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06 FdSc Bioscience - Metabolism and Enzymology > Week 1 - enzymes > Flashcards

Week 1 - enzymes Flashcards

1
Q

Describe the common nature of enzymes and what gives rise to the 3D structure of the enzyme

A
  • Enzymes are proteins with catalytic activity
  • They are polypeptides
  • The polypeptide chain “folds” which gives rise to the 3D structure of enzymes
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2
Q

Describe the active site of the enzyme

A
  • The active site is part of the tertiary structure and is responsible for the catalytic activity of the enzyme
  • It is generally a hydrophillic cavity containing amino acids with side chains which are arrange to interact specifically with the substrate in an attractive manner
  • The indentation on the surface of the enzyme is complementary in shape to the substrate
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3
Q

Outline the specificity of enzymes

A
  • Enzymes are highly specific both in binding to substrate and catalysing reactions
  • Stereospecificity is due to a series of specific non-covalent enzyme-substrate binding interactions
  • The chirality of the active site (proteins consist mainly of L-amino acids) ensures that it is able to bind to one enantiomer of substrate
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4
Q

What is the chirality of the active site due to?

A
  • The chirality occurs because the amino acids in the polypeptide are mainly L-amino acids
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5
Q

What are the types of enzyme-substrate interactions used by enzymes?

A
  • There are four types of interactions used by enzymes
  1. Electrostatic interactions
  2. Hydrogen bonding
  3. van der Waals interactions
  4. Hydrophobic interactions
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6
Q

Name the different classes of enzyme structural types

A
  1. Metallo-enzymes
  2. Membrane-associated enzymes
  3. Glycosylated enzymes
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7
Q

Describe the metallo-enzyme structural type

A
  • Metallo-enzymes are enzymes that bind metal ions, which typically have a metal cofactor located at the active site of the enzyme
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8
Q

Describe the membrane-associated enzyme structural type

A
  • Membrane-associated enzymes are enzyme associated with biological membranes and consists of two classes:
  1. Extrinsic membrane proteins which are situated on the surface of the membrane
  2. Intrinsic (or integral) membrane proteins which are situated within the membrane bilayer
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9
Q

Describe the glycosylated enzyme structural class

A
  • Glycosylated enyzmes arise due to the attachment of carbohydrates to the peptide backbone of the protein
  • The attachments of the carbohydrate residues are typically required for full activity of the enzyme rather than participating in active site catalysis
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10
Q

Describe enzyme function

A
  • Once the substrate is bound the enzyme starts to catalyse its specific chemical reaction using active site catalytic groups and releases its product back into solution
  • They function to catalyse biochemical reactions
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11
Q

Describe the hallmarks of enzyme catalysis

A
  • Speed - enzymes catalyse reactions several folds faster than uncatalysed reactions
  • Selectivity - once the enzyme is bound to its substrate via specific binding interactions at the active site, they are able to recognise subtle changes in substrate structure
  • Specificity - enzymes select unique site of action within the substrate and therefore catalyse the reaction with precision
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12
Q

What are cofactors commonly described as?

A
  • The enzymes “chemical teeth”
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13
Q

What enzymes might have cofactors?

A
  • Enzymes that participate in oxidation-reduction reactions and group transfer processes are mediated by cofactors
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14
Q

Gives some examples of cofactors

How may the cofactors be associated with the enzyme?

A
  • Metal ions such as Zn2+
  • Organic molecules such as coenzymes - e.g. NAD+ in yeast alcohol dehydrogenase (YADH)
  • They are transiently associated with the enzyme, effectively functioning as a co-substrate
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15
Q

What are cofactors called that are permanently associated with their protein?

A
  • Cofactors which are permanently associated with their protein are called prosthetic groups
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16
Q

What happens to a coenzyme during an enzymatic reaction and what must happen afterwards?

A
  • Coenzymes that participate in an enzymatic reaction are chemically changed during the reaction and must be returned to their original state upon completion of the catalytic cycle
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17
Q

How is the chemical change that occurs in a prosthetic group different to that of a coenzyme?

A
  • The chemical change that occurs for a prosthetic group can be returned to its original state in a separate phase of the enzymatic reaction
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18
Q

What are some coenzyme precursors?

