Please check your calculation again.
Hi,
Please sure what is the product enzymatic reaction, plot concentration of the enzymatic product result using unit; mM, M, ppm, mg/mL, mg/L etc. vs interval of time of enzymatic reaction. Enzymatic activity can be done measure of the substrate.
My experience activity of enzyme was known without standard curve but plotting that product of reaction and depend unit. The slope is the value of product per minute for example, follow Robyt and White.
Thank you for your clarification@Dominique Liger. I did the mistake in calculating part. As per the reading I received for e.g. for 2.889 (0,706 - 4 Dilution). I observing 6.5 mM of ammonium sulphate in standard graph If I convert in to micromolar I need to multiply with 1000, so I receiving 6500 micromolar. Now if I want to convert this in to micro moles. How do I do it?
Well, a concentration is converted into a number of moles by mutiplying the former with a volume.
In this assay, which would you suggest me as the volume.. either the whole reaction mixture or just the volume of substrate alone. This is the most confusing part. If its 6500 micromoles, Do I need to divide it with the molecular weight of the ammonium sulphate?
The question is how did you draw the standard curve you are using : as X-axis, is it the concentration of AS present in the whole sample submitted to OD measurement or is it the concentration of AS of the solution you actually introduce into the Nessler reaction mixture? Whatever the way you proceeded, the way to convert concentration into a number of moles is to multiply the concentration by a volume.
What should be the concentration of ammonium sulfate for standard curve preparation. I am using 0.25 mM to 1.5 mM concentration, but whenever I am adding Nessler reagent in to that, brown precipitate formation occur. So it is not possible to take absorbance. Yogeswaran Jagadeesan Dominique Liger
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