The catalytic mechanism of chymotrypsin & measuring activity Flashcards

1
Q

Explain the hydrolysis of peptide bonds by chymotrypsin

A
  • Chymotrypsin binds weakly to peptide chain upstream of the target amino acid
  • The targetted amino acid (Phe, Tyr or Trp) fits into the binding pocket, so substrate binds more tightly
  • If substrate binding makes a good fit, the targetted peptide bond lines up with the catalytic components
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2
Q

How does chymotrypsin catalyze hydrolysis?

A

X-CO-NH-Y + H2O —> X-COO- + +NH3-Y

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3
Q

Explain peptide hydrolysis by H2O, without catalyst

A
  • H2O acts as a nucleophile, lone pair donates to electron deficient C
  • Neutral O is not a good nucleophile
    • It tends to hold onto its own electrons
    • It makes unfavourable O+ transition state
  • C maintains 8 valence electrons by allowing upper O to take back a bond
  • This leads to the oxyanion transition state
    • Note that C is now sp3 tetrahedral
  • Transition state may break down with N as leaving group
    • Oxyanion O returns the bond to C
    • Carboxylate C must give up a bond to maintain 8 valence electrons
  • If N takes back the excess electrons, the peptide bond breaks
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4
Q

How can the transition state break down?

A

Transition state can also break down by returning electrons to O

  • Carboxylate C may give up the excess bonding electrons to the original nucleophilic O
  • The C–O bond breaks, and H2O is a good leaving group
  • Reactants are back to the starting point, and no net reaction has occurred
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5
Q

How can chymotrypsin do better?

A

How can chymotrypsin do better?

The reaction is broken into two easy steps instead of one difficult one

In Step 1 a nucleophilic group -X: in the enzyme attacks the peptide C=O to split off the C-terminal half of the substrate, but leaves the N-terminal half covalently bonded to the enzyme group-X (acyl-enzyme intermediate) - acyl means acid

Step 2 brings in H2O to release the N-terminal half, and restores the enzyme group-X: to its original state

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6
Q

What is the catalytic triad?

A
  • 40 reactions per second for chymotrypsin; 1 reaction in 10 years for H2O
  • Chymotrypsin uses a better nucleophile in the form of the cataltic triad, three amino acids that line up side by side in correctly folded chymotrypsin and cooperate for maximum effectiveness
    • Asp 102 – negative charge favours a positive charged partner
    • His 57 would be positive if it could capture H+
    • Ser 195 could give up H+ if it shares a lone pair with a suitable atom
  • Combined effect makes Ser 195 into a better nucleophile
  • Transition state is stabilized by the oxyanion hole
    • Backbone N-H groups of Gly 193 and Ser 195
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7
Q

What is the catalytic reaction step one?

A
  • His 57 removes H+ at the moment of reaction to help Ser
  • –ve charge of Asp helps His act as base
  • The oxyanion hole helps pull O– into transition state by H- bonding to backbone NH groups of Ser and Gly 193
  • H-bonds are aimed at location matching a tetrahedral C
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8
Q

Step 1: __________ _______

What does His act as?

Ser 195 becomes?

Negative charge on Asp 102…

A

Step 1: Nucleophilic Attack

a) His 57 acts as a general base, removing H+ from Ser 195.
b) Ser 195 becomes a better nucleophile & attacks peptide C=O.
c) Negative charge on Asp 102 delocalizes positive charge on His 57.

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9
Q

How is the first transition site formed?

A

Formation of the First Transition State

a) Oxyanion hole pulls the O- into the transition state.

  • Complementary to the transition state.
  • favours tetrahedral carboxyanion configuration.
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10
Q

Explain the breakdown of the 1st transition state formation of what immediate?

A

Breakdown of the 1st transition state Formation of Acyl-Enzyme Intermediate

a) NH group of substrate acts as the leaving group
b) His 57 acts as a general acid, donating H+ to the leaving group.
c) C-terminal peptide leaves.
d) N-terminal peptide remains covalently bound: the acyl enzyme intermediate

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11
Q

Explain step 2: Nucleophilic attack

A

a) Water now enters the catalytic site.
b) His-57 acts as a general base removing H+ from water.
c) Water becomes a better nucleophile, attacks the acyl-enzyme C=O

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12
Q

Explain the formation of the second transition state

A

a) Oxyanion hole stabilizes the transition state configuration.

