1.5 ‚âà√≠¬¨‚à´g of enzyme gives a Vmax of 3 ‚âà√≠¬¨‚à´mol product produced per minute. What is the turnover number for this enzyme?$

1.5 ‚âà√≠¬¨‚à´g of enzyme gives a Vmax of 3 ‚âà√�

1.	1.5 ‚âà√≠¬¨‚à´g of enzyme gives a Vmax of 3 ‚âà√≠¬¨‚à´mol product produced per minute. What is the turnover number for this enzyme?$
A.	100 sec<sup>-1</sup>
B.	200 sec<sup>-1</sup>
C.	1000 sec<sup>-1</sup>
D.	2000 sec<sup>-1</sup>
Answer» D. 2000 sec<sup>-1</sup>