Q5: As a non-beetle, I can't speak for male beetles, but I will try!
The passage tells us that position 6 favours (6R,7S) and position 8 favours (6S, 7R).
Since the male has a higher ratio of position 6, we expect the male beetle to INCORPORATE more (6R,7S).
The female has a higher ratio of position 8, so we expect the female beetle to INCORPORATE more (6S,7R).
Since males are attracted to females, we expect the male beetle to be attracted to the female beetle's (6S,7R) isomer more than their own isomer.
Therefore, B is the best answer.
Q6: An aldehyde/ketone by definition cannot be stereogenic. It has a double bond. When carbon is attached to two of the same thing, then it cannot be a stereocenter.
Q9: You're correct. Histidine isn't mentioned in the list in the CC, and typically isn't mentioned as an aromatic AA because its CHARGE is much more pronounced as a characteristic than the fact that it has an aromatic group attached to it. However, as you've seen in this question, you should know that it has an aromatic group attached.
Remember what the CC says: MEMORIZE THE STRUCTURE OF ALL OF THE AMINO ACIDS.
Also, the question is asking you "how many of the amino acid residues have an AROMATIC SIDE CHAIN", not "how many of these are aromatic AAs". The wording is very important.
Q12: You can essentially treat this system as one giant box, as opposed to an anchor + spring + mass. This is because they are all attached to each other, so they all (eventually) act together. They're one object.
The question tells us that there is a net acceleration to the right. So, based on Fnet = ma, if a is to the right, then F must be to the right (a and F always correlate in terms of direction).
This isn't a "Newton's third law" kind of question. That requires TWO separate objects.
Q19: You're right, this enzyme catalyses the equilibrium reaction (meaning "it does both"). However, the name of enzymes implies their MAIN function. Since the name is carbonic anhydrase, the ANhydrase implies that the main function is the REMOVE water. So, the primary reaction is the reverse of Reaction #1 (which yields water as a product, or "removes" it).
Q35: We are told to perform the reaction uner basic conditions, so only B and D are possible answers. Then, we know that Cl is a great leaving group. The nitrogen required in the final product is provided by the amino group in Step 6, so we don't need to use D. Based on process of elimination, we can determine that B is the correct answer.
Provided that we use excess reactant, two leaving groups can easily leave, especially ones that tend to WANT to leave.
NH2- is a very poor leaving group because it is a very good nucleophile (good base).
Q47: Let's do some math.
KIE = light isotope / heavy isotope
A: 13/15
B: 12/15
C: 12/14
D: 14/15
D is the largest ratio and is the combination of the largest C isotope and the largest N isotope
B is the smallest ratio and is the combination of the smallest C isotope and the largest N isotope
Let's only analyze the extremes.
From the graphs, we can see that PGHS-1 + AA (combination of smallest and largest isotopes) gives us the LARGEST values on the plot.
And we can see that PGHS-2 + D4-AA (combination of largest isotopes) gives us the LOWEST values on the plot.
So, we can see that larger isotopes give lower values. Therefore, D is the best answer.
Q54: We care most about the graph having the correct slope, not the orientation of the graph. They say which graph "best illustrates".
Another instructor can tackle your last question! 😀