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Topic: c1v1=c2v2 ... 2 seperate solutions?  (Read 2866 times)

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Offline Kloda

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c1v1=c2v2 ... 2 seperate solutions?
« on: April 12, 2008, 11:40:29 AM »
Hey

I'm trying to measure the activity of an enzyme so I'll first to a standard curve of the substrate so I know the absorbance for known concentrations of the substrate.

I have 0.1 mM p-NP (ml) of the substrate and 0.02M NaOH(ml) to dilute it down.  I also have an amount p-NP given as an example in umol

so first question is ... with 0.5 0.1 p-NP(ml) and 9.5 0.02 NaOH I have 0.05 p-NP(umol) *product I'm assuming?* ... does this mean the product in umol will always be a tenth of the origional substrate concentration?

k.. once I have my graph ...next solution ...   I have 0.02 M p-NPP(ml) substrate and a buffer (ml) and add .1ml of enzyme  ... once I get the absorbance I'll be able to read off my graph to get amount of p-NP (umol) *product*

I have C2=(C1*V1)/V2 to determine substrate concentration .. plus I have to fill in values for p-NP(umol/30min) and V0 (Umol p-NP/min)

if p-NP product values taken from graph are the initial velocity V0 are the values for p-NP(umol/30min) the V0 values * 30?

last question!

Using C1V1=C2V2 do I ignor my first solutions used to plot the standard curve?  and if so is V1 the 0.02M p-NPP(ml) values?  and C1 0.02M?  and V2 the value for 0.02M p-NPP(ml) + buffer(ml) + .1ml (enzyme) ?

I've been trying to figure it out on my own and thats what I've come up with but I've no way of knowing if I'm on the right track... any help is greatly appriciated!!!

Thanks  :)

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