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13
Q

Explain the second transition state formation of products

A

a) His 57 acts as a general acid, donating H+ to Ser 195.
b) Breaks the Acyl-enzyme bond.
c) N-terminal peptide leaves.
d) Catalytic triad is regenerated.

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14
Q

Chymotrypsin catalytic cycle

A
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15
Q

Actual structure of catalytic site of chymotrypsin

A
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16
Q

What is the chymotrypsin catalytic cycle

A
  • At the end, chymotrypsin is back to its original state: reaction cycle can repeat 40 times per second
  • This mechanism was determined by structural studies, and by mutation studies that replaced selected amino acids
    • Replace Asp with uncharged Ala: 10% of normal rate
    • Replace His with Lys: 0.1% of normal rate
      • His has pKa = 6.5, able to act as both general base and general acid at pH 7
      • Lys, pKa = 10.2 is a good base, but poor acid
    • Replace Ser with Ala: little catalytic reaction
  • Catalytic mechanism of trypsin and elastase is identical
17
Q

Explain enzyme assay and detection

A
  • Enzymes speed up reaction rate in proportion to amount of enzyme present
  • Enzyme assay is the process of measuring enzyme-catalyzed reaction rate
  • Enzyme kinetics is the mathematical analysis of how rate varies as a function of substrate concentration: kinetics can be used to test reaction mechanisms
  • Measure rates: rate of disappearance of reactant or appearance of product

Equation format: enzyme name placed above, because it’s not consumed

Measure volume of O2 released

Measure rise in pH as [H+] decreases

Not easy to measure in this form

18
Q

Direct analysis of trypsin reaction products is too time-consuming what provides a better way?

A

Direct analysis of trypsin reaction products is too time consuming - artificial substrates provide a better way: true for many enzymes

  • Artificial substrate is a molecular “look alike” for the real substrate
  • The reaction product is distinctly coloured – easy to measure
  • Trypsin recognizes Lys and the peptide chain upstream of the Lys, but peptide after the lysine is less critical
  • Trypsin accepts any primary amino group in position after Lys or Arg; won’t accept proline, which has secondary amino group
19
Q

Some natural substrates show UV absorbance change after what?

A

Some natural substrates show UV absorbance change after conversion to product

lactate dehydrogenase

pyruvate + NADH + H+ ⇔ lactate + NAD+

  • Lactate dehydrogenase uses 2 H• atoms (H: + H+) from the reduced form of nicotinamide adenine dinucleotide (NADH) and transfers them to pyruvate to form lactate
  • This uses up NADH, which absorbs ultraviolet light at 340 nm, whereas the NAD+ product does not absorb
  • Overall absorbance decreases as reaction proceeds
20
Q

What do coloured UV-absorbing molecules contain?

A

Coloured or UV-absorbing molecules contain chromophores

  • Chromophores are parts of molecules with conjugated double bonds e.g. N–C=C–C=O, or aromatic rings, which absorb visible or UV light
  • Coloured compounds absorb between 400-700 nm
  • Natural biochemical chromophores frequently absorb in the UV range, 200-400 nm
  • Larger chromophores absorb at longer wavelength
21
Q

How is absorbance measured?

A

Absorbance is measured with a spectrophotometer

  • Beam has intensity Io; after light is absorbed, measured intensity is I
  • Absorbance A = log10(Io / I)
    • so if I = 10% of Io, A = log 100/10 = 1
  • Most spectrophotometers read out directly in absorbance units
22
Q

The absorbance of a sample is a measure of…

A

The absorbance of a sample is a measure of its concentration

  • Absorbance is proportional to sample concentration (Beer’s Law) and sample thickness (Lambert’s Law)
  • Absorbance A = εcl Beer-Lambert Law

c = sample concentration in mol L-1

l = sample thickness in cm, usually 1.00 cm

ε = the extinction coefficient, a constant for any given substance; for NADH, ε = 6200 L mol-1cm-1 at 340 nm

Sample calculation using absorbance

  • Lactate dehydrogenase reaction showed decrease in absorbance of 0.357 at 340nm due to conversion of NADH to NAD+
  • Sample is 1.00 cm thick
  • For NADH, ε = 6200 L mol-1cm-1 at 340 nm
  • Decrease in [NADH] = A / (ε × l) = 0.357 / 6200 = 5.76x10-5 mol L-1