# Question 43225

Apr 10, 2015

The initial velocity for your enzyme will be ${\text{0.417 nmol L"^(-1)"s}}^{- 1}$.

Once again, use the MIchaelis-Menten equation to solve for ${V}_{0}$. Keep in mind, however, that you must convert $\mu {\text{mol L"^(-1)"s}}^{- 1}$, the units used for ${V}_{\text{max}}$, to ${\text{nmol L"^(-1)"s}}^{- 1}$, the units required for ${V}_{0}$.

500cancel(mu"mol")"L"^(-1)"s"^(-1) * (10^(-3)"nmol")/(1cancel(mu"mol")) = 500 * 10^(-3)"nmol L"^(-1)"s"^(-1)#

So, the initial velocity will be equal to

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

${V}_{0} = \left(500 \cdot {10}^{- 3} {\text{nmol L"^(-1)"s"^(-1) * 5.00cancel("mmol L"^(-1)))/((1.00 + 5.00)cancel("mmol L}}^{- 1}\right)$

${V}_{0} = \frac{5}{6} \cdot 500 \cdot {10}^{- 3} {\text{nmol L"^(-1)"s"^(-1) = "0.417 nmol L"^(-1)"s}}^{- 1}$

SIDE NOTE Since ${V}_{\text{max}}$ is given with one sig fig, while all other variables are given with three, I'll leave the answer rounded to three sig figs.

If, instead of 500 you'd have 500. for ${V}_{\text{max}}$, then the answer should indeed have three sig figs; if not, it should only have one.