Question

Question #0e124

Answer 1

The key to solving this problem is to realize that an uncompetitive inhibitor does not change the slope of the double-reciprocal plot that describes the enzyme-catalyzed reaction when no inhibitor is present.

![https://biochemanics.wordpress.com/2013/04/07/reversible-inhibition/]()

This is important because, for an uncompetitive inhibition, both `V_"max"` and `K_m` decrease by the same factor, which is why the slope remains unchanged.

More specifically, both `V_"max"` and `K_m"` decrease by

`1 + ([I])/K_I^(')`, where

`[I]` - the concentration of the inhibitor;
`K_i^(')` - the inhibitory constant.

However, this is actually equal to `alpha^(')`, the degree of inhibition.

As a result, `V_"max"^(')` will be equal to

`V_"max"^(') = V_"max"/(1 + ([I])/K_I^(')) = V_"max"/(alpha^('))`

Therefore,

`V_"max" = alpha^(') * V_"max"^(') = 3.00 * "9.00 nmol s"^(-1) = "27.0 nmol s"^(-1)`

SIDE NOTE Check out this great site on enzyme inhibition, it allows you to play with Lineweaver-Buk plots for various inhibitors

http://higheredbcs.wiley.com/legacy/college/voet/0470129301/guided_exp/guided_exploration_11/michaelis_menten.html