A
  • Many vitamins are coenzyme precursors
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19
Q

For the reaction given below, what is the order of the reaction and how do we define the instantaneous rate / velocity?

A
  • The reaction is first order
  • K1 is the rate constant for this reaction and the units are s-1
  • Rate / velocity (v) is defined in terms of the time (t) dependant production of protein P
    • The formation of P involves the loss of A
    • v is defined in terms of the time-dependant consumption of A
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20
Q

Describe the concentration-time graphs of a first order enzymatic reaction using the examples of A (substrate) and P (product)

A
  • Transformation of substrate to product is unaffected by concentration
  • Converstion of A to P results in the concentration of A decreasing exponentially with time
  • k1 is directly proportional of [A] so reaction is 1st order
  • [A] decreases exponentially with time so plotting ln([A]) versus t produces the straight line which is characteristic of a 1st order reaction
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21
Q

What is one of the characteristics of a 1st order reaction?

A
  • The half life (the time taken for A to decompose) is constant and independant from the initial [A]
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22
Q

Write the equation for a second order reaction and explain why it is considered to be second order

A
  • When two product react together to form a product the reaction is considered to be second order
  • The reaction is considered to be second order because the rate is proportional to the second power of the concentration
  • The rate of this reaction is proportional to the consumption of A and B and to the formation of P
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23
Q

Write the rate equation for a second order reaction using A and B as substrates and P as the product

A
  • The reaction is considered to be second order because the rate is proportional to the second power of the concentration
  • The units for this equation are s-1M-1
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24
Q

Describe the difference between a first order and a second order concentration time graph

How is the half life for a second order reaction defined?

A
  • The second order reaction graph has a steeper curve compared to the first order curve
  • The half life for the second order reaction is defind as t1/2 = 1/(k1[A0]) which is dependant on the initial reactant concentration
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25
Q

Define the term catalyst

A
  • A species which accelerates the rate of reaction whilst remaining unchanged at the end of the reaction
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26
Q

What does the transition state theory state?

A
  • As molecules collide, a chemical reaction occurs and they are momentarily in a less stable state than either of the reactants or the products
  • During the transition state, the activated complex causes the potential energy to increase which creates an energy barrier between the reactants and the products
    • The products can only be formed when the colliding reactants have enough energy to overcome the energy barrier (activation energy)
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27
Q

What is the energy barrier between the reactants and the products known as?

What is the significance of a large and small value of this?

A
  • Activation energy (Ea)
  • The greater the activation energy the lower the number of effetive collisions, the rate of reaction decreases as the Ea increases
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28
Q

What does ΔG refer to?

A
  • ΔG is a measure of free energy and defines the free energy difference between the products and the reactants
  • If ΔG is negative (less than 0) then the reaction can take place spontaneously and the reaction is exergonic
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29
Q

What is the current molecular model that is used to explain the catalytic nature of enzymes?

Explain this molecular model

A
  • The induced-fit hypothesis
  • The enzyme binds to its specific substrate to form an enzyme substrate complex which causes the structure of the enzyme to become distorted
    • This pulls the complex into a transition state conformation
  • The transition state conformation in turn causes a reduction in the energy required for the conversion of a specific reactant into a product and increases the rate of reaction by lowering the Ea
    • Therefore increasing the # of effective collisions resulting in the formation of the product
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30
Q

What is important regarding the reactions that enzymes catalyse?

A
  • Enzyme-catalysed reactions are multistep sequences involving one or more intermediates
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31
Q

Besides the induced-fit molecular model, what is the other model?

A
  • Enzymes can interact with substrates in the lock-and-key model
  • The enzyme is shaped to fit precisely (complementary) to the substrate which ensures that the correct substrate binds specifically to the active site
32
Q

How can enzyme kinetics be summaried?

A
  • Enzymes increase the rates of reactions but do not alter their equilibrium constants
33
Q

What is the Michaelis-Menten equation?

A
  • It is critical to enzyme kinetics and the equation is subject to two constants:
  1. Km - the Michaelis-Menten constant
  2. kcat - the catalytic constant
34
Q

Who initiated the study of enzyme kinetics, when and what reactions did they propose?

