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BB10235 > Enzyme Kinetics > Flashcards

Enzyme Kinetics Flashcards

1
Q

What is a basic rate equation for reaction aA + bB –> p

A
  1. Rate= k[A]^a [B]^b

2. K is the proportionality constant- rate constant

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2
Q

Describe what the order of reaction means

A
  1. The order of reaction corresponds to the molecularity of the reaction- the number of molecules that must simultaneously collide in the elementary reaction
  2. A–>P is first order or unimolecular reaction
  3. 2A–>P or A+B–>P are second-order and bimolecular
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3
Q

How can you experimentally determine the order of a reaction

A
  1. Measure the initial rate of either product (P) formation or substrate (S) disappearance.
  2. Rate v = dP/dt or –dS/dt (mmol/min)
  3. Observed “v” depends on amount of active enzyme. Specific activity = mmol/min.mg protein
  4. Enzyme Unit = amount which gives 1 mmol Product/min
  5. Katal (SI unit) = amount which gives 1Mol/sec
  6. First order reaction A–> p
    V=-d[A]/dt = k[A]
  7. Second order reaction A + B–> P
    v= -d[A]/dt = - d[B]/dt = k[A][B]
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4
Q

What is the transition state

A
  1. H(a) + H(b)–H(c) –> H(a)–H(b) + H(c)
    As atom and molecule approach each other they increasingly repel one another and therefore usually fly apart
  2. If the system has sufficient kinetic energy to continue its coalescence it will cause the covalent bond of the H2 molecule to weaken until ultimately, if the system reaches the saddle point there is an equal probability that either the reaction will occur or that the system will decompose back to its reactants.
  3. At this saddle point the system is said to be at its transition state and therefore to be an activated complex
  4. Transition state is the point of highest energy on the reaction coordinate
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5
Q

What is the transition state theory of enzyme binding

A
  1. The binding of the transition state to an enzyme with greater affinity than the corresponding substrates or products.
  2. Enzymes mechanically strain their substrates towards the transition state geometry through binding sites in which undistorted substrates did not fit properly.
  3. Reactions that preferentially bind the transition state increase the its concentration and therefore proportionally increase the reaction rate.
  4. Multiple weak interactions are optimised in stabilising the transition state
  5. Binding to an enzyme stabilises the transition state, thus reducing the transition state energy
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6
Q

What is an enzyme assay

A
  1. Enzyme assays are laboratory methods for measuring enzymatic activity.
  2. They are vital for the study of enzyme kinetics and enzyme inhibition.
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7
Q

What is a direct assay

A
  1. Assay has a way of detecting P or S directly due to some property of either.
  2. E.g. If P is coloured then colour change will be monitored
  3. This can be detected using a spectrophotometer.- Some substances absorb at a certain wavelength.
  4. Product might fluoresce or be a gas (e.g. CO2)
  5. All can be measured continuously
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8
Q

What is a discontinuous direct assay

A
  1. The reaction is stopped at set time points and P or S measured (e.g. stopped with acid or alkali).
  2. The assay to measure P or S may quench the enzyme
  3. E.g. Glucose measured with the Hardings test
  4. reduces CuSO4 to Cu2O when boiled in alkaline solution
  5. oxidation of Cu2O by molybdate compex gives a blue or green
  6. molybdate complex measured by spectrophotometry
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9
Q

What are coupled assays

A
  1. Sometimes neither P nor S can be measured
  2. P can be consumed in another reaction and product of that reaction can be measured.
  3. 2nd enzyme must be in excess so that the rate limiting step is the one to be measured
  4. e.g. a-glycerokinase (GK) assay
  5. glycerol + ATP–>alpha-glycerol-phosphate–> Dihydroxy-acetone phosphate
  6. first reaction calatlysed by GK, second by a glycerol- phosphate dehydrogenase and NAD+ –>NADH + H+
  7. The production of NADH (from the cofactor NAD+) can be measured at 340nm
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10
Q

What are the 3 main assumptions of the Michaelis-Menten equation

A

E + S ES –> P + E

  1. Step ES to E+P is irreversible
  2. [ES] is in steady state- one can assume it is constant with a reasonable degree of accuracy
  3. 𝑘2≪𝑘(−1) rate of dissociation of substrate is much faster than the rate of formation of product
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11
Q

What is a hyberbolic substrate realtionship

A
  1. When you plot initial rate of reaction over substrate concentration you get a hyperbolic relationship between the rate of reaction and the concentration of substrate
  2. Starts of steep then plateaus
  3. Data fits to Michealis-Menten model
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12
Q

What can you derive from a Michaelis-Menten curve

A
  1. Vmax= velocity at plateau
  2. Km = susbtrate concentration at 1/2 Vmax
  3. When [S]»kM ; 𝑣≈𝑉𝑚𝑎𝑥, The initial rate is independent to [S]
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13
Q

What is Km

A
  1. Km is the concentration of a substrate at which the reaction occurs at half rate
  2. It is different between enzymes and for different substrates
  3. It alters with temperature and pH
  4. The higher the Km the higher the [S] needs to be to reach Vmax
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14
Q

Name the double reciprocal method used to determine the parameters of the Michaels-Menten equation

