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Biochemistry > Module 2 > Flashcards

Module 2 Flashcards

1
Q

Molecular Motor

A

A protein that uses ATP to produce cyclic conformational changes.

Ex: Myosin, Kinesin

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2
Q

Muscle Fiber

Myofiber

A

Muscle Cell

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3
Q

Myosin

A

A motor protein that comprises the thick filaments of sarcomeres.

Myosin forms thick filaments that (along with thin filaments) mediate muscle movement via myofiber contraction.

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4
Q

Actin

A

A protein that polymerizes within muscle cells to form the major component of thin filaments.

Actin polymers form thin filaments (that mediate muscle movement via myofiber contraction) and microfilaments (that are critical components of the cytoskeleton).

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5
Q

What molecular action leads to muscle contraction?

A

Myosin Conformational Change

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6
Q

How many subunits does Myosin contain?

A

Six Subunits

  • Two Heavy Chains
  • Four Light Chains (2 Regulatory Chains + 2 Essential Chains)
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7
Q

Myosin Subunit Interactions

A
  • Two light chains are bound to each heavy-chain “head” (at the heavy-chain “neck”).
  • Two heavy chain “tails” coil around one another (as extended α-helices).

  • Heavy-Chain “Head” = N-Terminus of Subunit
  • Heavy-Chain “Tail” = C-Terminus of Subunit
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8
Q

Which region of the Myosin molecule serves as an ATPase?

A

Myosin Head

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9
Q

Which region of the Myosin molecule binds to the thin filament?

A

Myosin Head

The actin-binding domain of the Myosin molecule binds to the thin filament to initiate muscle contraction.

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10
Q

Length: Thick Filament

A

~ 325 nm

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11
Q

G-Actin vs. F-Actin

A
  • G-Actin: Monomer of Actin
  • F-Actin: Polymer of Actin (Comprised of Polymerized G-Actins)

  • G-Actin = Globular Actin
  • F-Actin = Filamentous Actin
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12
Q

G-Actin

A

A monomer of Actin comprised of 375 amino acids.

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13
Q

F-Actin

A

A filamentous polymer of Actin comprised of numerous G-Actin subunits.

The polymerization of Actin requires ATP.

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14
Q

Titin

A

A large protein that imparts flexibility to the sarcomere and connects the Z disk to the thick filament.

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15
Q

Muscles contract when ____________________.

A

thick filaments and thin filaments slide past one another.

The cyclical attachment, detachment, and reattachment of Myosin to thin filaments causes muscular contraction.

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16
Q

How does Ca2+ control muscle contraction?

A

The binding of Ca2+ ions to Troponin induces a Troponin/Tropomysin conformational change that exposes Myosin-binding sites on Actin.

  • Relaxed Muscle: Myosin-binding sites on Actin are blocked by Tropomyosin.
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17
Q

Muscle Contration Cycle

5 Steps

A
  1. The binding of Ca2+ ions to Troponin induces conformational changes that expose Myosin-binding sites on Actin.
  2. The binding of Myosin and release of Pi induces a power stroke to pull the thin filament across the thick filament.
  3. The release of ADP from the Myosin head empties the nucleotide-binding sites on Myosin.
  4. The binding of ATP to the Myosin nucleotide-binding site causes Myosin to dissociate from the thin filament.
  5. The hydrolysis of ATP on Myosin induces the “recovery” position of the Myosin head.
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18
Q

5 Major Functional Classes of Proteins

A
  • Metabolic Enzymes
  • Structural Proteins
  • Transport Proteins
  • Cell-Signaling Proteins
  • Genomic Caretaker Proteins
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19
Q

Metabolic Enzymes

Enzymes

A

Proteins that catalyze biochemical reactions involved in energy conversion pathways (e.g. the synthesis/degredation of macromolecules).