A
  • Adrian Brown initiated the study of enzyme kinetics in 1902
  • He proposed that an overall reaction is composed of two elementary reactions:
    1. The substrate (S or A) forms a complex with the enzyme (E)
    2. The enzyme substrate complex (EA) decomposes to products (P) and the enzyme (E)
35
Q

What is another word for the enzyme-substrate complex?

A
  • EA complex is also known as the Michaelis complex
36
Q

In the equation below, how is the formation of the product defined?

How does this relate to the [enzyme-substrate complex]?

A
  • The formation of the product is defined as the dissociation rate (k2) of the ES
    • k2 is more commonly referred to as kcat
  • The [EA] is defined in the equation:
    • v = kcat [EA]
37
Q

Regarding the equation below, what is assumed about k1 , k-1 and k2 ( aka kcat)?

A
  • It is assumed that the dissociation rate (kcat) if the ES complex is slow compared to k1 (association) and k-1 (redissociation)
38
Q

Draw a concentration / time graph showing [P], [A], [EA] & [E]

Describe what the graph shows

A
  • Graph shows the consumption of the substrate (A), the formation of product (P) and the concentration of the free enzyme (E) and the Ea complex change over time
  • After the initial period the concentration of EA remains relatively constant (steady state) until the substrate is nearly fully consumed
39
Q

Write an equation which describes the steady state in an enzyme reaction

What is the significance of the steady state?

A
  • See equation below
  • Therefore the association of the ES complex and the dissociation/redissociation reactions are equal:
    • k1 [E][S] = k-1 [ES] + kcat [ES]
40
Q

How can k1 [E][S] = k-1 [ES] + kcat [ES] be rearranged to give [ES] on the left?

How can this equation be simplified?

A
  • The equation can be rearranged to give [ES] = k1 [E][S] / k-1 + kcat
  • It can be simplified to [ES] = [E][S] / Km where km is the Michaelis-Menten constant
  • This simplification can occur because k1 , k-1 and kcat are combined into one term called the Michaelis-Mentent constant (km) - see below
41
Q

Outline km what it is used for and what its limitation is

A
  • The Michaelis-Menten constant (km) is a parameter and its units are mol l-1 or M-1
  • km is the concentration of substrate at which half-maximal rate is observed
  • It is commonly used as a guide to how tightly the enzyme binds to its substrate
    • A substrate weakly bound by an enzyme will have a larger km than an enzyme with a high affinity for the substrate
  • km is not a true dissociation constant for a substrate because it is dependent on the rate constant kcat
42
Q

What is kcat (S-1) referred to as?

A
  • It is referred to as the turnover rate of the enzyme, this is a measure of the maximum catalytic formation of the product under saturating substrate conditions per unit time per unit enzyme
  • The higher the kcat the faster catalytic events occur
43
Q

Write an equation that describes the total amount of enzyme found over the course of a reaction

A
  • The amount of free enzyme (E) and ES varies over the course of the reaction, however the total amount of enzyme (Et) reamains constant
    • E = Et – ES
44
Q

How is the equation E = Et – ES rearranged using the Michaelis-Menten constant?

How is this used to derive the Michaelis-Mentent equation?

A
  • [ES] = [Et][S] / Km + [S]
  1. The max rate of reaction (vmax) occurs where all available enzyme is bound to substrate, i.e.
    • [ES] = [Et]
  2. Therefore saturating conditions produces the equation
    • vmax = kcat [E]
  3. Leading to the Michaelis-Menten equation:
    • v = (vmax [A]) / (km + [A])
45
Q

What is vmax?

A
  • It is a theoretical value describing the maximum rate of reaction which occurs when all available enzyme is bound to substrate in a catalytic reaction
  • It is theoretical because at a given time it requires all enzyme to be firmly bound to substrate
    • Therefore vmax is approached at high [S] but it is never reached
46
Q

How is enzyme efficiency defined?

A
  • Enzyme efficiency (s-1 (mol l-1)-1) is defined by the ratio of kcat / km and can be taken as a measure of substrate specificity
    • When kcat is significantly greater than k-1 the catalytic activity of the enzyme is extremely fast and the enzyme efficiency depends on the ability to bind the substrate
47
Q

When defining enzyme efficiency (s-1 (mol l-1)-1), how do we determine values to kcat and km?

