Name:				Total test points:			Course grade:

Exam score is on the exam. 17 = A, 13 = B, 9 = C. (ignoring +/-) 
Mean = 17.5, median = 17.5. Top scores: 3 at 20.5.

In figuring your course grade, I look at several things. First, I average your four test letter grades. For most people, this will give a proper course grade. But I do look to see if one of your early tests was unusually low (and I may consider that, discounting the early low score a little). After that, I generally look over the whole picture, trying to make sure everything looks fair and consistent. 

The following point cutoffs work for most (ignoring +/-): A = 65, B = 49. (possible = 86; high = 78; mean = 67.5, median = 66.) Your test total is shown above; check for accuracy of my records.

If there are any questions that might reasonably affect your course grade, please contact me ASAP. I’ll mail in the official grades Tues 9/9. (Of course, legitimate problems can be fixed later, but it is easier before submission.) Beyond that, feel free to call/e-mail anytime to discuss other issues, or to chat. If you want to do further work from Ouellette, I am happy to discuss it with you, or to read any written work. 

Additional feedback/suggestions always welcomed; e-mail is fine. Comments/suggestions on the web pages are particularly welcomed. If you haven’t already done so, look at the Supplemental Information page at the web site for some things you might comment on. (You can also send me anonymous messages by using the message form at the web site.)

A brief answer key follows. Alternative answers are reasonable in some cases. I have avoided drawing structures here as much as possible. Note that several questions here (and some on earlier tests) asked you to apply what you had learned to the real world; important, and in general you did well with these.

1. Draw glycine, and put two -CH2CH2OH groups on the N (replacing the two H there).

2. a. threonine (Ala has no reactive functional groups on its side chain.)

b. The hemiacetal group of the sugar joins with the alcohol group of the Thr side chain to form an acetal. (You know the Thr -OH is involved, because there is no other choice. The sugar might contribute another alcohol group or a carbonyl group -- or the cyclic form could contribute a hemiacetal. Since the question specified a stable linkage, the answer given here is the only likely one, based on course material. It produces a common glycosidic linkage, an acetal.) 

B Goodman, Where ice isn’t nice. BioScience 48:586, 8/98

3. Lipids. They contain a long hydrophobic tail, the long alkyl chain, which leads to insolubility. Lipids are a key part of biological membranes. They provide a hydrophobic barrier between the inside and outside of the cell.

4. a. It has distinct polar (hydrophilic) and non-polar (hydrophobic) regions. There is a long non-polar tail -- the hydrocarbon chains shown on the right. The N and O atoms on the left side make this part polar (especially since some of these groups are ionizable).

b. They are all ether linkages. (This kind of linkage, incidentally, is found naturally in the membranes of Archaea.) (Don’t be distracted by the amide group; this does not involve the glycerol.)

c. In common membrane lipids, the polar group is attached at a terminal -OH of the glycerol (usually shown as the bottom -OH). Here, the polar group is attached to the middle -OH. (How it is drawn -- e.g., whether groups are shown pointing to right or left -- is irrelevant.)

d. From left to right: tertiary amine, secondary amide. (Don’t call the amide here a peptide bond -- a biochemical term for the linkage between amino acids in proteins.)

e. The tertiary amine does not have an -H, so cannot be an H-bond donor. But it does have the usual amine N lone pair, so it can accept H-bonds. The secondary amide has an H+, so can serve as a H-bond donor -- just as with the peptide bonds of proteins. However, the N lone pair one might expect is pulled toward the >C=O, and is unavailable as an H-bond acceptor.

One way to think this through is to start with ammonia, :NH3 (where the colon denotes the lone pair). It is both an H-bond donor and acceptor, and can serve as a reference point when discussing other N. The tertiary amine lacks any H on the N, and amides effectively lack the lone pair.

f. The amide (at the bond between the N and the carbonyl C)  amine + acid

g. They are all at the extreme upper left: three -COOH groups are acidic, and the amine is basic. (Assuming independence, all of these groups would be ionized at pH 7, thus giving the molecule a net charge of -2.) (Remember, amides are not basic -- for the reason discussed for part e.)

C Richard et al, Supramolecular self-assembly of lipid derivatives on carbon nanotubes. Science 300:775, 5/2/03.

5. a. -ketoglutarate or oxaloacetate
b. isocitrate or malate
c. none. (In fact, ethers are quite uncommon in basic metabolism. This is presumably due to their general inertness. Lignin, a stable structural material, has lots of ether linkages. Some of you marked succinyl CoA. No, that is a thioester. First, the apparent O-S bond should be a big tip off; ethers are C-O-C. But perhaps more importantly, the C of “C-O-S” in the figure is shown with no H, so it clearly is a carbonyl group.)
d. isocitrate or malate; note that the former has two stereocenters.

6. The micelle would be “inside-out” -- polar heads in the middle, nonpolar tails sticking out, into the nonpolar gasoline.

7. a. The two amide linkages, between the carbonyl C and the attached -NHR.
b. The isopropyl group sticking up, near the left hand amide bond. (Given the complexity of the molecule, especially on the right hand end, it is not entirely clear what parts of this molecule should be called “side chain”. But that really doesn’t matter here; the isopropyl group is the only reasonable candidate to be considered hydrophobic, and it is obviously a side chain.)
c. Zero; there are no acidic or basic groups in this chemical. (The major acidic and basic groups are carboxylic acids and amines; this molecule has neither.)

K Anand, Coronavirus main proteinase (3CLpro) structure: Basis for design of anti-SARS drugs. Science 300:1763, 6/13/03.

This paper is very topical, and is a good lesson about drug design. We noted how insulin is made as a “super-protein”, which must be cleaved to make the active hormone. Similarly, these viruses require certain protease actions in order to replicate. So, blocking the viral protease is potentially a useful way to control the virus. The trick, of course, is to find an inhibitor that inhibits the viral protease without doing anything to you. To some extent, finding such an inhibitor is a matter of trial and error -- and luck. However, since detailed 3D structures are known for more and more proteases, some parts of the drug design process are now done by “rational” analysis, trying to design a good fit between the drug and the enzyme. (Protease inhibitors are used against HIV.)

8. a. none
b. Any of the reactions that reduces one of the H-carriers, i.e., which makes NADH+H+ or FADH2. Those are reactions 3, 4, 6 or 8 from the test handout. The H of these reduced cofactors is later oxidized in the oxidative phosphorylation (electron transport) pathway, to produce 2-3 ATP per molecule of cofactor.

cq\bq4a	(Summer 2003)
9/1/03