Lecture 6 - Enzyme Reactions Flashcards
Michaelis-Menten mechanism and an example
• Enzymes act by binding to a substrate and ‘activating’ it.
• e.g. chymotrypsin binding to a polypeptide substrate.
Like lock and key formation:
E (enzyme) binds to S (substrate) forming ES (enzyme-substrate complex) - is reversible
P (product) is formed and then enzyme is released
Michaelis-Menten mechanism
- general equations using E,S and P
- rate equations
- what is the intermediate
- how do you work out [ES]?
ka kb
E + S ⇌ ES and ES —> E + P
ka’
Rate formation of product P: d[P]/dt = kb[ES] (1)
The reactive intermediate is the ES complex
Therefore applying the steady-state approx:
d[ES]/dt = ka[E][S] - ka’[ES]- kb[ES] = 0
[ES] = ka[E][S] / (ka’ + kb) (2)
Michaelis-Menten mechanism
What is [E]?
How can we apply this to eq 2 before
• [E] is the concentration of free enzyme: we generally
know the total (free + bound) concentration [E]0.
[E]=[E]0 - [ES] (3)
Sub (3) into (2) gives you
[ES] = (ka[E]0 x [S]) / (ka’ + kb + ka[S]) (4)
How does equation (4) link to rate of formation of product
d[P]/dt = kb[ES] = kb x (ka[E]0 x [S]) / (ka’ + kb + ka[S])
The rate is often referred to as the velocity, v, in enzymolysis (substrate reacting with enzymes)
How do v, k, [E]0 link together?
V = k[E]0
How do k, kb, kM and [S] link together
k = (kb[S]) / (kM + [S])
How does kM, ka’, kb and ka link together
kM = (ka’ + kb) / ka
Where kM is the Michaelis constant
How is v influenced by high substrate concentration, [S]»_space; kM]
And low substrate conc [S] «_space;kM
v = kb[E]0
v = kb/kM x [S][E]0
Ie velocity increases with increasing [S]
Rate is limited by concentration of ES
More maths on slides
Enzyme inhibition
Competitive vs non-competitive
Competitive:
Substrate cannot bind to enzyme at same time as an inhibitor
Non-competitive:
Inhibitor does not prevent enzyme from binding substrate but prevents formation of product
More maths on slides
