Respuesta :
Explanation:
According to the given data, we will calculate the following.
Half life of lipase [tex]t_{1/2}[/tex] = 8 min x 60 s/min
= 480 s
Rate constant for first order reaction is as follows.
[tex]k_{d} = \frac{0.6932}{480}[/tex]
= [tex]1.44 \times 10^{-3}s^{-1} [/tex]
Initial fat concentration [tex]S_{o}[/tex] = 45 [tex]mol/m^{3}[/tex]
= 45 mmol/L
Rate of hydrolysis [tex]V_m_{o}[/tex] = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration (S) = [tex]S_{o} (1 - X)[/tex]
= 45 (1 - 0.80)
= 9 [tex]mol/m^{3}[/tex]
or, = 9 mmol/L
It is given that [tex]K_{m}[/tex] = 5mmol/L
Therefore, time taken will be calculated as follows.
t = [tex]-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)][/tex]
Now, putting the given values into the above formula as follows.
t = [tex]-\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)][/tex]
= [tex]-\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s }{K_{M} ln (\frac{45 mmol/L }{9 mmol/L }) + (45 mmol/L - 9 mmol/L )][/tex]
= [tex]1642.83 s \times \frac{1 min}{60 sec}[/tex]
= 27.38 min
Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.
