Question

Score: 3318/3/00 15 of 37 > An enzyme catalyzes a reaction with a km of 8.50 mM and a Vmax of 2.45 mM.s. Calculate the reaction velocity, to, for each substrate concentration. S = 3.75 mm mMs-1 S = 8.50 mM mM.s-1 [S] = 10.5 mM about us Careers Privacy policy terms of use contact us help

Answer 1

Rate of enzymes catalysis reaction will be given as follows,

V_{​​​​​​o} = [S]V_{​​​​​​max} /([S] + K_{​​​​​​m} )

given Value,

V_{​​​​​​max} = 2.45mM.s^-1

K_{​​​​​​m} = 8.50mM

Part (a)

[S] = 3.75mM

Putting the all value in formula

V_{​​​​​​o} = (3.75mM)(2.45mMs^-1)/(3.75mM + 8.50mM)

V_{​​​​​​o} = 9.1875mMs^-1/(12.25)

V_{​​​​​​o} = 0.75mMs^-1

part (2) similarly

[S] = 8.50mM

Putting the all value in formula

V_{​​​​​o} = (8.50mM)(2.45mMs^-1)/(8.50mM+ 8.50mM)

V_{​​​​​​o} = 1.225mM

Part(3) Similarly

[S] = 10.50mM

Putting the all value in formula

V_{​​​​​​o} = (10.50mM)(2.45mMs^-1)/(10.50mM +8.50mM)

V_{​​​​​​o} = 1.35mMs^-1