1. The Michaelis-Menton Model
For non-allosteric enzymes, the most widely used kinetic model is based upon work done by Leonor Michaelis and Maud Menton
For a given enzyme-catalyzed reaction:
Substrate --> Product E + S ES P k1
k-1 k2
Note: There is no appreciable back reaction of Product to substrate!

2. k1 = Rate constant for the Formation of the enzyme/substrate complex + S ES P k1
k-1 k2
k1 = Rate constant for the Formation of the enzyme/substrate complex
k-1 = Rate constant for the Dissociation of the enzyme/substrate complex
k2 = Rate constant for the formation of product and its release
The enzyme is specifically (or explicitly) described in the mechanism
This means it appears in the rate equation

3. When we carry out an experiment to characterize the kinetics of an enzyme, we do so at several concentrations of substrate
We want to mix the solution and then immediately measure the initial velocity (V0)
Why?
The [P] is low, so we do not have to be concerned with the back-conversion of product to substrate
We plot V0 versus the [S]

4. We can see two distinct portions in the curve:
First Order Kinetics:
Increasing [S] yields an increased V0
Zero Order Kientics:
The enzyme molecules are saturated with substrate and increasing [S] does not yield an increased V0
Two useful terms can be extracted from this plot:
Vm or Vmax: The maximum velocity of the enzyme (where do we find it?)
Km: The concentration of substrate that gives (1/2)Vm

5. Deriving Km and the Steady State Assumption+ S ES P k1
k-1 k2
In the first step of the reaction, 2 competing processes are taking place.
Rate of Formation of ES:
Rate of Breakdown of ES:
Note: The conversion of ES --> E+P pulls ES out of the pool, so we have to include it in the breakdown rate!

6. The Steady State Assumption
When we measure the velocity of a reaction, there is no product present in the solution during the initial time periods
This means that E+P --> ES does not occur
We also know that enzymes are very efficient, so during the initial time periods, the lifetime of the ES complex is VERY short
Based upon these two facts, we can assume that within milliseconds (or microseconds or less), the reaction will reach a Steady State in which the rate of ES formation equals the rate of ES breakdown.
This is the Steady State Assumption

7. Km and the Steady State Assumption
From the steady state assumption
What is [E]? (We know [S] from the experiment?)
At any given moment in the reaction, the amount of free enzyme is:
[E] = [ET] - [ES]
Plug this into the equation:

8. Km and the Steady State Assumption
Multiply by [ES]
Multiply the left side through
Reorganize
Factor out [ES]
Isolate [ES]
Now we have a term to describe [ES], we can use it to describe V0 in terms of Km and Vm

9. V0, Km and Vm
Part of our steady state assumption is that the E+P --> ES back reaction doesn’t occur
We also know that the rate limiting step will be the conversion of ES to E+P going through the transition state
Does this make sense to you?
If the rate limiting step (energetically most expensive) step was the formation of the ES complex, would the reaction even occur?
Based upon these observations, we can say that the observed velocity of the reaction is:
V0=k2[ES]

10. Deriving the Michaelis-Menton Equation
With the mathematical definition of [ES], V0, Vm and Km, we are ready to derive the Michaelis-Menton equation
If we are in the zero order part of the V0 vs. [S] curve, then V0=VM=k2[ET]
Substituting Vm for k2[ET], we get:

11. The Michaelis-Menton Equation
With the Michaelis-Menton equation, we can relate the observed enzymatic velocity at any substrate concentration to the value of Vm
This allows us to analyze enzymes and:
Identify differences in enzymatic function between species
Compare the same enzyme from different tissues of the same organism
Analyze the efficiency of newly discovered enzymes
Identify the effect of mutations on enzymatic function

12. Working with / Thinking about the Michaelis-Menton Equation
If [S] >>> Km, what does V0 equal?
If [S] = Km, what does V0 equal?

