Please Show All Work Please Show All Work The following kinetic data were obtain
ID: 592270 • Letter: P
Question
Please Show All Work Please Show All Work The following kinetic data were obtained for an enzyme in the absence of any inhibitor, and in the presence of two different inhibitors (Inhibitor 1 and 2) at 5 mM concentrations. The concentration of enzyme used in all experiments was 10 x 106 M. 1. Vo with no inhibitor o with Inhibitor 1 Vo with Inhibitor 2 umol mL-I sec-1 4.3 IS (mM) (umol mL1 sec sec) 5.5 12 20 29 35 40 14 21 26 13 16 18 4 12 a) Using a Lineweaver-Burk plot, determine Km and Vmax for the uninhibited enzyme. b) Determine the type of inhibition for each inhibitor and explain your choice. c) Determine the Ki for each inhibitor.Explanation / Answer
Lineweaver-burk plot is V= VmaxS/(KM+S)
1/V= (KM+S)/VmaxS
1/V= (KM/Vmax)*1/S + 1/Vmax
in the presence of inhibitor, Vmax becomes Vmaxapp and KM becomes Kmapp
so a plot of 1/V vs 1/S gives straight line whose slope is 1/Vmax and intercept is KM/Vmax. the data points along with the plot are shown below.
from the plots, for no-inhibition, 1/Vmax= 0.019, Vmax=1/0.019= 52.63 umol/ml.sec
slope is =KM/Vmax= 0.063, KM=0.063*52.63=3.32 mM
for the case of inhibitor-1, 1/Vmaxapp=0.019, Vmaxapp= 1/0.019= 52.63 umol/ml.Sec same as the case of no inhibitor. This is the case of competitive inhibition where the inhibitor competes with the substrate in binding to the enzyme. KMapp= KM*(1+I/KI)
I= concentration of enzyme = 5mM,
from the slope of curve for inhinbitor-1, KMapp/Vmax=0.213, KMapp= 0.213*52.63=11.21
KMapp= KM*(1+I/KI), 11.21= 3.32*(1+5/KI)
1+5/KI= 11.21/3.32= 3.4, 5/KI=2.4, KI= 5/2.4 =2.1mM
2. for the case of 2- inhibitor, !/Vmaxapp= 0.043, Vmax= 1/0.043= 23.3 umol/L.sec
KMapp/Vmaxapp= 0.137, Kmapp= 23.3*0.137 mM=3.2 mM
Kmapp = Km for no-inhibition, this is the case of non-competitive inhibition where in the inhibitor reacts with the enzyme-substrate complex. for this case
Vmaxapp =Vmax/(1+I/KI)
1+I/KI= 52.63/23.3=2.6
i/KI= 1.6
KI= I/1.6= 5/1.6 mM=3.125 mM