  • Enzymes are NOT consumed during a chemical reaction.
  • Enzymes increase the reaction reate without altering the equilibrium concentration of products and reactants.
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20
Q

How is an enzyme able to increase the rate of product formation?

A

An enzyme lowers the activation energy of a reaction.

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21
Q

Active Site (Enzymes)

A

The region of an enzyme where the catalytic reactions take place.

The shape and chemical environment of enzyme actives sites are determined by amino acid side chains.

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22
Q

Examples: Metabolic Enzymes

A
  • Malate Dehydrogenase
  • Pyruvate Dehydrogenase
  • Phophofructokinase–1
  • Acetyl-CoA Carboxylase
  • Thymidylate Synthase
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23
Q

Examples: Structural Proteins

A
  • Actin
  • Tubulin
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24
Q

Structural Proteins

A

Proteins that function as the architectural framework for individual cells, tissues, and organs.

Structural proteins are the most abundant proteins in living organisms.

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25
Q

Cytoskeletal Proteins

A

Structural proteins that are responsible for cell shape, cell migration, and cell signaling.

Ex: Actin, Tubulin

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26
Q

Thin Filament

A

An Actin polymer that forms part of a network to control cell shape, cell migration, and muscle contraction.

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27
Q

Intermediate Filaments

A

A type of cytoskeletal protein complex that is critical to cell structure and cell function.

Ex: Vimentin, Laminin, Keratin

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28
Q

Transport Proteins

A

Proteins that span the width of a cell membrane and allow polar/charged molecules to enter/exit the cell.

The two classes of membrane transport proteins are passive transporters and active transporters.

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29
Q

Passive Transporter

Passive Tranport Protein

A

A membrane transport protein that allows specific molecules to enter/exit the cell while moving down their concentration gradient.

Ex: Porins, Ion Channels

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30
Q

Active Transporter

Active Transport Protein

A

A membrane transport protein requiring energy to induce protein conformational changes that open/close a gated channel (and pump molecules against their concentration gradient).

  • Active transporters obtain energy from either ATP hydroylsis or ionic gradients.
  • Ex: Ca2+-ATPase
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31
Q

Types of Cell-Signaling Proteins

A
  • Membrane Receptors
  • Nuclear Receptors
  • Intracellular Signaling Proteins
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32
Q

Membrane Receptor

A

A transmembrane protein that changes conformation upon binding of a cognate ligand molecule.

Ex: Erythropoietin Receptor (Growth Hormone Receptor); Insulin Receptor (Receptor Tyrosine Kinase); Adrenergic Receptors (G Protein-Coupled Receptor)

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33
Q

Nuclear Receptor

A

A transcription factor that regulates gene expression in response to ligand binding.

Ex: Glucocorticoid Receptor, Vitamin D Receptor, Estrogen Receptor, Progesterone Receptor

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34
Q

Intracellular Signaling Protein

A

A protein that functions as a molecular switch by undergoing conformational changes in response to incoming signals (e.g. receptor activation).

E: Protein Kinases, Phosphatases, Intermolecular Adaptor Proteins, Site-Specific Proteases

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35
Q

Genome Caretaker Proteins

A

Proteins that function to ensure the integrity of genomic DNA throughout a cell’s lifespan.

Ex: DNA Synthesis Enzymes, DNA Repair Enzymes, DNA Recombination Enzymes, RNA Synthesis Enzymes

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36
Q

Myoglobin

A

A monomeric globular transport protein concentrated in muscle tissue that functions in Oxygen storage.

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37
Q

Hemoglobin

A

A tetrameric globular transport protein that transports Oxygen from the lungs to the tissues via the circulatory system.

Hemoglobin is the major protein in red blood cells (i.e. accounts for 35% of the cells’ dry weight).

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38
Q

Heme

A

An Fe2+ porphoryin complex that functions as a prostethic group to bind Oxygen.

The Fe ion must be in the 2+ oxidation state for Oxygen binding to occur.

