A single chain, single active site enzyme has been purified to homogeneity. The

Question

A single chain, single active site enzyme has been purified to homogeneity. The molecular weight of this enzyme is 40 kilodaltons. It converts compound A to compound B. To examine the kinetics of this reaction, the data show in the table below was obtained in a reaction mixture of 1 mL that contained 0.1 mug of enzyme. What are the K_M and V_max values for this reaction? Determine these values from a Lineweaver-Burk plot. Has a good range of substrate concentrations been chosen for this substrate to obtain accurate values of K_m and V_max? Why or why not? What is the turnover number for this enzyme? Note that by convention, turnover numbers are expressed in units of sec^-1 (or s^-1). What is the catalytic efficiency of the enzyme for this substrate? The kinetics of the same enzyme described in Problem 1 were also determined in the presence of an inhibitor. Shown below are the values obtained for a reaction of one mL containing 1 times 10^-6 moles of inhibitor. What type of inhibitor is this? Be sure to show the plot! What is the K for this inhibitor?

Explanation / Answer

Lineweaver- berk plot is 1/V = (KM/Vmax)*1/S + 1/Vmax. the plot of 1/V vs 1/S is shown drawn.

from the plot 1/Vmac= intercept = 2*109, Vmax= 1/(2*109) = 5*10-10 M/min

Km/Vmax = 5*106, KM= 5*106*5*10-10 = 0.0025M. The substrate concentration choosen is adequate

since at V= Vmax/2,

S= KM 0.0025 the concentration choosen falls in this range

comcentration of enzyme, ET = 0.1*10-6/(40*1000) /(1/1000)=2.5*10-9 M

Kcat= Vmax/ET = 5*10-10/2.5*10-9 =0.2/min

Kcat/KM= catalytic efficiency = 0.2/0.0025=80%

for the inhition, data points and the plot is shown below

Vmax= 5*10-10 M/min since Vmax is same as in the absence of inhibitor, it is competitive inhibtion.

KMapp/Vmax= 1*107 , KMapp = KM*(1+I/KI)

KM*(1+I/Ki)= 1*107*5*10-10= 0.005

1+I/Ki= 0.005/ 0.0025= 2

i/KI= 1

Ki= I = 1*10-6/(1/1000) Moles/L = 10-3 M

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