[시1발버러지새끼들아]

Hi guys

I was on learning the enzyme inhibitor.

But I cant understand,
"an uncompetitive inhibitor lowers KM"

The KM is [E][ S]/[ES] <--this is an equilibrium.
But If uncompetitve inhibitor decline density of ES
I think KM must be higher
what happened???

Help me plzzzzzz

[Babcock_Hall]

There are some nontrivial points that need to be addressed IMO.  One, K_M is not a thermodynamic dissociation constant; it can be greater than, equal to, or less than a dissociation constant in magnitude.  Two, it is more fruitful to think in terms of what effect an inhibitor has on V_max and V_max/K_M than it is helpful to ask what effect an inhibitor has on K_M.  (Leaving out subscripts) competitive inhibitors lower V/K but not V, whereas uncompetitive inhibitors lower V but not V/K.  The apparent effects of the inhibitors on K are just by-products of these two effect.  I may have more time this weekend to discuss further.

[Yggdrasil]

The inhibitor does not actually change the intrinsic K_M of the enzyme, but rather the inhibitor changes the shape of the enzyme activity vs substrate concentration graph in a way that makes it looks as if the K_M of the enzyme has decreased (i.e. the concentration for half-maximal activity is lower in the presence of the inhibitor).  Therefore, it is more accurate to say that the inhibitor lowers the apparent K_M of the enzyme.