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39
Q

Structure: Myoglobin vs. Hemoglobin

A
  • Myoglobin: Single Polypeptide w/ 1 Heme Group
  • Hemoglobin: Four Polypeptides w/ 4 Heme Groups

  • Less than 20% of amino acids are identical between Myoglobin and Hemoglobin.
  • Myoglobin and Hemoglobin share the globin fold conformation.
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40
Q

Structure: Hemoglobin

A
  • Two α-Subunits + Two β-Subunits (Each Possess 1 Heme Group)
  • Dimer of Heterodimers (α1β1 + α2β2) = Heterotetramer
  • Eight α-Helices (Globin Fold) per Subunit
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41
Q

Globin Fold

A

A protein folding pattern that contains eight α-helices.

Ex: Myoglobin, Hemoglobin

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42
Q

How many coordination bonds does the Heme Fe2+ possess?

A

Six

  • Four coordination bonds are with Nitrogens in the plane of the porphyrin ring.
  • One coordination bond is above the plane of the porphyrin ring.
  • One coordination bond is below the plane of the porphyrin ring.
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43
Q

Proximal Histidine

His F8

A

A Histidine residue in globin proteins that coordinates with the Fe2+ of the porphyrin ring (either above or below the plane of the ring).

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44
Q

Distal Histidine

His E7

A

A Histidine residue in globin proteins that forms a hydrogen bond with O2 (when O2 is bound to the porphyrin ring) to stabilize its intereaction with the Heme group.

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45
Q

How does O2 bind to the globin protein Heme group?

A

The O2 forms a coordination bond with the Fe2+ of the porphoryin ring (either above or below the plane of the ring).

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46
Q

Conformation: Bound O2 vs. Unbound O2

Heme Group of Globin Proteins

A
  • Unbound O2: The Heme group is puckered (i.e. Fe2+ is NOT in the plane of the porphyrin ring) because of the larger Fe2+ ionic radius.
  • Bound O2: The Heme group is planar (i.e. Fe2+ IS in the plane of the porphyrin ring) because of the smaller Fe2+ ionic radius.

The binding of O2 to the Heme group moves His F8 (and the entire F Helix) toward the Heme group.

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47
Q

Why is the binding of O2 to Heme Fe2+ reversible?

A

The structural changes of Hemoglobin/Myoglobin under different physiological conditions results in altered affinities for O2.

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48
Q

Equation: Protein-Ligand Binding

A

P = Protein
L = Ligand
PL = Protein-Ligand Complex

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49
Q

Association Constant

Ka

A

An equilibrium constant for the binding of two molecules (e.g. a protein and a ligand to form a protein-ligand complex).

Units: M–1

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50
Q

Equation: Protein-Ligand Ka

A
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51
Q

Dissociation Constant

Kd

A

An equilibrium constant for the dissociation of two molecules (e.g. a protein-ligand complex unbinding to yield a protein and a ligand).

Units: Kd

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52
Q

Equation: Protein-Ligand Kd

A

The Kd is the inverse of the Ka equation.

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53
Q

What does a higher Kd value indicate?

A

A lower affinity between the two molecules (i.e. more of the dissociated protein and ligand are present).

A lower Kd value indicates a higher affinity between the two molecules.

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54
Q

What does the Kd value represent?

A

The ligand concentration at which 50% of potential ligand binding sites are occupied.

Kd: [L] at θ = 0.5

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55
Q

Fractional Saturation

θ

A

The fraction of protein binding sites that are occupied by ligands.

56
Q

Equation: θ

Fractional Saturation

A
57
Q

Equation: Relationship of θ and Kd

A
58
Q

What is the shape of a typical protein-ligand binding graph?

θ vs. [L]

A

Hyperbolic

59
Q

What does P50 represent?

A

The partial pressure of a ligand at which 50% of potential binding sites are occupied.

