The catalytic mechanism of chymotrypsin & measuring activity Flashcards
Explain the hydrolysis of peptide bonds by chymotrypsin
- 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
How does chymotrypsin catalyze hydrolysis?
X-CO-NH-Y + H2O —> X-COO- + +NH3-Y
Explain peptide hydrolysis by H2O, without catalyst
- 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
How can the transition state break down?
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
How can chymotrypsin do better?
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
What is the catalytic triad?
- 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
What is the catalytic reaction step one?
- 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
Step 1: __________ _______
What does His act as?
Ser 195 becomes?
Negative charge on Asp 102…
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.
How is the first transition site formed?
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.
Explain the breakdown of the 1st transition state formation of what immediate?
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
Explain step 2: Nucleophilic attack
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
Explain the formation of the second transition state
a) Oxyanion hole stabilizes the transition state configuration.
Explain the second transition state formation of products
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.
Chymotrypsin catalytic cycle
Actual structure of catalytic site of chymotrypsin
What is the chymotrypsin catalytic cycle
- 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
Explain enzyme assay and detection
- 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
Direct analysis of trypsin reaction products is too time-consuming what provides a better way?
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
Some natural substrates show UV absorbance change after what?
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
What do coloured UV-absorbing molecules contain?
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
How is absorbance measured?
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
The absorbance of a sample is a measure of…
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
