5. Protein Function and Enzymes Part II

Question 1

Enzyme Kinetics

Define the terms initial velocity (V0), maximum velocity (Vmax) and Michaelis constant (KM).

Answer 1

initial velocity (Vo) = Vo is the initial rate of a reaction (at a given [S])
Maximum Velocity (Vmax) = is the maximum reaction velocity when all the enzyme is saturated with substrate and additional substrate changes the rate of the reaction less and less
Michaelis Constant (KM) = is the [S] when Vo is 1/2V max

Question 2

Enzyme Kinetics

• Define KM and Vmax in terms of rate constants (k) and enzyme concentration.

Answer 2

E + S ⇌(k1 / k-1) ES ⇌ (k2/k-2) E + P
Km = (k-1 + k2)/k1 (breakdown of ES/Formation of ES)
- Michaelis Constant reflects the balance between formation and breakdown of ES complex.

Vmax = k2[Etotal]
- Maximum velocity is determined by the total concentration of enzyme in the system.

Question 3

Enzyme Kinetics

Explain the shape of the graph of V0 versus substrate concentration in terms of formation of an enzyme-substrate complex (ES).

Answer 3

The shape of Vo vs. [S] plot resembles a hyperbola, as [S] increases, the [ES] increases as well.
The point (Vmax) at which [S] causes less and less change in Vo, is when [ES] has reached its maximum. (enzymes are all saturated)

Km = [S] @ Vmax/2

Question 4

Enzyme Kinetics

Describe the Michaelis-Menten equation and the assumptions in its derivation.

Answer 4

Vo = Vmax[S]/(Km+[S])
- Reflects the balance between formation and breakdown of ES complex
- One substrate,
- enzyme catalyzed reaction,
- quantitative relationship between Vo and Vmax and initial [S], all reacted through Km

Question 5

Enzyme Kinetics

• Manipulate the Michaelis-Menten equation (eg. derive Lineweaver-Burk equation).

Answer 5

Vo = Vmax[S]/(Km + [S])
1/Vo = (Km + [S]) / (V max[S])
1/Vo = Km/(V max[S]) + [S]/V max[S]
1/Vo = Km/ (V max[S]) + 1/(Vmax)

Question 6

Enzyme Kinetics

Determine KM and Vmax from a Lineweaver-Burk plot.

Answer 6

Lineweaver-Burk equation is useful for a double-reciprocal plot of enzyme reaction rates, 1/Vo = Km/Vmax[S] + 1/Vmax Slope of this linear relationship is = K m/V max. Y-Intercept = 1/V max on 1/V o axis X-Intercept = 1/Km on 1/[S] axis.

Question 7

Enzyme Kinetics

• Define the terms turnover number (kcat) and specificity constant (kcat/KM) and calculate them from kinetic data.

Answer 7

Turnover Number (kcat) - the ability of an enzyme to catalyze product when enzyme is fully saturated with substrate.
- Normalized value of Vmax

kcat = V max/[E total]
↑Kcat → ↑Specificity constant → better enzyme

specificity constant = kcat/Km
- Reflection of efficiency of BOTH binding and conversion

Specificity constant Kcat/Km can idicate “catalytic perfection”

Question 8

Enzyme Kinetics

Describe kcat and KM and how they may be complex combinations of multiple rate constants, depending on the enzyme mechanism being studied.

Answer 8

kcat and Km vary among different enzymes because Km depends on rate constants and each enzyme may have a different mechanism or series of reactions.

Question 9

Enzyme Inhibition

Define the terms α and KI, α’ and KI’.

Answer 9

α:
- Reflects how much substrate concentration must be changed to overcome inhibition
- the factor by which Km increases in the presence of a competitive inhibitor
- Vmax is not affected
- diagnostic of a competitive inhibitor:
α = 1 + [Inhibitor]/KI

KI:
- competitive inhibitor binding equilibrium constant
- Inhibitor dissociation constant (Kd for inhibitor)
KI = [E][I]/[EI]

α’:
- the factor by which Vmax is affected by a uncompetitive inhibitor.
- α’ = 1 + [I]/K’I

K’I:
- uncompetitive inhibitor binding equilibrium constant
- K’I = [ES][I]/[ESI]

Uncompetitive: Inhibitor binds site distint from substrate binding site and only binds ES complex

Competitive Inhibitor: Inhibitor resembles substrate and competes with it for binding to enzyme active site (binds E)

Question 10

Enzyme Inhibition

• Write the equations relating the values of α to KI and α’ to KI’.