P50: pX at θ = 0.5

60
Q

P50 vs. Kd

A
  • P50: Used when the ligand is a gaseous compound (e.g. O2)
  • Kd: Used when the ligand is a solid/liquid compound (e.g. Drug Compounds)
61
Q
A
62
Q

What are the three primary mechanisms of regulating enzyme catalytic efficiency?

A
  • Binding of Regulatory Molecules
  • Covalent Modification
  • Proteolytic Processing
63
Q

Reversible Inhibition

A

An enzyme regulatory mechanism involving the noncovalent binding of biomolecules/proteins to an enzyme subunit.

The effect of reversible inhibition can be decreased by diluting a reaction (with more substrate).

64
Q

Irreversible Inhibition

A

An enzyme regulatory mechanism involving the covalent binding of an inhibitory molecule to catalytic groups on an enzyme’s active site.

Irreversible inhibition is not affected by diluting a reaction with more substrate (since the covalent bonds remains intact regardless of enzyme/inhibitor concentration).

65
Q

Classes: Reversible Inhibitors

A
  • Competitive Inhibitors
  • Uncompetitive Inhibitors
  • Mixed Inhibitors
66
Q

Competitive Inhibitor

A

A molecule that decreases enzymatic efficiency by inhibiting/blocking substrate binding at the enzyme’s active site.

Competitive inhibitors bind to the free enzyme at the active site (or at a location overlapping with the active site).

67
Q

Uncompetitive Inhibitor

A

A molecule that decreases enzymatic efficiency by binding to the enzyme-substrate complex and altering the active site conformation.

Uncompetitive inhibitors bind to the enzyme-substrate complex at a location other than the active site (i.e. an allosteric site).

68
Q

Mixed Inhibitor

A

A molecule that decreases enzymatic efficiency by altering the active site conformation (before or after substrate binding).

Mixed inhibitors bind to the free enzyme or the enzyme-substrate complex a location other than the active site (i.e. an allosteric site).

69
Q

KI

A

Equilibrium Dissocation Constant for Enzyme-Inhibitor Complex

KI describes the affinity of an inhibitor for an enzyme.

70
Q

KI‘

A

Equilibrium Dissocation Constant for Enzyme-Substrate-Inhibitor Complex

KI describes the affinity of an inhibitor for an enzyme-substrate complex.

71
Q

Equation: KI

A
72
Q

Equation: KI‘

A
73
Q

Equation: Km-app

A
74
Q

How does noncompetitive inhibition effect the vmax and Km-app of a reaction?

A
  • vmax: Decreases
  • Km-app: Unaffected
75
Q

How does uncompetitive inhibition effect the vmax and Km-app of a reaction?

A
  • vmax: Decreases
  • Km-app: Decreases
76
Q

How does mixed inhibition effect the vmax and Km-app of a reaction?

KI’ > KI

A
  • vmax: Decreases
  • Km-app: Increases
77
Q

How does mixed inhibition effect the vmax and Km-app of a reaction?

KI > KI‘

A
  • vmax: Decreases
  • Km-app: Decreases
78
Q

Michaelis-Menten Equation: Noncompetitive Inhibition

A
79
Q

How does noncompetitive inhibition impact the Lineweaver-Burk plot?

A

Slope (Km/vmax): Increases
Y-Intercept (1/vmax): Increases
X-Intercept (-1/Km): Unaffect

Noncompetitive inhibition causes the Linewaver-Burk plot to pivot leftward about the x-intercept.

80
Q

How does uncompetitive inhibition impact the Lineweaver-Burk plot?

A

Slope (Km/vmax): Unaffected
Y-Intercept (1/vmax): Increases
X-Intercept (-1/Km): Increases

Uncompetitive inhibition causes the Linewaver-Burk plot to shift right.

81
Q

How does competitive inhibition impact the [S] required to reach vmax?

A

Increases [S] Requirement

Competitive inhibition shifts the Michaelis-Menten graph rightward.

