## 15.81

$V_{0} = \frac{V_{max}[S]}{K_{M} + [S]}; K_{M} = \frac{[E][S]}{[ES]}$

Ardo 2K
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### 15.81

Can someone please show me how to approach problem 15.81 from the book. How are we suppose to account for the loss of the unreacted substrate from the active site?

ZoeHahn1J
Posts: 63
Joined: Sat Jul 22, 2017 3:01 am
Been upvoted: 1 time

### Re: 15.81

We are given that:
K = 326
the attachment/loss of substrate to or from enzyme is 2nd order
k of attachment = 7.4 x 10^7 L/(mol*s)
and instructed to find the rate constant of the loss of substrate from the enzyme. Recall that K = k/k', if k is the rate constant of the forwards rxn (in this case, the attachment) and k' is the rate constant of the backwards rxn (in this case, the disattachment of substrate from the enzyme). Note: I am not sure if K = k/k' is on the constants/equations sheet, so you may have to remember/derive? this for the final if we get a question like this one. Using this relationship, it should be straightforward to find k'.

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