Hi Natasha,

1. What is the difference between E1 and E2 reactions, and how can you tell if they occurred?

Can you refer to the question where this comes up? E1 and E2 are relatively low yield so often there is some explanation in the question.

E1 is an elimination reaction that proceeds through a carbocation intermediate (analgous to Sn1 but with the deprotonation of a H forming an alkene at the end). You can tell it occurs because you started with an Sn1 type substrate and formed the most stable alkene possible. E1 reactions tend to have to be in strong acid with no nucleophiles in solution.

E2 is an elimination reaction that proceeds in one step by deprotonating an alpha proton and eliminating a good leaving group. You can tell an E2 reaction occured because the leaving group and the proton have to be on opposite sides (one wedge one dash etc) in the substrate. E2 reactions tend to occur in good sn2 substrates and in base with good Sn1 substrates.

2. What is the difference between allosteric enzymes and enzymes that follow MM kinetics, and how can you tell if an inhibitor is acting on an allosteric enzyme? (Question 44 in the Chemistry and Physics section)

For the graph interpretation of this equation, the important way to identify allosteric inhibition is if you can see the shape of the curve (or they'll tell you it is allosteric and expect you to know the shape of the curve from that). If the curve is sigmoidal, the interaction is allosteric.

For michaelis menten kinetics, you get a logarithmic type curve which will shift up and down with Vmax.

However, for allosteric interactions, you get a sigmoidal cure (it looks kind of like an S like in the question stem), which mostly shifts left or right. For allsoteric inhibition, a rightward shift means that substrate concentration needs to be higher for the same velocity, which corresponds to inhibiton.

3. For question 48 in the Chemistry and Physics section, were we supposed to have memorized the structure of GHB? How else could we solve this?

You did not need to have memorized GHB for this. In the question they mention that GHB stands for gamma-hydroxy butanoic acid. From this, you should be able to infer that it is a carboxylic acid which is 4 carbons long and has an alcohol. If you know alpha beta gamma numbering for chemistry, you could also infer that there is an alcohol 3 carbons away from the carboxylic acid. Y

You would then oxidize this alcohol twice to wind up with the dicarboxylic acid species. In short, this section did not expect you to memorize GHB but is just testing a bit of IUPAC nomenclature and some knowledge of oxidation chemistry.

Please let me know if any of this is unclear.

Katt