82
Q

How does competitive inhibition impact the Lineweaver-Burk plot?

A

Slope (Km/vmax): Increases
Y-Intercept (1/vmax): Unaffected
X-Intercept (-1/Km): Increases

Competitive inhibition causes a leftward pivot about the y-intercept of the Linewaver-Burk plot.

83
Q

How does mixed inhibition impact the Lineweaver-Burk plot?

KI’ > KI

A

Slope (Km/vmax): Increases
Y-Intercept (1/vmax): Increases
X-Intercept (-1/Km): Increases

84
Q

How does mixed inhibition impact the Lineweaver-Burk plot?

KI > KI‘

A

Slope (Km/vmax): Increases
Y-Intercept (1/vmax): Increases
X-Intercept (-1/Km): Decreases

85
Q

Michaelis-Menten Equation: Mixed Inhibition

A
86
Q

How does competitive inhibition effect the vmax and Km-app of a reaction?

A
  • vmax: Unaffected
  • Km-app: Increases
87
Q

Equation: Lineweaver-Burk Plot

A
88
Q

Michaelis-Menten Equation: Competitive Inhibition

A
89
Q

Michaelis-Menten Equation: Uncompetitive Inhibition

A
90
Q

Michaelis-Menton Graph

A
91
Q

Which type(s) of inhibition can be overcome by increasing [S]?

A

Competitive Inhibition

92
Q

Noncompetitive Inhibition

KI + KI‘

A

A type of mixed inhibition in which the inhibitor compound has equal affinity for the free enzyme and the enzyme-substrate complex.

  • Noncompetitive inhibitors bind to the free enzyme or the enzyme-substrate complex at a location other than the active site (i.e. an allosteric site).
93
Q

Feedback Inhibition

A

An enzymatic regulatory mechanism in which a metabolic pathway’s end product serves as an inhibitor of the first enzyme in the pathway.

94
Q

Why do cooperative enzymes NOT follow standard Michaelis-Menten kinetics?

A

Substrate-binding to one subunit of a cooperative enzyme changes the substrate affinity at other enzyme subunits.

95
Q

What does sigmoidally shaped v0 vs. [S] graph indicate?

A

The enzyme’s catalytic activity can increase/decrease significantly over a small range of substrate concentrations.

96
Q

Structure: ATCase

A

C6R6 (Association of Two C3R3 Complexes)

  • C = Catalytic Subunit
  • R = Regulatory Subunit
97
Q

Zymogen

A

An inactive enzyme precursor that is activated by a proteolytic cleavage (either auto-cleavage or trans-cleavage) reaction.

98
Q

Model: Lock and Key

A

A model of enzyme catalysis in which rigid physical/chemical complementarity between the substrate and the enzyme are required for the reaction to occur.

99
Q

Model: Induced Fit

A

A model of enzyme catalysis in which an enzymatic conformational change will occur upon binding of the substrate.

100
Q

Conformational Selection

A

The process of stabilizing a particular enzyme conformation that is preferred for ligand binding.

101
Q

How do enzyme active sites provide a chemical environment that enables catalytic reactions to occur?

A
  • Active sites exclude excess solvents from the reaction environment.
  • Active sites bring reactive functional groups in close proximity to the substrate.
  • Active sites facilitate optimal/reactive orientations of the substrates.
102
Q

Do enzymes alter the free energy of a reaction?

A

No

Enzymes decrease the activation energy of a reaction, not the ground energy states of the reactants and products.

103
Q

Do enzymes alter the equilibrium position of a reaction?

A

No

Enzymes change the rate at which a reaction proceeds, not the ratio of products and reactants at equilibium.

104
Q

How do enzymes alter the reaction rate while not affecting the reaction equilibium?

A

Enzymes increase the reaction rate in both directions by the same amount, so the overall product-to-reactant ratio remains unchanged.

105
Q

How does adding an enzyme catalyst to a reaction increase the reaction rate?

