Question

An enzymatic experiment was conducted on the enzyme A with and without a potential inhibitor. The double reciprocal plot for the uninhibited enzyme yielded an equation of y = 0.778x + 0.4139. Whereas the inhibited plot yielded y = 0.3915x + 0.4625.

Is the inhibitor effective? Explain your answer.

Determine the Vmax of the uninhibited enzyme. Your answer must include units.

Determine the km of the inhibited enzyme. Your answer must include units.

Answer 1

1/Vo = (Km/Vmax)1/S + 1/Vmax

Comparing this equation with Y=aX + c

where a is the slope and c is the intercept

1/Vmax = intercept

so V max = 1/intercept

For Uninhibited State.

y = 0.778x + 0.4139.

1/Vmax = 0.4139

Vmax = 1/0.4139= 2.41

Km/Vmax is equal to slope

Km/2.41 = 0.778

Km =2.41*0.778

= 1.87

For inhibited state

0.3915x + 0.4625.

1/Vmax =0.4625

Vmax = 1/0.4625= 2.16

Km/Vmax is equal to slope

Km/2.16 = 0.3915

Km =2.16*0.3915

= 0.84

Inhiitor is effective as it decrease the Vmax and Km.

As the inhibitor is lowering the Km and Vmax inhibitor is an uncompetitive type of inhibitor.

For units use the Unit for Subsbate concentration for Km and Unit of Vo from Vmax.

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