# Question c8e8d

Apr 24, 2017

See detailed workout below

#### Explanation:

First things first, we have to write out what is given to us.

Given
$\left[E\right] = 4 n M$

${K}_{m} = 2.8 \mu M$

${V}_{m} = \frac{60 \mu M}{\min}$

Next, we need to use two formulas to figure this problem out

$\textcolor{w h i t e}{a a a a a a a a a a a}$color(red)((k_(cat))/K_m color(white)(aaaaaaaa) k_(cat) = V_m/[[E]]

We are going to figure out our ${k}_{c a t}$ or the turnover number of the enzyme, but first, our answer is supposed to be represented in $\mu M$ so we will first convert $4 n M \to \mu M$.

$\frac{4 \cancel{n M}}{1} \cdot \frac{1 \cdot {10}^{-} 9 M}{1 \cancel{n M}} = \frac{4 \cdot {10}^{-} 9 \cancel{M}}{1} \cdot \frac{1 \mu M}{1 \cdot {10}^{-} 6 \cancel{M}} = \downarrow$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a} \left(4 \cdot {10}^{-} 3 \mu M\right)$

Plugin

$\textcolor{red}{{k}_{c a t} = {V}_{m} / \left[\left[E\right]\right]} \to \frac{\frac{60 \cancel{\mu M}}{\min}}{\frac{4 \cdot {10}^{-} 3 \cancel{\mu M}}{1}} = \frac{15000}{\min}$

Now that we have our ${k}_{c a t}$, we will use it to find out the $\text{catalytic efficiency}$, which basically tells us how good the enzyme is at catalyzing the reaction at low substrate concentration.

$\textcolor{red}{\frac{{k}_{c a t}}{K} _ m} = \frac{\frac{15000}{\min}}{\frac{2.8 \mu M}{1}} = \frac{5357.14}{\min \cdot \mu M}$

Take this answer and convert $\min \to s$

(5357.14)/(cancelmin*muM) * (1 cancelmin)/(60 s) = color(blue)((89)/(s*muM)#

$A n s w e r : \frac{89}{s \cdot \mu M}$