A

The enzyme catalyst will decrease the activation energy of the reaction by (1) properly orienting the substrates for a reaction to occur and (2) providing an alternative path to product formation.

  • Enzymes stabilize the transition state of a reaction (to lower the activation energy barrier).
  • The entropy change of a reaction is reduced by properly orienting the substrates for a reaction.
106
Q

Cofactor

A

A small inorganic molecule (often a metal ion) that assists with the catalytic reaction mechanism within an enzyme’s active site.

Ex: Fe2+, Mg2+, Mn2+, Cu2+, Zn2+, Ni2+, K+, Se, Mo

107
Q

Holoenzyme

A

The catalytically active form of an enzyme that possesses a bound cofactor.

108
Q

Apoenzyme

A

The catalytically inactive form of an enzyme that lacks its cofactor.

109
Q

Coenzyme

A

An type of enzyme cofactor that possesses organic components.

Ex: NADH, FADH2, TPP, Biotin, CoA,THF, PLP, Lipoamid, Cobalamin

110
Q

Prosthetic Group

A

A type of coenzyme that is permanently associated with an enzyme.

Ex: Heme Group (Globin Proteins)

111
Q

Co-Substrate

A

A loosely bound molecule that is transformed into a co-product during the course of an enzymatic reaction.

Ex: NAD+/NADH, NADP+/NADPH

112
Q

Serine Protease

A

An enzyme containing a nucleophilic serine amino acid within the active site that functions to cleave the peptide backbone of dietary proteins.

Ex: Chymotrypsin

113
Q

Structure: Chymotrypsin

A

Chymotrypsin consists of three individual polypeptide chains that are covalently linked by disulside bonds.

  • The three individual polypeptide chains were formed from the cleavage of a single nascent polypeptide chain.
  • The enzyme active site sites on the enzyme’s surface (to allow easy access to the polymer substrates).
114
Q

Catalytic Triad

A

A set of three amino acids in serine proteases that form a hydrogen-bonded network required for catalysis.

115
Q

Catalytic Triad: Chymotrypsin

A
  • Ser195 (C Chain)
  • His 57 (B Chain)
  • Asp102 (B Chain)
116
Q

Reaction: Chemotrypsin-Catalyzed Proteolytic Cleavage

A

Polypeptide Substrate ⟶ Carboxyl-Terminal Fragment + Amino-Terminal Fragment

  • Phase 1: The polypeptide substrate is cleaved to release the carboxyl-terminal fragment.
  • Phase 2: The enzyme active site is regerated via release of the amino-terminal fragment.
117
Q

Mechanism: Chemotrypsin-Catalyzed Proteolytic Cleavage

(Detailed)

A
  1. The polypeptide substrate binds to the enzyme active site.
  2. The aromaticHis57 Nitrogen deprotonates Ser195 to enable the Ser195 Oxygen to nucleophilically attack the polypeptide’s carbonyl Carbon (to form an Oxyanion intermediate).
  3. Electron rearrangment at the Oxyanion intermediate causes the polypeptide substrate to deprotonate His57 (and cleave its Ctetrahedral—N bond).
  4. The carboxyl-terminal fragment is released from the enzyme active site.
  5. An H2O molecule enters the enzyme active site.
  6. The aromaticHis57 Nitrogen deprotonates the H2O molecule to create an OH– nucleophile.
  7. The OH– nucleophile attacks the Acyl-enzyme intermediate’s carbonyl Carbon (to form an Oxyanion intermediate).
  8. Electron rearrangment at the Oxyanion intermediate causes the polypeptide substrate to deprotonate His57 (and cleave its Ctetrahedral—O bond).
  9. The amino-terminal fragment is released from the enzyme active site (to regenerate the functional catalytic triad within the enzyme active site).
118
Q

Mechanism: Chemotrypsin-Catalyzed Proteolytic Cleavage

(Brief)