13. Working with Real Data to Calculate Km and Vm
Using V0 vs. [S] plots to calculate Vm and Km can be difficult and inaccurate because of the asymptotic nature of the plot
Two scientists, Hans Lineweaver and Dean Burk, realized that if you plotted the reciprocal of the Michaelis-Menton equation, you’d get a straight line:
This has the form: y = mx + b

14. Lineweaver-Burk Plots
This has the form of a linear equation: y = mx + b
If we plot the data as 1/V0 vs 1/[S]:
The slope = ?
The y-intercept = ? (This occurs when 1/[S] = ?)
The x-intercept = ? (This occurs when 1/V0 = ?)
This is the common method to analyze non-allosteric Enzymological Data

15. Plot the data on p.156 of Campbell and Farrell on your own
Also: Check the Useful Links page on the course website!

16. What do Km and Vm Really mean?
We know that V0=Vm/2 when [S] = Km
We can use this to say that when [S] = Km, 50% of the active sites of the population of enzymes are occupied
If an enzyme has a low Km, the enzyme achieves maximum efficiency at low [S]
Km is unique for each and every enzyme-substrate pair
Some enzymes catalyze reactions that utilize 2 substrates:
E + A EA + B EAB E + P + Q
There is a unique Km value for A and one for B

17. Thinking about Km
As long as k2 << k-1 , the Km can be viewed as a measure of the affinity of the enzyme for the substrate
If k-1 is high, what does that tell you?
The Km value is also a function of pH and temperature.
Why?

18. Turnover number and kcat
In order to better understand the catalytic efficiency of enzymes, we need to define another term: Turnover number or kcat
The catalytic efficiency of an enzyme will be defined as:

19. kcat and Catalytic Efficiency
kcat/KM results in the rate constant that measures catalytic efficiency.
This measure of efficiency is helpful in determining whether the rate is limited by the creation of product or the amount of substrate in the environment.
If the rate of efficiency is:
HIGH, kcat is much larger than KM, thus the enzyme complex turns more enzyme-substrate into product that the enzyme encounters substrate, which is defined as k1.
LOW, this means that the rate of product turnover is much lower than the substrate concentration, thus the enzyme and substrate have high affinity for each other.

20. kcat and Catalytic Efficiency
kcat/KM is only useful when the substrate concentration is much less than the KM. Looking at the enzyme/substrate catalytic reaction equation:
And knowing that k2 could be considered to be equal to kcat, it is evident from
that even if kcat is much greater than k-1, the equation will still be limited by k1, which is the rate of ES formation.
This tells us that kcat/KM has a limit placed on efficiency in that it cannot be faster than the diffusion controlled encounter of an enzyme and its substrate (k1). E + S ES P k1
k-1 k2
kcat = Vm/[Et]
Vm = k2[Et], therefore kcat = k2
Km=(k-1 + k2)/k1

21. kcat and Catalytic Efficiency
Therefore, enzymes that have high kcat/KM ratios have essentially attained kinetic perfection because they have come very close to reaching complete efficiency only being limited by the rate at which they encounter the substrate in solution.

22. Enzyme Inhibition
An inhibitor of an enzyme slows the V0 by sequestering enzyme molecules from the reaction pathway
We will be concerned with 3 types of Reversible Inhibitors: Competitive, Noncompetitive and Uncompetitive

23. Competitive Inhibition
The simplest kind of inhibition and the easiest to understand:
The Inhibitor COMPETES for the active site with the substrate

24. Competitve Inhibition
In this process, we can visualize the role of the inhibitor on the reaction pathway as: E + S ES P I EI
We can write a dissociation constant, KI for the EI complex E + I EI K =
[E][I]
[EI]

25. Competitive Inhibition
The removal of a certain fraction of the free enzyme molecules from the reaction pool scales the apparent Km value down
The scaling factor for the Km value in competitive inhibition is called 

26. Competitive Inhibition
By plugging Km into the uninhibited equation, we get the modified Lineweaver-Burk equation for a competitive inhibitor

27. Graphically Evaluating Competitive Inhibition
Note: In a competitively inhibited reaction, the slope term changes, but the y-intercept term doesn’t (Vmax remains the same)
Vmax is a measure of V0 when [S] ≈ ∞
The x-intercept also changes!

