[Emma]

Hello,

we have a question to determine Turnover number of an enzyme-catalyzed reaction, so I need to get Vmax and Et. The data given is the weight of the enzyme in kDa and
* when 20microgram of the enzyme catalyzes the phosphorylation of 100 mg glucosoe in 2 minutes in 25mL in presence of excess Mg and ATP. And they asked for turnover point.

what i did was to convert the weight from kDa to grams so i can get the Vmax in units of 1micromol/mL/seconds. I got number around 2x10-5 which is unreasonable!


any help ?

[Yggdrasil]

Can you show how you set up the calculations so that it is easier to figure out where the error occured?  The Vmax is the change in substrate concentration over time.  Perhaps it is easier to break the problem into small chunks, and first ask what was the change in substrate concentration in µmol/mL?

[Emma]

Can you show how you set up the calculations so that it is easier to figure out where the error occured?  The Vmax is the change in substrate concentration over time.  Perhaps it is easier to break the problem into small chunks, and first ask what was the change in substrate concentration in µmol/mL?

The enzyme weight is 150kDa so i found the moles of the e zymes to get Vmax. I didnt use the volume given in the question

[Yggdrasil]

You're using the wrong molecular weight for the substrate.  The substrate is glucose (and you're using the molecular weight of the enzyme).  Otherwise, your approach looks correct.

[Emma]

You're using the wrong molecular weight for the substrate.  The substrate is glucose (and you're using the molecular weight of the enzyme).  Otherwise, your approach looks correct.

The teacher provided us with the volume of the buffer solution where the reaction is taking place. I didn't use it, is that fine ?

[Yggdrasil]

Yes. You could calculate the turnover number by calculating the change in glucose concentration per second then dividing by the enzyme concentration, but the volume would end up canceling in the final answer.