Biochem Lec 10-Enzymes II: Selectivity and Kinetics

What does Michaelis-Menten Kinetics define?

Defines in quantitative terms how enzymes behave and catalyze reactions under cell conditions, specifically how enzyme activity depends on substrate concentration.

The rate of formation and breakdown of the ES (enzyme-substrate) complex are governed by what?

k₁, k_-1, and k_cat

What equation/relationship expresses the rate of ES formation?

Rate of ES formation= k₁[S][E]

What equation/relationship expresses the rate of ES breakdown?

Rate of ES breakdown= k_-1[ES]+k_cat[ES]

k_-1[ES] is rate of ES dissociation back to E and S

k_cat[ES] is rate of ES conversion to E and P

What is the Steady State Assumption?

At any given time in an enzyme reaction:

• Rate of ES formation=Rate of ES breakdown

• The [ES] remains constant

Necessary for M-M kinetics

From this, the M-M equation can be derived

What is K_M?

The Michaelis constant is an important component of enzyme behavior and measures the affinity (strong/weak binding) of an enzyme for its substrate.

([E][S]/[ES])=(k_-1+k_cat)/k₁=K_M

What is V₀?

The rate of product formation.

V₀=k_cat[ES]

What is the relationship between [ES] and [E]_t when all enzyme is bound to substrate?

[ES]=[E]_t

[E]_t=total enzyme=[E]+[ES]

What is V_max and how is it determined mathematically?

Maximum rate of reaction when the enzyme is fully saturated with substrate.

V_max=k_cat[E]_t

What is the Michaelis-Menten Equation and what do each of the variables represent?

1.V_o is the initial rate of the reaction (moles P produced/sec)

2.V_max is the maximum possible rate when all E is occupied with S (i.e., when [ES] = [E]total )

3. [S] is the initial substrate concentration

4. K_M is the Michaelis constant, it is the [S] that gives ½ Vmax

What is the shape the plot of V_o vs. [S] from the M-M equation?

Hyperbolic

When does V_max=V_o? Where is this represented on the plot?

V_max=V_o when all enzyme molecules are bound with substrate. Represented by the curve’s asymptote.

K_M is the concentration of substrate ([S]) where _______

V_o= ½ V_max

At very high [S] all enzyme molecules become saturated with S molecules. How does this impact the rate and reaction order? How does this relate to the equation for V_max?

The rate no longer increases with more substrate

Zero order conditions

The equation for V_max is V_max = k_cat [E]_T → [S] is not part of the equation, so V_max is not impacted by [S]

What is k_cat and what does it represent?

“catalytic rate constant'“ or “turnover number”

It is the number of molecules of product P produced per sec per enzyme active site when all enzyme is saturated with substrate [S]

What is the inverse of k_cat (1/k_cat)?

The time needed for one enzyme active site to produce one product.

K_M is a measure of the range of ____ where the enzyme is _____. It is often close to biological or cellular _____.

K_M is a measure of the range of [S] where the enzyme is effective. It is often close to biological or cellular [S].

The ____ the K_M the more effective the enzyme at low [S]

The lower the K_M the more effective the enzyme at low [S]

How are V_max and K_M determined?

An enzyme kinetic experiment:

• keep enzyme concentration constant

• vary [S]

• measure V_o (P produced/sec)

Plot data:

• asymptote=V_max

• [S] that is ½ V_max is K_M

What is an alternative way to obtain K_M and V_max?

Use the Lineweaver-Burk form of the M-M equation (“double reciprocal”)

How does a plot using the Lineweaver-Burk form of the M-M equation differ? What is this plot called?

Lineweaver-Burk plot:

plot 1/V_o vs. 1/[S]

Describe a Lineweaver-Burk plot. What is its shape and why? What do the X/Y intercepts and the slope correspond to?

It is a plot of 1/V_o vs. 1/[S]. It has a linear shape because the Lineweaver-Burk equation—1/V_o=(K_M/V_max)*(1/[S])+(1/V_max)—follows the y=mx+b format of linear functions.

X-intercept → -1/K_M

Y-intercept → 1/V_max

Slope → K_M/V_max

How do k_cat and K_M differ?

k_cat → how quickly product is turned over once substrate is bound to the active site

K_M → the effective substrate concentrations at which the enzyme works

What is the specificity constant? What two variables are used to determine it?

• Describes the catalytic efficiency of an enzyme → how good the enzyme is at converting substrate into product

• Describes the “catalytic efficiency” of an enzyme and kinetics when    V_o < V_max

• Can be used to compare different substrates for enzymes

Determined by k_cat/K_M

What is an example of how the specificity constant demonstrates enzyme efficiency/efficacy?

Chymotrypsin (a protease):

When R₁ was glycine=0.13 (sec*M)^-1

When R₁ was phenylalanine= 100,000 (sec*M)^-1

Chymotrypsin is more effective at peptide bond cleavage in the presence of a large/bulky/hydrophobic side chain than a small/flexible one → enzymes are specific to substrates

What is catalytic perfection?

• Happens in extreme cases

• Absolute efficiency in which every encounter with a substrate results in conversion to product (k_cat/K_M ≈ 109 sec^-1M^-1)

What limits the rate of reaction when catalytic perfection is reached?

The rate of reaction is limited by the time it takes for a S molecule to diffuse into an enzyme active site (i.e., diffusion limited)

M-M key concepts: What variables does the plot show a relationship between? What is its shape?

Hyperbolic relationship between rate of reaction (V_o) vs. [S]

M-M key concepts: describe V_max and how it is determined mathematically.

• Approached asymptotically

• Rate when enzyme is saturated with substrate

• V_max=k_cat[E_t] → E_t= E+ES

M-M key concepts:what is K_M ?

M-M constant: K_M = [S] at V_max/2

M-M key concepts: what is k_cat?

• Turnover number → substrate molecules converted into product per second

• k_cat = V_max / E_t

M-M key concepts: what is the specificity constant and what limits it?

• k_cat/K_M is catalytic efficiency. Better reflects rate under typical cellular conditions.

• k_cat/K_M cannot be faster than diffusion-controlled encounter of enzyme and substrate (limit of ~10⁹ s^-1 M-¹)

From the plot of velocity versus substrate concentration shown below ([E]T = 10^-3 μM), obtain the following parameters: a) V_max, b) K_M, c) turnover number (k_cat), d) specificity constant (k_cat/K_M)

a) asymptote → 6 μM min^-1

b) point where [S]=1/2 V_max → 0.5 mM

c) V_max = k_cat [E]_T → k_cat= V_max/[E]_T → k_cat= 6 μM min^-1/10^-3 μM= 6×10³ min^-1= 100s^-1

d) k_cat/K_M= 100 s^-1/0.5 mM= 200 s^-1 mM^-1= 2×10⁵ s^-1M^-1