GLAD YOU ASKED! 

Please e-mail me questions related to the course. I'll answer them. I'll also post (without your name, etc.) any questions/answers that may help everyone.

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15. (3/16/04)
QUESTION: {from my e-mail: Just an update, so far all of the review questions resemble ones one [sic. should have read "on"] the old exams. If that changes (if I cook up a totally new question), I'll let you know. I'd bet that there will be no such questions on the final--the weather's too nice!} Being that I am sort of stressed about the test, i need some clarification in your last email...did you mean to say that so far all of the "final question" instead of "review questions" resemble one on the old exam?
ANSWER: All of the "old" material on the final covers things already tested on in exams 1-3. Of course there will be plenty of questions that deal with heterocycles and nucleic acids too.

14. (3/15/04)
QUESTION: In going over the old exams this weekend, I noticed what I think might be an error: In question 17a.), I drew a picture of the epimer of D-idose at C-2 and listed its name as "D-mannose". I got the picture of the sugar correct but the name marked wrong, and the key lists the answer as "D-gulose". According to the notes, however, D-gulose is the C-3 epimer of D-idose and the drawing that I had done correctly was in fact of D-mannose. (D-idose has C-1 and C-3 on the wrong side, D-mannose has only C-3 on the wrong side.) This also relates to question 4.), where I missed points by saying that alpha-D-sorbofuranose would interconvert between D-idose and D-mannose when placed in base. The key says it will interconvert between D-idose and D-gulose, but the interconversion would be at C-2 and not at C-3, so that the -OH on C-2 would interconvert from being on the right and wrong sides of the sugar. So D-mannose should be the correct answer.
ANSWER: OK, I sat down with this--finally! Here's what I did: I drew D-idose. Remember that it codes ("1,3"), which means that, in the Fischer, the OH's at carbons 2 and 4 are on the left. The epimer at C-2 of this sugar would code ("3"). That's D-gulose. The confusion came for you in not remembering, I assume, that the "codes" refer to the undesignated sites in the sugar, not the carbon numbers. Maybe I should change this in the future so that the codes for the aldohexoses would be "0," "2," "3," "2-3," "4," "2-4," "3-4," and "2-3-4." Then both sets of numbers would refer to actual carbon numbers.

13. (3/15/04)
QUESTION: Yes, a lifetime; maybe two. I guess I'm confused. I thought the systemic names were the IUPAC names.
ANSWER: The systematic, in this case, are sort of a middle ground and not quite IUPAC. For sugars, the IUPAC gets very strange, so we stay away from strict IUPAC.

12. (3/15/04)
QUESTION: In naming amino or deoxy sugars, do we need to know just the IUPAC name (and not the common name)?
ANSWER: We've never learned the IUPAC names, actually. However, we did learn systematic names that seem a bit bulky, like 2-amino-2-deoxyglucose." THESE you have to know! The shorter 2-aminoglucose won't do the job on the final. Seems like a long time ago, doesn't it?

11. (3/1/04)
QUESTION: By the way, I keep checking the virtual office hours but there are no additions since about 1/22.
ANSWER: I've been way too lazy lately--Your e-mails are the only ones I've received, so I've assumed that no one else cared. I'll post them now.

10. (2/28/04)
QUESTION: Should N5 in N5,N10-methenyl-THF have a + charge?
ANSWER: Yes, my mistake.

9. (2/28/04)
QUESTION: My notes lack any mechanism going from a shiff base of an amino acid (#2 aa in transamination) to aa + PLP. Should we know the mechanism?
ANSWER: That's just transimination, a simple, but tedious, mechanism. We have agreed NOT to bother with any mechanism for tranimination. It does, however, go from aa-PLP + lysine to aa + lys-PLP.

8. (2/28/04)
QUESTION: When adding the head of di-methyl pyrophosphate to the tail of isopentyl pyrophosphate (pi bond acting as nucleophile) it looks like we're simply replacing one bond with another ... so why does a radical form at that C?
ANSWER: It's not a radical, it's a cation (carbocation). The pi bond was shared by the two carbons of the double bond, but now it's used to form a bond from one of those carbons to the other compound, leaving the other carbon electron deficient by 1.

7. (1/24/04)
QUESTION: So we need to learn how to write out all the mechanisms for all the coenzymes. Are you going to at least give us a picture of what the substrate looks like on these problems?
ANSWER: You'll get the blanks from the lectures!

6. (1/24/04)
QUESTION: I was looking over the mechanism of aldolase. When the Lysine forms the protonated imine with carbon 2 of D-fructose 1,6 diphosphatase, where does the oxygen go? It simply disappears in the book, as well as in your notes. Do I need to worry about how it is removed.
ANSWER: Look on page 748 of Bruice 4th--this is the mechanism for the formation of an imine. The oxygen becomes water.

5. (1/23/2004)
QUESTION: Does the mix of ring form anomers (eg, for glucose, 64% beta, 36% alpha pyranose form) mean that chair flipping does NOT occur, or that it does and spends 64% of its time in the beta position?
ANSWER: The %'s represent the dynamic mixture as they flip back and forth continuously.

4. (1/22/2004)
QUESTION: There does not seem to be a second pratice test for exam 1 in the packet. Just thought I would let you know.
ANSWER: It's there--look deeper in the exams. It may mean turning pages one at a time until it shows up. Sorry.

3. (1/21/2004)
QUESTION: In the problem on page 48 of your notes that we worked on today, I get two disulfide bonds as follows: from composition 3: the C in AVCR to the C in CS from compostion 4: the C in TGCK to the C in CFR Compostions 5, 6 and 7 yield no information not available in 1 and 2. Am I doing this right?
ANSWER: Yes!

2. (1/25/2003)
QUESTION: I guess my "chiral-eyes" aren't working that well, or maybe they're just blurry from studying, but, I see THREE amino acids with chirality centers in their side chains: proline, isoleucine, and threonine. Am I seeing things?
ANSWER: Yes. Amines don't form chirality centers because they rapidly interconvert between the two forms--it's referred to as tunneling. So poor old proline is stuck with one chirality center.

1. (1/22/2003)
QUESTION: I was working on exam 1 for 2001, and realized that I've got a serious problem recognizing the difference between a keto vs. aldopentose. The problem for me occurs when the penultimate carbon is separated by the ring oxygen by a CH₂. Where is the identifying characteristic?
ANSWER: Look at the anomeric carbon to tell an aldose from a ketose. If it's an aldose, the anomeric will have an -OH and a hydrogen, but we won't show the hydrogen. If it's a ketose, there'll be the -OH AND a -CH₂OH attached (a cluttered anomeric carbon means ketose).
         Then. When you look on the other side of the ring oxygen, you'll either see -CH₂- or you'll that -CH₂OH is attached to that carbon. If you see -CH₂-, it means that the last carbon's oxygen closed the ring. This means go one more carbon to the left to find the penultimate and see which side the -OH is on. This one is easy. Right-Down, Left-Up. If the -OH is down, it's D.
         If you see a -CH₂OH group on that carbon just counterclockwise to the ring oxygen, then the penultimate's oxygen has closed the ring. This means that there was the 120 degree twist on the penultimate carbon, putting the -CH₂OH group off the penultimate carbon UP for a D sugar. Hope this helps!