# If I = Ki = Ki‘, then what will happen to Vmax and Km when inhibitor acts non-competitively?

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If I = Ki = Ki‘, then what will happen to Vmax and Km when inhibitor acts non-competitively?

(a) Lowers to 0.5 Vmax and 0.5 Km

(b) Vmax is unchanged and Km increases 2Km

(c) Lowers to 0.5 Vmax and Km remains unchanged

(d) Lowers to 0.67 Vmax and Km increases to 2Km

I got this question in semester exam.

The question is from Fundamentals in division Enzyme Kinetics Fundamentals of Enzyme Technology

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Right option is (c) Lowers to 0.5 Vmax and Km remains unchanged

Best explanation: Non-competitive inhibitors are those inhibitors which bind at site away from the binding site, causing reduction in catalytic activity. It is a rare case mixed inhibition and the rate equation is given by, $V=\frac{\left (\frac{V_{max}}{1+\frac{[I]}{K_i}}\right ) [S]}{K_m+[S]}$. When If I = Ki = Ki‘, Vmax is lowered to 0.5 Vmax and Km remains unchanged.

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