A plot (1/v vs. 1/[S]) for a type of inhibition is shown below. What is the sign

Question

A plot (1/v vs. 1/[S]) for a type of inhibition is shown below. What is the significance of the x- and y- intercepts and the slope for this plot of enzyme activity? What type of inhibition does it indicate? Steady state kinetics of MF_1 in the presence of quin-acrine with varying [ATP] and [Mg^2+] fixed at 3.0 mM. The enzyme assays were carried out at pH 7.0 an described under "Experimental Procedures." The concentration of MgSO_4 was 3.0 mM and the ATP concentrations were as indicated. The concentrations of quinacrine (QA) were: none (), 0.25 mM (), 0.50 mM (), and 0.75 mM ().

Explanation / Answer

The enzyme kinetics is represented by

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

as the inhibition concentration has increased, the intercept 1/Vmax has increased and there by Vmax decreases. Since all the lines are paralle to each other ( and hence have constant slope, KM also decreases). This is the case ofuncompetitive inhibitor where the binds to the enzyme and enhances the binding of substraet as a result the formed enzyme-inhibitor-substrate complex undergoes reaction to form the product slowly.

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