28. Overcoming Competitive Inhibition
By flooding the reaction mixture with substrate, we can overcome Competitive Inhibition
Why?
This is the basis for treating methanol and ethylene glycol poisoning: Give the patient ethanol!
In the liver, the enzyme responsible for oxidizing alcohols is Alcohol Dehydrogenase
The enzyme converts alcohols to aldehydes
Methanol is converted to formaldehyde which can cause blindness or death
By administering ethanol, the methanol outcompeted for the active site of ADH and is instead excreted in the urine

29. Noncompetitive Inhibition
Also known as mixed inhibition
A noncompetitive can bind to the free enzyme OR the enzyme/substrate complex
Noncompetitive inhibitors bind to sites in enzymes that participate in both substrate binding AND catalysis
Metal ions are frequently noncompetitive inhibitors

30. Noncompetitive Inhibition
Several equilibria are involved:
The maximum velocity Vmax has the form:

31. Lineweaver-Burk Plots and Noncompetitive Inhibition
Because the inhibitor does not interfere with the binding of substrate by the enzyme, the apparent Km is unchanged
Increasing [S] won’t overcome Noncompetitive Inhibition

32. Lineweaver-Burk Plots of Noncompetitive Inhibition

33. Uncompetitive Inhibition
The binding of an Uncompetitive inhibitor completely distorts active site rendering the enzyme catalytically inactive
In Uncompetitive inhibition, the inhibitor affects the catalytic function of the enzyme BUT NOT substrate binding

34. Uncompetitive Inhibition

35. Uncompetitive Inhibition
Both Km and Vm change BUT the slope doesn’t!

36. Summary of Enzyme Inhibition

37. How Do We Measure Enzymatic Rates?
Perhaps the easiest way is to use Spectrophotometry
Certain electronic configurations of molecules allow them to absorb radiation in the UV, Visible or Infrared spectra
Proteins typically absorb UV radiation at 260 nm
DNA absorbs UV radiation at 280 nm
Aromatic rings typically absorb UV radiation at 310nm
Colored dyes absorb Visible radiation at the wavelength corresponding to their color
For example, certain copper dyes (Blue dyes) absorb at 595 nm

38. Spectrophotometry: A Vital Method of Analysis
Many solutions of compounds will absorb light at a specific wavelength(s)
This phenomenon is a function of the unique arrangement of electrons in the compound
Most compounds have a set of wavelengths at which they absorb maximally. This set of wavelengths is a kind of “fingerprint” of the compound

39. Transmittance and Absorbance
Transmittance is the ratio of the amount of light that came through the sample (‘Transmitted Light’) to the amount of light present before it hit the sample (‘Incident Light’)
Absorbance is defined as the negative logarithm of the transmittance
This means that, as the concentration of the absorbing species increases, the amount of transmitted light decreases, and therefore, the absorbance increases

40. The Beer-Lambert Law Let’s think about this for a bit…
If we increase the distance the light travels through the solution, the amount of light absorbed should increase. This distance is called the PATHLENGTH
Same concentrations, different pathlength
Different concentrations, same pathlength

41. The Beer-Lambert Law
We can summarize the depndence of Absorbance on pathlength AND sample concentration with the Beer-Lambert law
This shows that there is a linear relationship between the sample’s absorbance and concentration for a given pathlength. VERY USEFUL!

42. How do we use the Beer-Lambert Law?
Standard Curve for KMnO4
A=y-axis
l = 1 cm
C = x-axis
ε = slope
We know that the Absorbance = ? When the KMnO4 = 0M
A=εlC