A
  1. The polypeptide substrate binds to the enzyme active site.
  2. His57 removes a proton from Ser195 to enable the Serine Oxygen to nucleophilically attack the polypeptide’s carbonyl Carbon.
  3. His57 protonates the polypeptide’s amino group to cleave the polypeptide.
  4. The carboxyl-terminal fragment is released from the enzyme active site.
  5. An H2O molecule enters the enzyme active site.
  6. His57 deprotonates the H2O molecule to created an OH– nucleophile that attacks the Acyl-enzyme intermediate’s carbonyl Carbon.
  7. His57 protonates Ser195 to cleave the acyl-enzyme intermediate and relase the amino-terminal fragment from the enzyme active site.
  8. The functional catalytic triad is regenerated within the enzyme active site.
119
Q

What purpose does the Oxyanion hole serve in Chemotrypsin-catalyzed cleavage reactions?

A

The Oxyanion hole stabilizes Oxyanion (O–) intermediates formed during the proteolytic cleavage reaction.

Oxyanion Hole: Terminal Hydrogens of Gly193 and Ser195.

120
Q

Rate Constant

k

A

A numerical constant that reflects how quickly a substrate is converted into product as a function of time (under a defined set of conditions).

  • First-Order Reaction: s–1
  • Second-Order Reaction: M–1s–1
  • Third-Order Reaction: M–2s–1
121
Q

Equation: Velocity of Reaction

A
122
Q

Steady-State Condition

A

A state of a reaction in which [ES] remains relatively constant over an intial reaction period.

123
Q

Michaelis-Menten kinetics are only valid for enzyme reactions under which conditions?

A
  • Steady-State Conditions
  • First-Order Reaction
  • Single Enzyme
124
Q

Km

A

The substrate concentration at which the reaction rate is one-half its maximum value.

Km: [S] @ 0.5(vmax)

125
Q

What does a low Km value indicate?

A

The enzyme has a high catalytic activity (i.e. is highly effective) at lower substrate concentrations.

(A high Km value indicates that the enzyme has high catalytic activity only at higher substrate concentrations.)

126
Q

Equation: Michaelis-Menten Curve

A
127
Q

How does increasing [S] (while keeping all other variables constant) impact the Lineweaver-Burk plot?

(No Inhibition)

A

The plot pivots rightward about the x-intercept to yield a lower y-intercept and a less steep slope.

  • Km: Remains Constant
  • vmax: Increases
128
Q

Which types of enzymes cannot be modeled with Michaelis-Menten kinetics?

A
  • Enzymes with Cooperative Binding
  • Enzymes that Slowly Release Products
129
Q

Turnover Number

kcat

A

The maximum catalytic activity of an enzyme under saturating levels of substrate.

The kcat value is the number of conversions of substrate to product at a single active site per unit time (when the enzyme is saturated).

130
Q

Equation: kcat

A

[Et] = [E] + [ES]

131
Q

Can the kcat value provide information on the efficiency of an enzyme’s catalytic function?

A

No

An enzyme’s catalytic efficiency is best described by the specificity constant (kcat/Km).

132
Q

Specificity Constant

A

A ratio that measures the catalytic efficiency of an enzymatic reaction (to compare two different enzyme reactions or to compare the same enzyme reaction with two different substrates).

133
Q

How efficient is an enzyme with a high kcat and a high Km?

A

Moderately Efficient

134
Q

How efficient is an enzyme with a high kcat and a low Km?

A

Very Efficient

135
Q

Hemoglobin: Negative Allosteric Effectors

A
  • CO2
  • H+
  • 2,3-Biphosphoglycerate
136
Q

Heterotropic Allostery

A

An allosteric regulation mechanism in which the regulatory molecule affects the binding affinity of a different molecule(s).

Heterotropic allostery involves the regulatory molecule binding to a location distinct from the protein’s primary site (i.e. a secondary site).

137
Q
A

Biochemistry (7 decks)

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