00:01
So in this part, this problem is part of the concept where your enzymatic reaction has derived from the formation of the predominant form of the enzyme and the substrate saturation, wherein your k1 is greater than your k2.
00:19
So we can write the formation of your es, that is, d times es over dt, which is equal to 2.
00:30
Kd e times s so because your es is intermediate so when we break down of your es this would become equal to k2 es plus k1 es by steady state assumption rate of formation is equal to rate of breakdown your kd e .s will be equal to k2, es plus k1, es for this first equation.
01:09
According to enzyme conservation, for the total of your e, that would be e, t is equal to e plus es, and e is equal to e t minus es.
01:26
Then from your equation one and equation two we have kd we have et t minus es so we have equal to k2 es plus k1 es so we have kd e s plus k1 es so we have kd et s is equals to kd e s plus k d e s plus k d s k2, es plus k1, es...