The missing quantity in your calculation is the molecular mass of the enzyme in g\mol. The moles of enzyme and moles of product will cancel.

U/mg is µmoles/(min-mg of enzyme). This is called specific activity.

To get from specific activity to kcat (also called turnover number), do the following calculation, converting units as you go:

µmoles product/(min-mg enzyme) x (min/60 s) x (1000 mg/g) x (____g/mole enzyme) x (mole/10^6 µmole).

After canceling like units in the numerator and denominator, you are left with s-1, the correct units for kcat.

Volume enters into the calculation of specific activity because you don't measure µmoles of product; you measure µmoles/L (i.e. concentration). When you multiply by the volume of the reaction, in L, you get µmoles. At some point you must also calculate the number of mg of enzyme in the reaction.

An important point: The specific activity can be calculated for any preparation containing the enzyme at any level of purity, as long as you can be sure that the product formation is due only to that enzyme. However, kcat should only be calculated for a pure enzyme, or for an impure preparation in which the concentration of the enzyme itself (rather than total protein concentration) is known. This is because you need to use the amount of the enzyme in the calculation.

I wonder Lorenzo whether I am getting well where is your brickwall.

I miss what "your" "U" is. U may be different from enzyme to enzyme despite the standard definition is that 1 Unit transforms 1 micromol of substrate/min, then:

If you use the standard definition the calculation is quite easy , for instance if you are finding 1 U using 0.1 ml of your preparation then your have an activity of

1 micromol substrate transformed/min 0.1 ml = 10 micromol substrate transformd/min ml

By the way forget about molecular mass of the enzyme. This is not needed.

Please be carefull

1- U/mg it the expression of specific ativity in a powder.

2-Vmax is a speed of product liberation or subsract consumption in a solution.

3-Simply you have to do this :

- Convert Vmax from (mol/l)/s to (micromol/l)/min by multiply the first with a factor of 6.10^7

- Now you have to divise Vmax with the ponderal concentration of enzym component (Cp=quantity use in mg/volume of solution in liter)

- you will have ((micromol/l)/min)/(mg/l), then liter will be eliminated

- You will have AS = X (micromol/min)/mg = X U/mg.

O agree with Adam's Shapiro answer.

Thank you all for your answers

if rate activity of enzyme given in umol/mg Chl h, how vmax can be calculated from this rate?

I am searching for Vmax of different enzymes of photorespiration.

Vmax is determined when an enzyme is saturated with substrate, or from a hyperbolic Michaelis plot (rate versus substrate concentration). It applies to the reaction of a single enzyme. Photorespiration involves several enzymes working in sequence, so the concept does not apply.

thank you for help.

I know vmax for each enzyme vary, actually i have come through an article where vmax of different enzymes of photorespiration have been reported as normalized value of rubisco (Vm/V1, where V1 is vmax of rubisco).

Upon checking the reference given for the different vmax values in article, I found an article in which specific activity of different enzymes have been reported in terms of umol/mg Chl h .

Q: I want to know how the vmax can be calculated from the specific activity?

Q2: how different vmax values be normalized w.r.t. rubisco?

I guess you can say that µmol/mg chlorophyll/h is a specific activity, although that term usually has units of µmol/mg protein/min. When dealing with chloroplasts or thylakoids, it may be difficult to do a protein assay because of the deep color from the chlorophyll. In any case, Vmax can be expressed in any number of ways. The specific activity is just the rate measured under a specified set of conditions divided by the amount of protein (or chlorophyll in this case). It can only be considered to be the Vmax if it is known that the specific activity was measured under conditions at which the rate is the Vmax.

For example, I can specify my assay conditions as reacting the enzyme with 1 mM substrate in a certain buffer at a certain temperature for a certain amount of time. The specific activity would then be the measured amount of product formed under these conditions divided by the time of the reaction and the amount of enzyme. But the rate may not have been maximal. Perhaps the Km for the substrate is 10 mM, for example. So the specific activity was not measured under Vmax conditions and the specific activity is not equivalent to the Vmax. In contrast, if you knew that the conditions being used for the assay were those under which the enzyme was running at the Vmax rate (e.g. the Km is 0.1 mM and you were using 10 mM substrate), the specific activity would be the Vmax. It doesn't matter that it is divided by the amount of chlorophyll, as a stand-in for the amount of enzyme.

Ideally, you would want to know the turnover number (kcat) of the enzyme, but you need to know the enzyme concentration in the reaction to calculate that from Vmax. Vmax = kcat[E]

I don't know what purpose is achieved by comparing the Vmax of each enzyme to that of RuBisCO, which is a very slow enzyme.

Hi

I agree with Adam