Answer 10

α = 1 + [Inhibitor]/KI
KI = [E][I]/[EI]

α’ = 1 + [I]/K’I
K’I = [ES][I]/[ESI]

α:
- Reflects how much substrate concentration must be changed to overcome inhibition
- the factor by which Km increases in the presence of a competitive inhibitor
- Vmax is not affected
- diagnostic of a competitive inhibitor:
α = 1 + [Inhibitor]/KI

KI:
- competitive inhibitor binding equilibrium constant
- Inhibitor dissociation constant (Kd for inhibitor)
KI = [E][I]/[EI]

α’:
- the factor by which Vmax is affected by a uncompetitive inhibitor.
- α’ = 1 + [I]/K’I

K’I:
- uncompetitive inhibitor binding equilibrium constant
- K’I = [ES][I]/[ESI]

Question 11

Enzyme Inhibition

• Describe the differences between competitive, non-competitive and mixed inhibition.

Answer 11

Competitive Inhibitor - competitive inhibitor binds enzyme at the active site, to compete with the substrate
- Increases Apparent Km
- Michaelis-Menton Equation: Vo = Vmax[S]/(αKm + [S])
- Apparent Vmax = Vmax
- Apparent Km = αKm

Uncompetitive Inhibitor - inhibitor binds at a site other than the active site.
- Binds only to the ES complex.
- Decreases Apparent Km and Vmax
- Michaelis-Menton Equation: Vo = Vmax[S]/Km + α’[S]
- Apparent Vmax = Vmax/α’
- Apparent Km = Km/α’

Mixed Inhibitors
- binds at sites distinct from substrate active sites, but it binds to either E or ES.
- Michaelis-Menton Equation: Vo = Vmax[S]/αKm + α’[S]
- Apparent Vmax = V max/α’
- Apparent Km = αK m/α’

α = 1 => No inhibitor [I]=0

Question 12

Enzyme Inhibition

Understand the difference between reversible and irreversible inhibitors and their effects on the Michaelis-Menten equation, the KM and Vmax values.

Answer 12

Reversible Inhibitors bind reversibly to the enzyme

Irreversible inhibitors bind covalently with the enzyme and destroy the functional group on an enzyme that is essential for the activity of an enzyme.

Question 13

Enzyme Inhibition

Explain why transition state-analogs can be effective inhibitors.

Answer 13

Transition state analogs can be used as inhibitors in enzyme-catalyzed reactions by blocking the active site of the enzyme. Theory suggests that enzyme inhibitors which resembled the transition state structure would bind more tightly to the enzyme than the actual substrate.

Question 14

Enzyme Inhibition

Use Lineweaver-Burk plots to distinguish kinetically among different types of inhibitors.

Answer 14

a > a’ = Competitive Inhibition
KI<KI’ = increase affinity for E over ES

a’>a = Uncompetitive
Increase affinity for ES

a=a’ = non-competitive

Question 15

Control of Enzyme Activity

List mechanisms used by organisms to regulate enzymes.

Answer 15

Allostery (R/T)
- Feedback inhibition

Covalent modification
- Phosphorylation
- Adenylylation (Tyr)
- ADP-ribosylation
- Palmitorylation (lipid anchor)

Regulatory Enzymes - enzymes which have increased or decreased catalytic activity in response to certain signals. In multi enzyme systems the first enzyme of a sequence is typically a regulated pathway to prevent unneeded products from forming unless the pathway is active.

Question 16

Control of Enzyme Activity

Outline general features of allosteric enzymes.

Answer 16

• Enzymes do not obey Michaelis-Menten kinetics
• Sigmoidal activity curves are typical
• Cooperative substrate binding (positive homoallostery)
• Two (at least) states: high and low activities (R and T states)
• Often associated with feedback inhibition

Features:
* Allosteric proteins have quaternary structure.
* Each oligomer can exist in either R or T state (high and low activity respectively).
* Ligands/substrates will bind to both states (with different affinities) and binding shifts equilibrium between T and R states.
* The overall protein symmetry is maintained (i.e. All subunits are in the same conformation).

Question 17

Control of Enzyme Activity

• Define the terms homotropic and heterotropic effector, positive and negative effector, feedback inhibition.

Answer 17

Homotropic Modulator - modulators of allosteric enzymes which are identical to the substrate.
Heterotropic Modulator - modulators of allosteric enzymes which differ from the substrate the enzyme acts upon.

Positive Modulators - modulators which are stimulatory. When bound to the enzyme results in increased binding of the substrate.
Negative Modulators - modulators which are inhibitory. When bound to the enzyme results in decreased binding of the substrate.

Question 18

Control of Enzyme Activity

• Describe general kinetic differences between allosteric and non-allosteric enzymes.

Answer 18

Allosteric Enzymes have sigmoidal saturation curves, whereas with non-allosteric enzymes they experience hyperbolic saturation curves

Question 19

Control of Enzyme Activity

Outline and describe features of the allosteric enzyme aspartate transcarbamoylase in terms of kinetics, effectors and structure.

