Question

Enzyme reaction and deactivation Lipase is being investigated as an additive to laundry detergent for removal of stains from fabric. The general reaction is: fats- fatty acids 1 glycerol The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat hydrolysis is carried out in a well-mixed batch reactor that simulates a top-loading washing machine. The initial fat concentration is 45 gmol m23. At the beginning of the reaction, the rate of hydrolysis is 0.07 mmol l21 s21. How long does it take for the enzyme to hydrolyse 80% of the fat present?

Answer 1

Answer:

t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]

Explanation:given values

Half life of lipase t_1/2 = 8 min x 60s/min = 480 s

Rate constant for first order reaction

k_d = 0.6932/480 = 1.44 x 10^-3 s-1

Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L

rate of hydrolysis Vm0 = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3

K_m = 5mmol/L

time take is given by

$t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]$

all values are given and putting these value we get

t=1642.83 secs

which is equal to

t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]