Your units are correct. Extinction coefficients tend to be very high ('000s), so when you divide your absorption (

I guess you are using Extinction Coefficient to calculate the amount of product formed during the reaction; it does not matter if you are using uM concentration of Substrate because in the formula "Aλ= ε *concentration *L" of determined substance (in your case the product), you will need to convert the result of concentration (c=Aλ/ε, consider 1cm pathlength) to the unit you would prefer.

Keep track of the units of all variables. As suggested by Ana, you also need to consider the pathlength. C=A/εb

Since A is unitless, the unit of C depends on the units of ε and b.

velocity = ∆P/∆t = - ∆S/∆t, assuming the stoichiometry is 1:1.

∆A=(A1-A2)= ε λ (c2-c2) = ε λ ∆C. Now the complication is that we usually do not give velocity in terms of concentration (m/V) change per time unit, but molar mass (∆m) per time unit. Usually,

In other words:

∆C = ∆(m/V) = ∆A/(ε λ) , where V is the volume of the cell (constant), λ the light path length (generally 1 cm) and ε your molar extinction coefficient.

Therefore:

∆m= ∆A V/(ε λ) now you just have to use consistent units

∆m [mol] = ∆A [dimensionless] V [l] /( ε [l/(mol cm)] λ [cm]

The most widely used (though not the standard) is the international unit of activity, which is 1 µmol/min. You can either change your extinction coefficient to units µM^(-1) cm^(-1) or change your result to µmol. If you change your extinction coefficient and your volume to ml the factors simplify and you get conversion factor of 0.001 for your result (these should summarize the previous answers).

Best wishes

Rogelio

Just to clarify the subject, one unit (U) refers to the amount of enzyme able to produce/release 1 micromol per minute; it should not be confused with velocity.

Well, just to add even more clarity. 1 IU is a unit of enzyme activity, which is a velocity of the catalyzed reaction under a given set of conditions. You can read it in more detail here:

http://www.iupac.org/publications/pac/2001/pdf/7306x0927.pdf

Now, the enzyme mass is proportional to velocity, i.e., in the simplest case , the MM equation states:

v= Et* (kcat S/(Km+S))

Therefore, one way to quantify enzymes is to measure the enzyme activity under a standard set of condition, specially when the molar mass is unknown due to lack of purity or insufficient data. This is widely used in the measurement of serum enzymes in humans for medical diagnosis, but also used by producers to label the enzyme content of commercial enzyme preparations.

The question being asked here is somewhat less complicated than some of the above discussions regarding enzyme concentration. Since that is a constant in the assay, it doesn't play a role in calculating Km and Vmax.

A critical measurement that must be examined is the entire absorption spectrum of the substrate being produced. Keeping in mind your Absorption band width, are there any overlapping peaks in the absorption spectrum with your such as your substrate, co-factors, etc? Any overlap in those spectra must either be subtracted from your overall Absorption value being used in collecting your data Or choosing a wavelength in which there are no overlaps.

The next critical issue to be determined is whether the reaction is a first order or at least a pseudo-first order reaction. This is easily determined by looking at a plot of ln(Abs) versus time. The slope of the line is -k, the rate constant = t1/2 and the y-intercept is ln[Ao]. If it is a first order or pseudo-first order, this plot will be linear. However, if it is a second order reaction, you will see that the data does not fit a straight line, hence, multiple rate constants. I've observed this as a slow onset rate at long times indicative of a low affinity binding site. Remember, also, that you can determine your rate constants from your plot of [A] vs time. The initial slope in your reaction is your rate constant. Depending upon your ability to ascertain initial times...msec, sec, min, etc. will determine if you are able to catch the first order accurately. This is the reason behind the ln[A] vs time plot because it allows you to do interpolations back to zero time.

The question posed here has to do with the Km and Vmax of a reaction catalyzed by an enzymatic reaction. Knowing the molecular mass of the enzyme is irrelevant to the discussion since the Km is simply the x intercept -(1/Km) of the line and Vmax the y-intercept (1/Vmax) as a function of increasing substrate concentration with constant enzyme concentration (as well as reagents such as salts (Mg+2 in particular), temperature, etc.) in a standard Lineweaver-Burke double reciprocal plot (1/velocity = product/unit time) and 1/[S].

As to the issue of M versus mM or uM for the absorption coefficient, this is simply divisions by factors of 1000. As long as you maintain your units throughout your calculations, you will come up with the correct results for Km and Vmax. To get back to the 'normally reported values of Km and Vmax and Molar Absorptivity, etc, you simply need to multiply by the appropriate value of 10e3.

The Extinction coefficient is not a variable, but it is a non-variable constant for a given substance. According to your concentrations of Substrate in your experiments you can use either the molar extinction coefficient, or the milliomolar extinction coefficient or even the micromolar coefficient fr easier calculations. For NADH for example, the molar extinction coefficient is 6300 L/Mol and cm, the millimolar extinction coefficient is 6,3 L/mmol and cm und the micromolar extinction is 0,0063 L/micromol and cm at a wavelength of 340 nm.

Sighart Golf

In addition to my answer concerning extinction coefficient: a millimolar extinction coefficient of 6,3 L/mmol and cm: This means, that 1 mmol of NADH, dissolved in 1 Liter of water gives an Extinction of 6,3 at 340 nm and 1 cm of cuvette-thickness. This means also for the complete reduction of for example 1 mmol pyruvate to lactate with lactate dehydrogenase and sufficient NADH, you will have an extinction decrease of 6,3 Units at 340 nm and 1cm cuvette-thickness.

Sighart Golf

You have not been specific in your question. Apparently, what you suggest might work or might not work.

In general, you do not need to do anything, if you can handle the units properly. However, I prefer to use always the SI units in any experiment and any equation; in such case you do not need to pay attention to the units only the numbers ((:-)).

simply answer the question: YES

C = A / ε l

l is in cm ; A is unitless ; ε is usually in M-1cm-1

if you want C in µM you can divide ε by 1 000 000 to have it in µM-1cm-1

By this time the answer is clear. Simply express the extinction coefficient according to your suitability and need for easy calculation and expression.