Answer 19

Aspartate transcarbamoylase (ATCase)
* EC2.1.3.2 (2 = Transferase)
* Quaternary structure
* 6 catalytic subunits
* 6 regulatory subunits
* Symmetry: D3 (12 subunits; 6 protomers)
* Catalyzes First step in CTP biosynthesis
* Feedback inhibited (CTP) (Heterotropic Inhibitor)
* Activated by ATP (Heterotropic Activator)
* Generally follows the symmetry model
* Aspartate is homotropic activator

Question 20

Control of Enzyme Activity

Explain the physical origin of the sigmoidal curve of [S] versus V0 for a homotropic allosteric effector.

Answer 20

Homotropic Allosteric Enzymes are generally multi-subunit proteins and the active site is also the regulatory site. Sigmoidal kinetics signifies that there is cooperative interactions between protein subunits. Changes in the structure of one subunit are translated into structural changes in adjacent subunits. Binding of substrate to one subunit enhances the next substrate to bind.

Question 21

Control of Enzyme Activity

List some of the possible effects of a heterotropic effector on the sigmoidal curve, K0.5 and Vmax for an allosteric enzyme.

Answer 21

Activator
* can cause curve to become nearly hyperbolic,
* decrease in K0.5
* no change in Vmax,
* resulting in an increased reaction velocity at a fixed [S].
* can cause an increase in Vmax with little change in K0.5

Inhibitor
can produce a more sigmoidal substrate-saturation curve with an increase in K0.5.

Question 22

Control of Enzyme Activity

Outline and describe features of the allosteric enzyme phosphofructokinase in terms of kinetics, effectors, structure and mechanism.

Answer 22

ATP:
- Binds at active site and regulatory site
- Homotropic inhibitor = decreases activity

Fructose-6-phosphate
- Binds to active site
- Homotropic Activator
- Sigmoidal Relationship

AMP:
- Activator (Heterotropic)
- Same regulatory site as ATP

Phosphoenolpyruvate
- Heterotropic Inhibitor

homotropic allosteric effector is a substrate for the enzyme, as well as a regulatory molecule

Question 23

Control of Enzyme Activity

Differentiate between reversible and irreversible covalent modification.

Answer 23

Reversible Covalent Modification - covalent modification can be added and removed,

irreversible covalent modification, group cannot be removed. e.g. Protelytic cleavage of Chymotrypsinogen

Question 24

Control of Enzyme Activity

• Outline and draw simple reactions for two reversible covalent modifications that use ATP as a substrate.

Answer 24

Regulation of Glycogen Phosphorylase by phosphorylation
- Glycogen phosphorylase is activated by phosphorylation by specific kinase
- Enzyme + ATP → Enzyme-P + ADP

Adenylation (Tyr)
- Enzyme + ATP → Enzyme-AMP + PPi (pyrophosphate)

Glycogen Synthase is regulated by multiple regulatory phosphorylations. This allows for extremely subtle modulation of enzyme activity. It is modulated by at least 5 protein kinases.

Question 25

Control of Enzyme Activity

Describe the physical interactions that may lead to conformational change when a phosphate group is added to a protein structure.

Answer 25

Active site can undergo changes in structure and change catalytic activity when phosphorylated vs. non-phosphorylated form. Phosphorylation can also affect the enzyme’s catalysis by affecting the substrate-binding affinity.

Question 26

Control of Enzyme Activity

Define the term “target sequence” in the context of protein kinases.

Answer 26

In the context of protein kinases, the term “target sequence” refers to a specific amino acid sequence in a protein substrate that is recognized and phosphorylated by the protein kinase. Protein kinases are enzymes that catalyze the transfer of phosphate groups from ATP molecules to specific amino acids (usually serine, threonine, or tyrosine) in target proteins.

Some kinases need to recognize neighbouring amino acid residues (consensus sequences) e.g. Proline
Amino acid sequence and 3-D structure giving availability to amino acid residues to be phosphorylated.
Hierarchical phosphorylation: certain residues are only phosphorylated when other residues are phosphorylated first.

Question 27

Control of Enzyme Activity

Describe the process of zymogen or proenzyme activation using chymotrypsin as an example.

Answer 27

Zymogens (unaltered enzymes) are activated by modification by other enzymes
* Digestive proteinases are modified by other proteinases
* Enzymes must then be destroyed or inactivated

Zymogens = inactive enzyme precursor
- require specific proteolytic cleavage to become active. chymotrypsinogen, the inactive precursor of chymotrypsin .
- Cleavage: Trypsin cleaves a small peptide fragment from the N-terminal of chymotrypsinogen, exposing an active site for subsequent cleavage.

Formation of a functional active site occurs once the peptide is cleaved after residue 15 (oxyanion hole forms).
a-chymotrypsin is the fully mature form

Autocatalysis: The newly exposed active site of chymotrypsinogen undergoes autocatalytic cleavage, resulting in the removal of a larger peptide fragment and the formation of the active enzyme chymotrypsin

Feedback inhibition: Chymotrypsin in turn can act on other chymotrypsinogen molecules to produce more active enzyme. However, once enough chymotrypsin has been produced, it can also inhibit the further activation of chymotrypsinogen, thus regulating its own activity.