Enzyme Kinetics Flashcards

1
Q

What is an Enzyme?

A

A substance (usually a protein) that increases the rate of a chemical reaction without itself being changed in the overall process.

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2
Q

What is Primary, Scondary, Tertiary and Quarternary Structure?

A
  • Primary Protein Structure: Sequence of chain of amino acids
  • Secondary Protein Structure: Hydrogen bonding of peptide backbone caues the amino acids to fold into a repeating pattern
  • Tertiary Protein Structure: Three-dimensional folding pattern of protein to side chain interactions
  • Quarternary Protein Structure: Prtein consiting of more than one amino acid chain
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3
Q

What are characterisitics of Enzymes?

A
  • Enzymes increase the rate of a reaction
  • Enzymes do not change the reaction equilibrium
  • Enzymes are specific for their substrates
  • Enzymes are regulated: Genetic (transcription/translation) & Allosteric
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4
Q

Why do labs measure enzymes?

A
  • During disease, tissue damage may lead to release of enzymes from tissues: e.g Release of aminotransferases (ALT and AST) indicate liver damage
  • Inborn errors of metabolism may be due to a deficiency of a particular enzyme: e.g PKU, due to deficiency of the enzyme phenylalanine hydroxylase.
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5
Q

Which factors affect enzymatic activity?

A
  • The rate of an enzyme-catalysed reaction is proportional to the concentration of the enzyme.
  • pH
  • Temperature
  • Type of buffer
  • [Cofactor]
  • [Substrate]
  • Incubation time
  • [Enzyme]
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6
Q

What are enzymatic rate measurement methods?

A

Fixed-time method: use enzymes with high affinities (low Km)

  • Measure substrate or product concentration at two time points and calculate the difference then divide by time period to obtain rate.

Continuous monitoring method: use enzymes with low affinities (high Km).

  • Reaction is monitored by continuously taking a rate measurement over a designated time period.
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7
Q

What are Units for Expressing enzymatic activity?

A
  • SI unit of activity – katal
    • The amount of enzyme which will convert 1 mole of substrate in 1 second.
  • International unit – U
    • The amount of enzyme which will convert 1 μmole of substrate in 1 Minute
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8
Q

What are steps involved in calculating enzyme activity?

A
  • Convert physical measurement (e.g. absorbance) made over a timed interval:
  • Divide change in absorbance by the time period of measurement(minutes) to give rate (ΔA/min).
  • Convert into substrate concentration units and derive rate for the enzyme-catalysed reaction: Divide by molar absorptivity (L/mol/cm) and cuvette path length (cm): [Beer-Lambert’s Law]
  • Convert to the concentration of enzyme units in the clinical sample: Multiply by 1,000,000 to convert from mol to μmol. Multiply by total reaction volume (mL) and divide by sample volume(mL) – ‘dilution factor’
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9
Q

What is the final equation for enzyme activity?

A

(ΔA/min x 1,000,000 x Total vol (mL)) / ε (L/mol/cm) x l (cm) x Sample vol (mL)

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10
Q

What is the function of enzymes?

A
  • Enzymes accelerate the approach to equilibrium between substrates and products by lowering the activation energy of the reaction.
  • No effect on position of equilibrium
  • The ‘standard free energy of activation’ (ΔG°‡) is the additional free energy that substrate molecules must have to attain the transition state.
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11
Q

What is the Michaelis-Menten Equation?

A
  • Describes effect of substrate concentration on rate of enzyme catalysed reactions.

E + S ⇌ ES ⇌ E + P

  • A few milliseconds after mixing enzyme (E) and substrate (S) the enzyme-substrate complex [ES] builds up and does not change
  • This is called the Steady State [ES] is constant)
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12
Q

What is Michaelis Menten Equation?

A

V = (Vmax [S]) / Km + [S]

v = initial reaction velocity at [S]

KM = the Michaelis constant

Vmax = the maximum possible initial reaction velocity

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13
Q

What is Km?

A
  • Km is the substrate concentration at half-maximal velocity
  • Km is a measure of the strength of the ES complex:
  • A high Km indicates weak binding; a low Km indicates strong binding
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14
Q

What happens at low concetrations of subtrate to enzymatic kinetics?

A
  • At low concentrations of substrate [S], the ini­tial velocity of the reaction (v) is directly propor­tional to the substrate concentration.
  • The equation approximates to v = Vmax [S] / Km which is a linear expression.
  • When reaction rate is proportional to a single concentration term it is said to follow first-order kinetics.
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15
Q

What happens at high concetrations of subtrate to the enzymatic kinetics?

A
  • At very high [S], for example when [S] >>> Km most of the enzyme exists as the ES complex.
  • The enzyme is saturated with substrate.
  • The equation approximates to v = Vmax [S] / [S] = Vmax
  • Rate becomes independent of [S] and constant (v = Vmax)
  • When reaction rate is independent of [S] it is said to follow zero-order kinetics.
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16
Q

What are other graphical solutions for the Michaelis-Menten Equation?

A

Lineweaver-Burk Plot

  • The double-reciprocal plot of Lineweaver-Burk. Plot of 1/v versus 1/[S] is linear

Plot of Hanes

  • [S]/v versus [S]

Plot of Eadie-Hofstee

  • v versus v/[S]

Direct linear plot of Eisenthal and Cornish-Bowden.

17
Q

What are the actions of inhibitors?

A

Inhibitors lower reaction velocity (v).

  • Decreasing Vmax
  • Increasing Km
18
Q

What are the main types of enzyme inhibtors?

A
  • Competitive inhibitors
  • Non-competitive inhibitors
  • Uncompetitive inhibitors
19
Q

What is the actions of Competitive inhibitors?

A
  • Usually structurally related to the substrate
  • Bind reversibly with the enzyme at or near active site
  • The Km is increased (this is a denominator in the Michaelis-Menten Equation, therefore v is reduced).
  • It is theoretically possible to overcome competitive inhibition at high [S].
20
Q

What is the actions of Non-Competitive inhibitors?

A
  • Bind reversibly to the enzyme at areas other than the active site.
  • Binding of inhibitor and substrate is independent, therefore can form complex between enzyme and inhibitor [EI] and alsobetween enzyme, inhibitor and substrate [EIS].
  • Inhibition cannot be overcome by increasing [S] since binding of the two sites is independent.
  • Km is unchanged, but Vmax is reduced (a numerator, therefore v is reduced)
21
Q

What is the actions of Uncompetitive inhibitors?

A
  • Uncompetitive inhibitor combines only with the ES complex.
  • The Vmax is reduced
  • Interestingly, the Km is also reduced, but this is insufficient to overcome the reduction in Vmax therefore overall v is reduced.