To study the dependence of rate of an enzyme – catalyzed reaction on the substra
ID: 101727 • Letter: T
Question
To study the dependence of rate of an enzyme – catalyzed reaction on the substrate concentration, a constant amount of enzyme is added to a series of reaction mixtures containing different concentrations of substrate. The initial reaction rates are determined by measuring the number of moles (or µmoles) of substrate consumed (or product produced) per minute. Using this experimental procedure, the data in Table 2.0 were obtained for an enzyme in 10 mL reaction mixtures. Use numerical and not graphical calculations and answering the following questions:
d. What are the initial reaction rates at [S] = 1.0 x 10-6 M and [S]= 1.0 x 10-1 M?
f. Suppose the enzyme concentration in each reaction mixture were increased by a factor of four. What would be the value of KM? of Vmax? What would be the velocity, v, at [S] = 5.0 x 10-6 M?
Explanation / Answer
Initial rate of an reaction – V0 = (Vmax x S) / (S +Km)
In the equation, v is the rate of reaction, Km is frequency of the production or Michaelis constant, and S is substrate concentration. The Vmax is the maximum rate at which half of the substrate covert into product. The Vmax = Kcat x E, wehre Kcat is catalytic constant and E is enzyme concentration.
= 1x 10-6 M initial rate V01 = Vmax x 10-6) / (10-6 +Km)
= 10-6 Vmax / (10-6 +Km) or Kcat x E x 10-6 / (10-6 +Km)
Initial Km for E enzyme concentration- Km = [E] [S] / [ES]
If we increase enzyme concentration by four times, the Km for an enzyme will remain constant because it is unique for particular enzyme. However rate of reaction is varying.
The value of new Vmax’ = Kcat x 4E = 4Vmax
The velocity of the reaction for 5.0 x 10-6 M substrate concentraiton= (Vmax’ x S) / (S +Km)
= 4Vmax x 5.0 x 10-6 / (5.0 x 10-6 +Km)
= 2 X 10-6 Vmax / (5.0 x 10-6 +Km)
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