lecture 47 - transition state stabilization and substrate recognition Flashcards
what is the primary way in which enzymes make reactions go faster?
stabilization of the transition state
give a general reaction for the conversion of a substrate into its products via an enzyme
then show the gibbs free energy graph
E + S = ES = EP* + Q = EP + Q = E + P + Q
- small bump from E+S–>ES
- bigger bump with two peaks (tetrahedral intermediate in the center of the peaks) from ES –> EP*
- another bigger bump with two peaks tetrahedral intermediate in the center of the peaks) from EP* –> EP
- small bump from EP –> E + P + Q
- recall: bump = transition state
are tetrahedral intermediates higher or lower in energy? why?
higher in energy due to the formation of charge
how do serine proteases stabilize the transition state and tetrahedral intermediates without stabilising ES too much? (2)
(1) the partial negative charge on the carbonyl O is stabilized by 2 h-bonds to 2 backbone NH groups (gly and ser ** same ser from the catalytic triad)
(2) the conformaiton of the carbonyl group to C-O(-) causes a change from planar to tetrahedral and from a short c=o bond to a longer c-o bond
describe in more detail the effects of the conformation of the C=O to C-O(-)
- the O from the carbonyl shift position as the nucleophile attacks to form the tetrahedral intermediate
- also a the same time, the O gains a negative charge (becomes an oxyanion)
- this leads to the formation of the oxyanion hole
- i.e. the oxyanion forms h-bonds only to the transition state and the tetrahedral intermediate
- this results in preferential stabilization of the transiton state/tetrahedral intermediates
- therefore, decreasing the ΔG of the transition state
what does the enzyme use to position the substrate for reactions?
binding energy
what does the enzyme provide to stabilize the transition state? (3)
- enzymes provide interactions that only occur in the transition state or in high energy intermediates
- enzymes provide electrostatic stabilization of the charges found in the transition state/tetrahedral intermediate
- some enzymes add strain to the substrate to distort its conformation towards the transition state structure, decreasing strain as the transition state forms
describe the specificity of chymotrypsin
- cuts after aromatics (F,W,Y)
- has a large, deep, and mostly hydrophobic pocket as the regocnition domain for aromatics
- when the peptide is bound to the enzyme it is in the beta conformation (extended backbone conformation)
- then the enzyme cleaves the peptide after the aromatic residue
- “R2” is not recognized, where R2 = the following aa side chain
describe the specificity of trypsin
- cuts after R, K
- has a large deep pocket with a negative charge at the bottom
- when the peptide is bound to the enzyme it is in the beta conformation (extended backbone conformation)
- then the enzyme cleaves the peptide after the pos residue
- “R2” is not recognized, where R2 = the following aa side chain
describe the specificity of elastase
- cuts after small reisdues (S, V, T)
- has deep but narrow pocket compared to trypsin
why are the recognition pockets different for serine proteases and elastase?
- because of an aa substitution
- pocket of serine proteases is made by 2 gly residues
- pocket of elastase subsititues gly for val and thr
why don’t trypsin and chymotrypsin cut after small aa residues?
because there are fewer optimal reactions (bc of smaller side chain):
- this decreases binding affinity (decreases recognition of the substrate)
- this also leads to a less optimal position in the transition state
all interactions between the enzyme and the substrate/transition state are important for what?
what does this mean for the transition state binding?
- positioning
- therefore the more precise the fit of enzyme and the transition state results in a decrease in activation energy
can proteases recognize longer sequences?
yes
what type of protease are TEV proteases?
cys proteases