A
  1. Double-reciprocal plot/ Lineweaver-Burk
  2. 1/V over 1/[S]
  3. 1/[S] intercept is -1/[Km]
  4. Gradient = Km/Vmax
  5. 1/V0 intercept= 1/Vmax
  6. Disadvantage- most experimental measurements involve high [S] so are crowded on the left side of the graph
  7. Disadvantage- Small values of [S], small errors in Vo lead to large errors in 1/Vo and therefore large errors in parameters.
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15
Q

How can you define the catalytic constant of an enzyme

A
  1. Kcat= Vmax/ [ET]
  2. Kcat= turnover number- Number of reaction processes that each active site catalyses per unit time
  3. Kcat/KM can be a measure of an enzymes catalytic efficiency
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16
Q

What is an inhibitor

A
  1. Substances which alter the activity of an enzyme by combining with it in a way that influences the binding of substrate and/or its turnover number
17
Q

What is competitive inhibition

A
  1. A substance that competes directly with a normal substrate for an enzymatic-binding site
  2. Usually resembles the substrate so that it specifically binds to the active site but differs so is unreactive
  3. Inhibitor (I) and substrate (S) compete
  4. Binding of I and S mutually exclusive
  5. Inhibition reduced by increasing substrate concentration
18
Q

Describe the kinetics of competitive inhibition

A
  1. If [S] increases to ∞ then all I displaced. Therefore Vmax remains unchanged- inhibitor doesn’t affect turnover number of enzyme
  2. At all [S], it will move equilibrium to E from ES
    [E] [S] will increase so Km will appear to increase to Kapp, Decreases apparent affinity for substrate.
  3. KI is the dissociation constant of the EI complex
    KI= [I][E]/[EI]
19
Q

Describe the uncompetitive inhibition

A
  1. The inhibitor binds directly to the enzyme-substrate complex but not to the free enzyme
  2. Occurs when the inhibitor binds only with the ES to make ESI.
  3. Cannot yield products
  4. Not reversed by increasing substrate concentration
  5. Usually found in enzymatic reactions with two or more substrates.
20
Q

Describe the kinetics of uncompetitive inhibition

A
  1. Increasing [I] diminishes Vmax and Km
    but Km/Vmax remains constant
  2. Vmax reduced as catalytic rate reduced
  3. Lineweaver-Burk plot consists of family of parallel lines
21
Q

Describe noncompetitive/mixed inhibition

A
  1. Does not have a structure like substrate
  2. Binds to the enzyme but not active site
  3. Changes the shape of enzyme and active site
  4. Substrate cannot fit altered active site
  5. No reaction occurs
  6. Effect is not reversed by adding substrate
  7. Affect Kcat but not Km
22
Q

Describe the kinetics of noncompetitive/ mixed inhibition

A
  1. I and S binds at a different sites
  2. KM is not affected
  3. Vmax reduced as catalytic rate reduced
  4. All lines in Lineweaver-Burk intersect to left of 1/Vo axis
23
Q

Describe allosteric inhibition

A
  1. Simple inhibition insufficient for metabolic need= Needs to be highly sensitive e.g 90% Vmax to 10% Vmax would need 80 x [I] – very unlikely
  2. Require activation as well as inhibition
  3. Effector needs to be structurally unrelated to S
  4. Solution ; Allos = other, Steros = solid
    the effector binds at another site on the enzyme
  5. Can be negative (inhibitor) or positive (activator)
  6. Much more sensitive than simple inhibition
  7. Relatively small amounts of effector needed
  8. Relatively large effects
  9. Usually these are “non competitive”
  10. Do no compete with S for the active site
  11. Effector binding alters protein conformation
  12. Change transmitted to active site
  13. Activator improves substrate binding
    Inhibitor reduces substrate binding
24
Q

Describe the kinetics of allosteric inhibition

A
  1. Allosteric proteins are oligomeric- Substrate binds one subunit and effector to the other.
  2. V against [S] gives a sigmoidal curve- demonstrates co-operativity
  3. Does not follow Michaelis-Menten kinetics.
  4. A small change in [I] gives a big inhibition- Reduction in V but no change in Vmax.
  5. A small change in activator concentration [A] gives a large increase in V with no change in Vm
25
Q

What are the 4 main elements of enzyme nomenclature

A
  1. (1) First number= one of the 6 main divisions/classes
  2. (2) Second number= subclass
  3. (3) Third number= sub-subclass
  4. (4) Fourth number= serial number
26
Q

What are the 6 main classifications of enzymes

A
  1. Oxidoreductases- oxidation- reduction reactions
  2. Transferases- transfer of functional groups
  3. Hydrolases- hydrolysis reactions
  4. Lyases- group elimination to form double bond
  5. Isomerases- isomerisation
  6. Ligases- bond formation coupled with ATP hydrolysis
27
Q

What is the effect of pH on enzyme activity

A
  1. Regulates catalytic efficiency
  2. Substrate ionisation effects
  3. Protein structural changes
  4. Typically 5-9
28
Q

What is the optimum pH for pepsin, arginase and salivary amylase

A
  1. Pepsin- 2
  2. salivary amylase-7
  3. Arginase- 9.5
29
Q

What is the effect of temperature on enzyme activity

A
  1. Increase in Temp increases
    a) flexibility (movement) in the protein backbone- enzyme tertiary structure
    b) 1- Increases activity
    c) 2- Active site effects (reversible)
    d) 3- Denaturation

BB10235 (16 decks)

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