It is not usual to refer to the concentration of substrate in an enzyme reaction as a dose, so it makes me wonder what you mean. Is the substrate being used as a drug or treatment? Will you be using it in cell culture or in vivo?

The substrate concentration you choose for an enzyme reaction in vitro depends on your purpose.

If you are trying to set up an assay to screen for competitive inhibitors of the enzyme, then it is desirable to keep the substrate concentration as low as possible in order to make the assay as sensitive as possible to competitive inhibitors. However, this approach has a cost: As you reduce the substrate concentrate, the amount of signal you can get before the reaction slows down decreases. It is often necessary to balance these 2 necessities, and a substrate concentration equal to the Km to 2 x Km is sometimes used (0.8 - 1.6 mM).

If you are not concerned about the assay being sensitive to competitive inhibitors but just want it to go as fast as possible using the smallest amount of enzyme, then you should use a saturating concentration of substrate. In practice, this might be 10 times higher than the Km, or 8 mM.

Dear Shiva,

The mechanism of enzyme catalyzed reactions is often studied by making kinetic measurements on enzyme-substrate reaction systems. These studies include measuring rates of the enzyme-catalyzed reactions at different substrate and enzyme concentrations. Here if we were to plot the rate of catalysis, V (also known as the reaction velocity), vs. the substrate concentration, [S] (at a fixed enzyme concentration) we would see a plot in which we see that V varies linearly with [S] for small [S]. As [S] increases, V “plateaus” indicating that V becomes independent of [S] at large values of [S]. The simplest model which accounts for this behavior is:

E+S↔ES→E+P 

where E is the enzyme, S the substrate, ES the enzyme-substrate complex, P the product of the enzyme-catalyzed reaction, k1 the rate constant of the forward reaction of E+S, k-1 the rate of the reverse reaction where the enzyme-substrate complex, ES, falls apart to E+S and k2 the rate constant of the forward reaction of ES forming E+P. In this model, it is assumed that none of the product reacts with the enzyme to form the enzyme-substrate complex, ES (this is true during the initial stages of the reaction when [P] is low, but towards the end of the reaction when [P] is high this may no longer be true).

Equation (1) can be re-arranged as:

1/V=1/Vmax+KM/Vmax1/[S]    (1)

If we were to plot 1/V vs. 1/[S] we would obtain a straight line with a y-intercept = 1/Vmax and a slope =KM/Vmax . This plot is called a Lineweaver-Burke plot.

Significance of KM

From equation 1, when [S] = KM, then V=Vmax/2. Hence KM is equal to the substrate concentration at which the reaction rate is half its maximum value. In other words, if an enzyme has a small value of KM, it achieves its maximum catalytic efficiency at low substrate concentrations. Hence, the smaller the value of KM, the more efficient is the catalyst. The value of KM for an enzyme depends on the particular substrate . It also depends on the pH of the solution and the temperature at which the reaction is carried out. For most enzymes KM lies between 10-1 and 10-7 M.

Based on the text depicted above for obtaining the dose (the appropriate substrate quantity), you should use different substrate concentrations and measure the corresponding velocity at a constant temperature, pH and enzyme concentration and then to plot 1/V vs. 1/[S]. From the linear plot and knowing the enzyme KM you can calculate the Vmax. The last step is to check in the first substrate concentration which gives the Vmax. This concentration is actually the substrate dose that you are looking for.

Hoping this will be helpful,

Rafik

It is not usual to refer to the concentration of substrate in an enzyme reaction as a dose, so it makes me wonder what you mean. Is the substrate being used as a drug or treatment? Will you be using it in cell culture or in vivo?

The substrate concentration you choose for an enzyme reaction in vitro depends on your purpose.

If you are trying to set up an assay to screen for competitive inhibitors of the enzyme, then it is desirable to keep the substrate concentration as low as possible in order to make the assay as sensitive as possible to competitive inhibitors. However, this approach has a cost: As you reduce the substrate concentrate, the amount of signal you can get before the reaction slows down decreases. It is often necessary to balance these 2 necessities, and a substrate concentration equal to the Km to 2 x Km is sometimes used (0.8 - 1.6 mM).

If you are not concerned about the assay being sensitive to competitive inhibitors but just want it to go as fast as possible using the smallest amount of enzyme, then you should use a saturating concentration of substrate. In practice, this might be 10 times higher than the Km, or 8 mM.

Dear Shiva,

All was said by the two previous contributors. I also, like Adam, wonder what you mean by "dose", your question is not very clear.

I'd like to add that when you use a 10xKm concentration of substrate to measure the reaction rate at a "saturating concentration" you do so because the Vmax (the maximal rate of reaction catalyzed by that enzyme) is theoretical since it could be observed at an "infinite" concentratios of susbtrate, a concentration that cannot be reached. A saturating concentration of 10 times the Km is used to get what is called the specific activity. Given the Michaelis-Menten equation:

v obs = (Vmax x [S]) / (Km + [S])

If the cocnentration in substrate [S] is equal to 10 x Km, then the previous equation simplfies to:

v obs = (Vmax x Km x 10) /  (Km + 10 x Km) = Vmax x (10/11) = 0.91 x Vmax.

Therefore, at a saturating concentration of substrate, the rate of reaction is roughly 90% of the Vmax, if your enzyme is Michaelian.

Best regards,

Philippe

Dear Shiva,

what has been said before is perfectly right. I like to draw your attention, however, to the fact that some enzymes do not approach Vmax with increasing substrate concentration. Theses enzyme are substrate inhibited an show - following an increase at low substrate concentrations - a clear declin in V at elevated substrate concentrations (see attached slide for appearance and the maths behind this behavior). In such cases it is contra productive to increase substrate concentration in order to achieve maximum turnover rates. Rather, it is neccessary to find optimum conditions (the possible "Vmax") empirically.

Best wishes

Thorsten

Thank you all for your answers.