Question

if a saturated enzyme is capable of producing 200 molecules of product each minute and there is 10 mmol of enzyme, what is the maximum rate of this enzyme?
20 00 mol/min
50 00 mol/min
2 mol/min
5 mol/min

Explanation:

Step1: Define given values

Turnover number $k_{cat}=200\ \text{molecules/min per enzyme}$, enzyme amount $[E]_t=10\ \text{mmol}=10\times10^{-3}\ \text{mol}$

Step2: Recall maximum rate formula

Maximum rate $V_{max}=k_{cat}\times[E]_t$ (1 enzyme molecule makes 200 product molecules, so total rate is turnover number times total enzyme moles)

Step3: Calculate total enzyme molecules

1 mole = $6.022\times10^{23}$ molecules, so $[E]_t$ in molecules: $10\times10^{-3}\times6.022\times10^{23}=6.022\times10^{21}$ molecules

Step4: Calculate product molecules per minute

Total product molecules = $200\times6.022\times10^{21}=1.2044\times10^{24}$ molecules/min

Step5: Convert to moles per minute

$\frac{1.2044\times10^{24}}{6.022\times10^{23}}=2\ \text{mol/min}$

Answer:

2 mol/min