A
  • These values can be determined for a specific enzyme by measuring the rate of the enzymatic reaction across a range of substrate concentrations
    • At high [S] ([S]>>km ) the rate equation is v = kcat [E0] and therefore the enzyme is fully saturated with S
    • At [S] = km the enzyme is 50% saturated with substrate and therfore the rate equation is v = k**cat [E0] /2
    • At low S concentration the rate equation is v = (kcat / km)[E][S]
      • Under these conditions the rate of reaction is dependent on the efficiency of the enzyme to bind to the substrate at that concentration
48
Q

How can the Michaelis-Menten parameters be approximated graphically?

What are the advantages and disadvantages for the methods discussed?

A
  • They can be approximated from a non-linear plot of rate versus [S] but a more accurate way is to use either a Lineweaver-Burk or Eadie-Hofstee plots which both product linear lines
  • A disadvantage of the two latter plots is that the errors in determining v at low [S] is magnified
    • Despite this the effect of enzyme inhibitors is apparent on linear plots
49
Q

Draw a Lineweaver-Burk plot

A
50
Q

Draw an Eadie-Hofstee plot

A
51
Q

When does enzyme inhibition occur?

What are the different types of enzyme inhibition?

A
  • It occurs if a molecule besides the substrate binds at the active site and prevents enzymatic reaction from occuring
  • There are two types of enzyme inhibiton:
    1. reversible
    2. irreversible
52
Q

Outline reversibile enzyme inhibition

A
  • An inhibitor is bound non-covalently at the active site which obscures substrates from binding
  • Competitive inhibition is the most common type of reversible inhibition observed where an inhibitor competes with a substrate for the same binding site on a specific enzyme
  • A non-competitive inhibitor binds to a second site on the enzyme and acts by reducing the turnover rate of the reaction
    • Therefore the binding of the inhibitor and substrate is independent and binding of the non-competitive inhibitor results in total inhibition of the catalytic reaction
53
Q

Outline irreversible enzyme inhibition

A
  • This occurs when an inhibitor binds at the active site, reacts with specific groups and forms a covalent bond
  • The active site becomes permanently inactivated because the active site is irreversibly blocked
  • An uncompetitive inhibitor binds only to the enzyme-substrate complex
54
Q

Draw Lineweaver-Burk plots for competitive, non-competitive and uncompetitive inhibition

A
55
Q

What is the difference between an uncompetitive and a non-competitive inhibitor?

A
  • A non-competitive inhibitor binds to a second site on the enyzme
  • An uncompetitive inhibitor binds only to the enzyme-substrate complex
56
Q

How are enzyme classified?

A
  • They are divided into 6 major classes based on the nature of the catalytic reaction
  • Nomenclature originally determined by the Enzyme Commission
    • All enzymes are assigned and EC number
  • Different classes of enzyme do not take into account the homology, protein structure or chemical mechanism
57
Q

Define homology

A
  • The similarity in structure, physiology or development of different species of organisms bases upon their descent from a common evolutionary ancestor
58
Q

What is the first class of enzymes?

A
  • Class 1: oxidoreductases
    • Catalyse oxidation reactions
    • Catalyse the transfer of hydrogen, oxygen or electrons from one substrate to another
  • Also called oxidases, dehydrogenases or reductases
  • Oxidoreductases are involved in redox reactions which require an electron donor/acceptor to complete the reaction
59
Q

What is the second class of enzyme?

A
  • Class 2: transferases
    • Catalyse group transfer reactions excluding oxidoreductases
60
Q

What is the third class of enzyme?

A
  • Class 3: hydrolases
    • Catalyse hydrolytic reactions
    • Include: peptidase/proteases, lipases, nitrilases and esterases
    • Catalyse hydrolytic cleavage of bonds including: C-C, C-O and C-N bonds
61
Q

What is the fourth class of enzymes?

A
  • Class 4: lysases
    • Catalyse non-hydrolytic removal of functional groups from substrates
    • Include decarboxylases, aldolases and synthases
62
Q

What is the fifth class of enzyme?

A
  • Class 5: isomerases
    • Catalyse isomerisation reactions
    • Includign racemisations and cis-trans isomerisations
63
Q

What is the sixth class of enzymes?

