1. Determine Km and Vmax for the uninhibited and inhibited enzymatic reaction fr

Question

1. Determine Km and Vmax for the uninhibited and inhibited enzymatic reaction from the linear regression equation produced from the Lineweaver burk plots.

2. is the inhibition competitive or mixed?

Substrate Concentration (mM) Inhibited (mM/Sec Uninhibited (mM/sec) 1/IS] Inhibited 1/Vo 1/[S] Uninhibited1/Vo 10 50 150 250 500 1000 0.12 0.44 0.9 1.21 0.3 0.93 1.28 1.48 1.54 1.53 0.1 0.02 0.00667 0.004 0.002 0.001 8.333 2.2727 0.02 0.00667 0.004 0.002 0.001 3.33 1.075 0.7812 0.675675 0.6493 0.6535 0.82644 0.7142 0.6711 1.49 Inhibited Uninhibited 3.5 y 77.631x +0.5919 R2 0.9995 y = 27.345x + 0.5849 R2 = 0.999 2.5 1.5 0.5 0.02 0.04 0.06 0.08 0.12 0 0.02 0.04 0.06 0.08 0.1 0.12 1/IS] 1/[S]

Explanation / Answer

Lineweaver-burke plot is given by

1/V= (KM+S)/VmaxS

1/V= (KM/Vmax)*1/S +1/Vmax

so a plot of 1/V vs 1/S gives a straight line whose intercept is 1/Vmax and slope is KM/Vmax

from the unhibited inhibiton equation

1/Vmax= 0.5849, Vmax=1/0.5849= 1.71mM/sec

Slope (KM/Vmax)= 27.345, Km= 27.345*1.71 mM =46.76 mM

for inhibited enzyme

1/Vmax= 0.5919, Vmax= 1/0.5919= 1.69, KM=1.69*77.631 =131.2 mM

since Vmax is almost same in both the types ( no inhibitor and in presence of inhibitor)

this is competitive inhibition.

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