Question

how does velocity of reaction varies in competitive inhibition and non competitive inhibition.
( please give explanation too)

Answer 1

Firsly, we need to know how competitive and non-competitive inhibitor works,then we can figure out how and why the velocity is different.


Please forgive my ugly drawing because it's so difficult to use right hand to control mouse for drawing and I'm a left-hander.

Competitive inhibitor is substrate analogue so it can bind to the enzyme's active site, that means both inhibitor and substrate can bind to active site and form 2 independent complex. However, only the enzyme-substrate complex can degrades into enzyme and product. The enzyme-inhibitor complex is like a 'dead end' because the inhibitor block the access of substrate to active site. The Vmax is same as uninhibited enzyme because at sufficiently high concentration of substrate, most enzyme is occupied by substrates. You can say, the more substrate you add, the less binding between enzyme and inhibitor, more binding between enzyme and substrate (increasing number of functional enzyme), more product you get.

Non-competitive inhibitor is not substrate analogue and it binds reversibly to some other part of enzyme other than the active site to change enzyme shape. Since it can bind at somewhere other than active site, it can bind independently of substrate i.e. it can bind both enzyme and enzyme-substrate complex. The binding of inhibitor change enzyme shape, so enzyme-inhibitor complex and enzyme-substrate-inhibitor complex are dead end. This reduce the number of functional enzyme. So, the Vmax of this enzyme is lower than uninhibited one.

i hope my opinion can help you.