A
  • Class 6: ligases
    • Catalyse the synthesis of a variety of bonds, which is coupled with the breakdown of energy containing substrates
64
Q

Outline all six classes of enzyme including their class, reaction type and important subclasses

A
65
Q

What is the significance of the regulation of metabolic processes and how are they controlled?

A
  • Controlling metabolic processes is dependent on regulating the enzymes responsible for mediating the reactions in the pathway
  • Regulating metabolic pathways is fundamentally important for living organisms, therefore the ability to regulate enzymatic acitivities is critical for survival
66
Q

What are the three ways in which enzymes can be regulated?

A
  1. Regulating the amount of enzyme
  2. Controlling the type of enzyme
  3. Modulating the activity of the enzyme
67
Q

Outline the regulation of the amount of enzyme

A
  • There are two main methods used to change the amount of enzyme present in a cell:
    1. Change the rate of enzyme synthesis
    2. Change the rate of enzyme degradation
  • Each of these rates is controlled by the cell
  • Changing the amount of enzyme changes the rate of reaction because vmax is directly proportional to the enzyme concentration and the velocity of reaction is directly proportional to the vmax
68
Q

Outline how the type of enzyme produced can be controlled

Give an example of this in action

A
  • Different isoenzymes are products of different genes which can catalyse a specific reaction
  • Isoforms are proteins that can be synthesised from more than one isoenzyme
  • Different isoenzymes may have different affinities for the same substrate
    • E.g. hexokinase and glucokinase both catalyse the phosphorylation of glucose but they differ in their affinity for glucose
69
Q

Outline how the activity of an enzyme may be modulated

A
  • Can be modulated directly through conformation or structural changes
  • Changing activity can involve changes in km or kcat
  • Possible mechanisms include:
    • Covalent modification - e.g. phosphorylation
    • Protein-protein interaction
    • Competitive inhibition
    • Non-competitive inhibition
    • Allosteric effectors
70
Q

Typically, what are regulatory enzyme?

A
  • Regulatory enzymes are usually allosteric enzymes
  • A small molecule known as an effector or modulator alters the activity of the allosteric enzyme
71
Q

How are regulatory enzymes controlled?

A
  • A small molecule known as an effector/modulator alters the activity of the allosteric enzyme
    • Effector uses non-covalent forces to bind reversibly to a regulatory site, distinct from the active site, causing a change in shape of the enyzme - as a result the activitiy of the active site is altered
72
Q

What are the different types of effector with regards regulatory enzymes?

A
  • A negative effector decrerases enzymatic activity or inhibits the enzyme
  • A positive effector increases enzyme activity
  • When the substrate serves as an effector, the effect is said to be homotropic
    • ​In these circumstances the presence of a substrate molecule at one site on the enzyme alters the catalytic properties of the other substrate-binding site
73
Q

With regards regulatory enzymes, what can effector-binding alter?

A
  • Effector-binding can alter the km or vmax of the enzyme
  • V-type effectors alter the vmax and K-type effectors alter the km
    • ​The majority of K-type effectors regulate activity as a result of affecting the cooperative behaviour of the protein
74
Q

How can changes in substrate-binding affinity affect the activity of an enzyme?

A
  • Rate of enzymatically catalysed reaction is directly proportional to the [ES], which in turn, changes with the enzyme concentration, substration concentration and the enzymes substrate-binding affinity
  • Enzyme catalytic activity can therefore be controlled by changes in the substrate-binding affinity.
75
Q

Identify and describe the environmental factors that affect enzyme activity

A
  • One of the most important environmental factors is substrate concentration
    • Low [S] enzymatic activity is reduced due to reduced contact with substrate
    • High [S] there is a greater number of successful collisions between the enzyme and the substrate, thus reaction velocity is greater
    • Further increases in [S] do not increase velocity becuase the available enzymes are occupied - the enzyme is saturated and operating at vmax
  • pH - enzyme functions most efficiently at optimum pH, when pH deviates from this the enzymatic activity is reduced and the enzyme may become damaged
  • Temperature - enzyms have optimum temperature, is temperature is greater the enzyme structure is disrupted and activity is lost - denaturation.
  • Denaturation may be caused by pH and temperature

06 FdSc Bioscience - Metabolism and Enzymology (8 decks)

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