Answers: 2
![v=V_{max}\times \frac{[S]}{(K_m+[S])}=k_{cat}[E_o]\times \frac{[S]}{(K_m+[S])}](/tpl/images/0480/3100/d521f.png)
![V_{max}=k_{cat}[E_o]](/tpl/images/0480/3100/617d3.png)
![[K_m]](/tpl/images/0480/3100/99bb2.png)
= Michaelis constant 
= Maximum rate achieved
= Catalytic rate of the system
= initial concentration of enzyme
![v=V_{max}\times \frac{[S]}{(K_m+[S])}](/tpl/images/0480/3100/6fc0b.png)
![\frac{V_{max}}{4}=V_{max}\times \frac{[S]}{(0.0050 M+[S])}](/tpl/images/0480/3100/40f70.png)
![[S]=\frac{0.005 M}{3}=1.7\times 10^{-3} M](/tpl/images/0480/3100/96501.png)
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At what substrate concentration would an enzyme with a k_cat of 30.0 s⁻¹ and a km of 0.0050 m operat...
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