Biochemistry 6th Edition by Reginald Garrett,Charles Grisham

Exercise 9

Quantitative relationships between rate constants to calculate K m , kinetic efficiency (k cat /K m ) and V max -I
Measurement of the rate constants for a simple enzymatic reaction obeying Michaelis-Menten kinetics gave the following results:
k 1 = 2 x 10 8 M -1 sec -1
k -1 = 1 x 10 3 sec -1
k 2 = 5 x 10 3 sec -1
a. What is K s , the dissociation constant for the enzyme-substrate complex?
b. What is K m , the Michaelis constant for this enzyme?
c. What is k cat (the turnover number) for this enzyme?
d. What is the catalytic efficiency (k cat /K m ) for this enzyme?
e. Does this enzyme approach "kinetic perfection"? (That is, does k cat /K m approach the diffusion-controlled rate of enzyme association with substrate?)
f. If a kinetic measurement was made using 2 nanomoles of enzyme per mL and saturating amounts of substrate, what would V max equal?
g. Again, using 2 nanomoles of enzyme per mL of reaction mixture, what concentration of substrate would give v = 0.75 V max ?
h. If a kinetic measurement was made using 4 nanomoles of enzyme per mL and saturating amounts of substrate, what would Vmax equal? What would K m equal under these conditions?