section bank cp #38

[goodtime]

pic attached
I don't understand A, why would there be a repulsive force and what is it between the variant and substrate?

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[theonlytycrane]

When swapping out Glu for Asp, the enzyme melts with less heat applied (it's less stable) and doesn't exhibit any activity. The correct answer choice will say something about the protein being less stable and messed up.

This is independent of the substrate.

 

[bobeanie95]

You can answer this question via process of elimination

B- False, both glutamate and aspartate will have a negative charge so it won't affect the charge
C- False, aspartate is smaller so there would actually be less steric hinderance
D- False, both are acidic residues so it won't affect its capacity to act as a base.

A- not a great answer but the only logical one. According to the passage "A computer modeling of the side chain variant showed that the side chain of D147 was placed in close proximity of D348." In other words, the substitution at residue 147 caused the aspartic acid to be placed close to another aspartic residue (residue 348), resulting in electrostatic repulsion.

 

[deleted647690]

You can answer this question via process of elimination

B- False, both glutamate and aspartate will have a negative charge so it won't affect the charge
C- False, aspartate is smaller so there would actually be less steric hinderance
D- False, both are acidic residues so it won't affect its capacity to act as a base.

A- not a great answer but the only logical one. According to the passage "A computer modeling of the side chain variant showed that the side chain of D147 was placed in close proximity of D348." In other words, the substitution at residue 147 caused the aspartic acid to be placed close to another aspartic residue (residue 348), resulting in electrostatic repulsion.

Wouldn't there also have been electrostatic repulsion with the E residue? Why would there be any more repulsion when switching D for E? They are almost the same amino acid, minus one carbon.

 

[letsgetstarted1234]

bumping this instead of bringing up a new thread.

Why wouldnt C work in terms of steric hindrance?

If it's shorter now, wouldn't it crowd around other side chains?

if you saw it as steric hindrance from the active site then it makes some sense i guess since shorter means its further away from the active site.

also still confused on how it would be repulsions (A)?

 

[sapientnarwhal]

bumping this instead of bringing up a new thread.

Why wouldnt C work in terms of steric hindrance?

If it's shorter now, wouldn't it crowd around other side chains?

if you saw it as steric hindrance from the active site then it makes some sense i guess since shorter means its further away from the active site.

also still confused on how it would be repulsions (A)?

Again, bobeanie95 provided an excellent explanation. The passage offers the info necessary to solve this question. The passage states that the E147D substitution results in a novel placement of said residue near D348. There is simply no reason to believe that it would crowd around other side chains, because the passage explicitly mentions the new interactions/placement.

The steric hindrance explanation also does not account for all "behaviors" of the variant, specifically the reduced melting temp. This explanation only accounts for the absence of catalytic activity due to lower substrate affinity.

The repulsive interaction likely reduces dimerization (if I remember correctly, the active site resides at the dimer interface), is consistent with reduced melting temps (instability), and also accounts for reduced catalytic activity (if enzyme denatures / is destabilized at normal physiological temperatures - i.e. the optimal temp for enzyme activity - then we expect impaired catalytic activity).