Use this data to answer the following questions IS] (mMO 802 (mM) 0.5 Initial ve

Question

Use this data to answer the following questions IS] (mMO 802 (mM) 0.5 Initial velocity v (mM min) 1.43 1.85 1.25 1.56 1.96 2.86 0.5 2.27 3.23 6.67 1.25 5 a) What type of inhibition is mediated by [], as determined by evaluation of a double- reciprocal plot (Lineweaver-Burke)? How do you know? b) Conceptually, how is the inhibitor interacting and preventing catalysis? (Does it compete with substrate, deplete functional enzyme, other ways...?) How does it effect the KM and Vmax? c) What is the Ki (or Ki') for this [Il? What type of plot do you use to find this?

Explanation / Answer

since V= VmaxS/(KM+S)

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

so a plot of 1/V vs 1/S gives straight line whose slope is KM/Vmax and intercept is 1/Vmax. This is also known as double reciprocal plot.

for the case of inhibition, KM becomes KMapp and Vmax becomes Vmaxapp

the data points of 1/V vs 1/S along with the plots for all the threee cases are generated and shown below

from the plots the slope KM/Vmax is identical for all the three plots. Hence the inhibition is uncompetitive.

from the plot for no inhibition, 1/Vmax= intercept = 0.1038, Vmax= 1/0.1038= 9.63 and slope KM/Vmax=0.247

KM= 9.63*0.2462=2.371 mM

for the cae of uncompetitive inhibition, the substrate binds to enzyme before binding to inhibitor.

hence 1/V= (KM/Vmax)1/S + (1+I/KI)/Vmax

for the case of 0.5mM inhibition. (1+I/KI)/Vmax= intercept =0.2038

1+I/KI= 0.2038*9.63= 1.96

I/Ki=0.96

KI= I/0.96= 0.5/0.96 mM=052

for the case of 1mM inhibition

1+I/KI= 0.1038*9.63=1.000632

KI= 1/0.000632= 1582.3 mM

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