Question 0e156

Apr 10, 2015

The initial velocity for the reaction will be ${\text{350 nmol s}}^{- 1}$.

Start from the Michaelis-Menten equation

${V}_{0} = \left({V}_{\text{max" * [S])/(V_"max}} + \left[S\right]\right)$

When an uncompetitive inhibitor is introduced, the equation takes this form

V_0 = (V_"max" * [S])/(K_m + [S] * underbrace((1 + ([I])/K_I))_(alpha)#

But $\left(1 + \frac{\left[I\right]}{K} _ I\right)$ is actually equal to $\alpha$, the degree of inhibition, which implies that the equation becomes

${V}_{0} = \frac{{V}_{\text{max}} \cdot \left[S\right]}{{K}_{m} + \left[S\right] \cdot \alpha}$

Now plug your values and solve for ${V}_{0}$

${V}_{0} = \left({\text{950 nmol s"^(-1) * 350cancel(mu"mol L"^(-1)))/((175+ 350 * 2.20)cancel(mu"mol L}}^{- 1}\right)$

${V}_{0} = \frac{350}{945} \cdot {\text{950 nmol s"^(-1) = "351.85 nmol s}}^{- 1}$

Rounded to two sig figs, the number of sig figs given for ${V}_{\text{max}}$ and $\left[S\right]$, the answer will be

${V}_{0} = \textcolor{g r e e n}{{\text{350 nmol s}}^{- 1}}$