Download Steady State Kinetics - Biochemical Laboratory Technique - Lecture Notes | BIOC 463A and more Study notes Biology in PDF only on Docsity!
Enzyme Steady-State Kinetics and Competitive Inhibition Steady-State Kinetics: Determination of Km, Vmax, and kcat
Objectives: We will emphasis the methods and techniques employed in steady-state kinetic analysis of AP. We will not cover Transition State Theory in this class. Rather we will try to fully understand how data is obtained and analyzed in the context of the Michaelis-Menten equation.
Historical Perspective: In a typical chemical reaction: k A + B Æ products
Velocity of reaction vs. [A] ([B] being held constant) is linear. If a catalyst is added, the velocity increases, but plot remains linear.
So, rate of reaction: v = kobs[A], where kobs = k[B].
And, kuncatalyzed << kcatalyzed.
Enzyme Steady-State Kinetics and Competitive Inhibition At beginning of 1900’s it was understood that enzymes were catalysts involved in biological reactions.
For enzymes, a different plot of v vs. [S] was observed:
Michaelis and Menten suggested that the approach of v to a maximal level (Vmax) meant E and S were forming a complex of distinct lifetime.
Assumption 1: E and S react rapidly and are in equilibrium: k 1 E + S <=> ES k- and Ks = k-1/k 1 = [E][S]/[ES] = dissociation constant for ES!
Enzyme Steady-State Kinetics and Competitive Inhibition
Km = (k-1 + k 2 ) = [E][S] (look familiar?) k 1 [ES]
- Initial Velocity Assumption:
at t = 0, [P] = 0
this does two things: Since many P are very effective competitive inhibitors, eliminates math complications due to the back reaction: E + P Æ ES
Taking ALL of these assumptions into consideration, B&H derived the beloved M-M eqn:
v 0 = Vmax[S] where Vmax = kcat[Et] Km + [S]
There are three [S] conditions to consider: [S] v 0 << Km Vmax[S] = kcat[E]t[S] Km Km = Km Vmax 2
Km Vmax = kcat[E]t
Enzyme Steady-State Kinetics and Competitive Inhibition
Two Important Definitions of Km:
- Km = (k-1 + k 2 )/k 1
- Km is the [S] that gives v 0 = Vmax/2!
The units of Km are CONCENTRATION. Km is a CONCENTRATION!!!!!!! (Sound familiar?)
Km is primarily a measure of S binding, with a little catalysis thrown in (if k 2 is large relative to k-1).
Vmax is a measure of catalysis only (kcat and [E]t).
Usually, when people compare Km values, it is implied they are comparing the binding of different substrates.
Enzyme Steady-State Kinetics and Competitive Inhibition How are Steady-State experiments performed:
Typically, you have S and buffer in a cuvette, you add some E and follow an absorbance change at some wavelength for either decrease in [S] or increase in [P]. Find the slope of the line tangential to data as close to t = 0 as possible. This gives a vo value for that [S]. Then you repeat using a different [S].
Data analysis: problem here is same as seen in ligand binding:
- Extrapolate to infinite [S].
- Determine Vmax.
- Determine Km.
Enzyme Steady-State Kinetics and Competitive Inhibition Linear Transform: Lineweaver-Burk Plot.
Starting with:
v 0 = Vmax[S] Km + [S]
Take inverse of both sides:
1 = Km ( 1 ) + ( 1 ) vo Vmax ([S]) (Vmax)
Enzyme Steady-State Kinetics and Competitive Inhibition Competitive Inhibitors: Determination of Ki.
Inhibitor binds to active site of E:
- Effect can be overcome at high [S] by mass action, so Vmax is the same.
- Km, however, is APPARENTLY changed:
Km,app > Km
Km,app = Km ( 1 + [I] ) Ki
Note, Km is not actually changed! Km is a property of the binding of S to E, and that is not altered by I !!!!!!
M-M eqn for Competitive Inhibitor:
vo = Vmax [S] Km(1 + [I]/Ki) + [S]
Enzyme Steady-State Kinetics and Competitive Inhibition Lineweaver-Burk form of eqn:
1 = Km ( 1 + [I]/Ki) ( 1 ) + ( 1 ) vo Vmax [S] Vmax
(As [I] increases or Ki decreases, how will slope change?)
It is important to note that [I], Ki, Vmax, and Km are all CONSTANTS. 1/[S] is the VARIABLE. This is not only true for the plot, it is also true for the way the experiment is done. [I] is held constant at ALL [S] throughout experiment!
This is a Lineweaver-Burk METHOD of studying Inhibitors!
Enzyme Steady-State Kinetics and Competitive Inhibition
(Km) [I]+ 1 (Km/[S1]+1) = (Km) [I]+ 1 (Km/[S2]+1) (VmaxKi[S1]) Vmax (VmaxKi[S2]) Vmax
rearranging and eliminating terms:
Ki + [I] = Ki + [I] [S1] [S2]
If S1 ≠ S2, how can this equation be true?
Answer:It is true only if Ki = - [I]! So the pt of intersection represents – Ki!
Ki = 4 uM
In Summary:
Method Constant Variable L-B [I] [S] Dixon [S] [I]
