0
$\begingroup$

In Uncompetitive inhibitors , the Km decrease . With uncompetitive inhibitors , the affinity of the enzyme for the substrate (E+S) increases .

I've searched up on the net and it says that the substrate has now been taken up to form E-S and E-S-I , therefore consuming more substrate than the original inhibitor-free equilibrium .

E + S -> ES (equilibrium position shifts to the right)

I've read up on all these and more doubts form in me .

Why is substrate needed to form ESI ?

I understood what it meant by equilibrium position will shift to the right . It is harder to form E + P as more substrate will form ESI.

Thanks !

Improve this question
$\endgroup$
1
  • $\begingroup$ u should add the link where u read that the substrate is required for the intermediate complex formation otherwise the question is unclear. $\endgroup$
    – user 33690
    Aug 14, 2017 at 15:49

1 Answer 1

Reset to default
1
$\begingroup$

An uncom-petitive inhibitor doesn't compete with the natural substrate of the enzyme for active site i.e. it binds at a site distinct from the substrate active site and binds to the $ES$ complex.

enter image description here

$K_I$ is the equilibrium constant for inhibitor binding to $E$ and $K_I^{'}$ is the equilibrium con- stant for inhibitor binding to $ES$. Uncompetetive inhibitor lowers the value of $V_{max}$ and $K_m$.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .