§ More substitution gives the __, which is attained when the __—that is, when [ES]=[E]_{T}. Thus: V_{max}= k_{2}[E]_{T}. More substitution into another equation gives __
· At very low substrate concentration, when [S] is much less than __, V_{0}= __; that is, the reaction is __ with the rate directly proportional to the __. At very high substrate concentration, when [S] is much greater than __, V_{0}=__; that is, the rate is __. The reaction is __, independent of substrate concentration.
- maxima rate, Vmax
- catalytic sites on the enzyme are saturated with substrate
- V0=Vmax ([S]/([S]+KM)
- KM
- (Vmax/KM)[S]
- first order
- substrate concentration
- KM
- Vmax
- maximal
- zero order