Calculating Vmax using Beer-Lamberts? watch

1. Hey all, I have a question requiring me to calculate Vmax in terms of Micromoles guaiacum/second. The information i'm given is 4ml volume in the cuvette and that oxidised guaiacum's molar absorption coefficient is 2000M-1cm-1 at 600nm. Path length is 1cm.

I'm guessing that the Beer-Lamberts law will have to be used? But I just don't see how I would derive the Vmax from that.

Thanks.

Edit: Okay, I see Guaiacum is in the reaction cocktail for the experiment I carried out. "Mix 5.0ml of “Reaction” cocktail (containing glucose oxidase/ peroxidase/ guaiacum dye) with 2.0ml of acetate buffer. " Any help please?

So using A=lEC it's 2000=1x600xC, this gives me 3.3 for C, I think in M/cm? Any help from here.
2. My Vmax for the experiment was 0.360mM/min, I have no idea if I'm supposed to be using beer-lamberts or not :/
3. I'm quite confused. You say that you have a Vmax value yet you do not know how you obtained this value? What was your experimental set-up? Have you generated your own set of absorption values?

Beer-Lambert law is A=Ecl

To obtain a concentration you need to divide the absorbance by the product of the molar absorption coefficient and the cuvette length: c =A/(El). In your original post you have divided the molar absorption coefficient by the wavelength, but this is not correct.

You should have generated a range of absorption values at different time points. You can use this to calculate the rate of change of absorbance with respect to time. If you incorporate the rate of change of absorbance into the Beer-Lambert law, then you will obtain a change in concentration over time (i.e. rate of reaction).

To obtain a Vmax, you would have to obtain the initial rate of reaction for different substrate concentrations. Once you obtain this, you there are various graphical plots that you can use (comparing substrate concentration and rate) that would allow you to obtain Vmax.
4. (Original post by Eddict)
Hey all, I have a question requiring me to calculate Vmax in terms of Micromoles guaiacum/second. The information i'm given is 4ml volume in the cuvette and that oxidised guaiacum's molar absorption coefficient is 2000M-1cm-1 at 600nm. Path length is 1cm.

I'm guessing that the Beer-Lamberts law will have to be used? But I just don't see how I would derive the Vmax from that.

Thanks.

Edit: Okay, I see Guaiacum is in the reaction cocktail for the experiment I carried out. "Mix 5.0ml of “Reaction” cocktail (containing glucose oxidase/ peroxidase/ guaiacum dye) with 2.0ml of acetate buffer. " Any help please?

So using A=lEC it's 2000=1x600xC, this gives me 3.3 for C, I think in M/cm? Any help from here.
You must be busy doing the same lab file as me, I have went through it all and i seem to have gotten an answer of 0.000008um produced per litre, I really don't know where to go from here!?
5. (Original post by Eddict)
Hey all, I have a question requiring me to calculate Vmax in terms of Micromoles guaiacum/second. The information i'm given is 4ml volume in the cuvette and that oxidised guaiacum's molar absorption coefficient is 2000M-1cm-1 at 600nm. Path length is 1cm.

I'm guessing that the Beer-Lamberts law will have to be used? But I just don't see how I would derive the Vmax from that.

Thanks.

Edit: Okay, I see Guaiacum is in the reaction cocktail for the experiment I carried out. "Mix 5.0ml of “Reaction” cocktail (containing glucose oxidase/ peroxidase/ guaiacum dye) with 2.0ml of acetate buffer. " Any help please?

So using A=lEC it's 2000=1x600xC, this gives me 3.3 for C, I think in M/cm? Any help from here.

Hi, did you ever find out the method of doing this, got this stupid question in a lab file and we're all very stuck!
C

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