Question

The following data was obtained in an enzyme-substrate reaction involving enzymatic oxidation of AICAR by ATPase at different substrate concentrations. S (microg/) 20 30 40 60 7090 100 120 140 150 160 v (microg/l) | 5 | 75 | 10 | 12.5 | 13.7 | 14 | 14 | 12.5 | 9.5 | 7.5 | 5.7 a. What type of inhibition is this? b. Determine Vm, Km and Ks. Determine the oxidation rate at [S] = 50 microg/l.

Answer 1

when plotting v vs S graph from given data it suggests that this is an example of substrate inhibition.

now, the expression for substrate inhibition reaction is represented as

v=Vmax[s]/(Km + [S](1+[S])/K_si))

rearranging this equation we will get

1/v= (K_m/Vm)(1/[s]) + 1/V_m + [s]/V_mK_si--------(1)

It should be noted that [s] appears in the denominator of the equation as a substrate and in the numerator as an inhibitor. The result is that an upward curvature is observed in double-reciprocal plots at higher concentrations of substrate.

1/v= (K_m/V_m)(1/[s]) + 1/V_m -------(2)

Over the high range of substrate concentrations, the equation 1 reduces to

1/v= 1/V_m + [s]/V_mK_si ----------(3)

use S value up to 90 to plot a graph between 1/v and 1/[s] to calculate K_m and V_m

when you will plot a graph as i said you will find an equation as

y = 3.3846x + 0.0248 compare this equation with equation 2 you will get

Vm = 40.322 and Km= 136.474

Now use S value from 100 to rest to plot a graph between 1/v and s to find out K_si

plotting a graph will give you equation as y = 0.0016x - 0.1051 when comparing this equation with equation 3 you will get

Vm for inhibition = -9.153 (negative sign indicates the inhibition state)

and K_si= 68.284

Now for oxidation rate for [s] = 50ug/ L, as this value does not fall in inhibition zone so we can use equation 2 to find out v for this substrate value, which will come out as = 10.811ug/L