To study the dependence of the rate of an enzyme-catalyzed reaction on the subst

Question

To study the dependence of the rate of an enzyme-catalyzed reaction on the substrate concentrate>n, a constant amount of enzyme it added to a sehes of reaction mixtures containing different concentrations of substrate (usually expressed in mol/L). The initial reaction ratos are determined by measuring the number of moles (or mumoles) of substrate consumed (or product produced) per minute. Using the experimental procedure, the data in the below Table was obtained for an enzyme in 10-ml reaction mixtures. Initial rates at venous substrate concentrations tor a hypothetical enzyme-catalyzed reaction. There are 5 questions to this problem. To answer roach question click on the page. Once you submit, you can come back to this Page to answer the next question. Once you finish answering all the questions you can dick on next question to go to the next question in the homework The 5 questions are:

Explanation / Answer

I will answer your first 3 questions per now, cause I have a doubt in the last two.

In order to get Vmax and Km we need to plot this data, but in order to get a linear equation, we just plot the inverse of the equation

The general equation is V = Vmax [S]/[S]+Km
1/V = [S]+Km/[S]Vmáx
1/V = 1/Vmáx + Km/Vmáx[S]
1/V = 1/Vmáx + Km/Vmáx (1/[S])

So we plot 1/V vs 1/S ans from the slope and y intercept we can obtain vmáx and Km. The data obtained is the following:

y = 3.989044 + 0.050093x
r2 = 0.99999

From y intercept we can get vmáx:

3.989044 = 1/Vmáx
Vmáx = 1/3.989044 = 0.25069 umol/min

From the slope we can get km:
0.050093 = Km/0.25069
Km = 0.25069 * 5x10-5 = 0.01256 L/umol

The innitial rate at [S] = 1x10-6 M would be
1/v = 1/0.25069 + 0.01256/0.25069*1x10-3 = 54.0907
v = 0.01848 umol/min

Hope this helps

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