Answer: Phospholipid: the glycolipid has the same structure as a phospholipid except that it has a different hydrophilic (polar) group. It has 3 groups attached to glycerol as in a P-lipid. The  polar group attached to the glycerol is (galactose) instead of the phosphate or phosphate ester group of the phospholipid; the nonpolar groups attached to the glycerol (2 fatty acids) are the same as in a phospholipid. 

2. Hints:   
        A. What differences in properties are exploited by each of these methods?
        B. How do the properties of asp (aspartic acid) and gln (glutamine) compare? Which method is most likely to take advantage of these differences?
        C. Where does the carbon in aspartate (aspartic acid)* come from (in minimal medium)?

* The terms 'aspartate' and 'aspartic acid' are used interchangeably by most biochemists. Strictly speaking, 'aspartate' is the ionized form and 'aspartic acid' is the un-ionized form. In solution, at most pH values, you have a mixture of the two.

    Answers:
    2A. Both -- you would use the method of fingerprinting (meaning a 2D separation, using chromatography in one direction and electrophoresis in the other). This is the best bet because it combines the separation power of both paper chromatography and paper electrophoresis. Next best: paper chromatography, since it separates on the basis of small differences among the amino acids in their solubility in organic solvents. (One dimensional chromatography won't separate all the amino acids, but 2 dimensional chromatography, using two different conditions, will. See answer to 2B.)

    Note: The term "fingerprinting" usually refers to the 2D separation of peptide fragments, that is, the separation of the products of partial hydrolysis of a protein. "Fingerprinting" usually does not refer to the separation of amino acids.
    Why? If you hydrolyze proteins completely to amino acids, and separate the amino acids, almost every protein will give the same pattern or "fingerprint" -- almost every protein will give the same 20 spots corresponding to the same 20 amino acids. On the other hand, if you partially hydrolyze a protein and separate the peptide products, each protein will usually give a unique pattern of spots corresponding to a unique mixture of peptides.
    What is meant by "method of fingerprinting," in this case, is the application of the fingerprinting procedure (that is, a 2D separation) to a mixture of amino acids.

    2B. Paper electrophoresis at pH 12 or paper chromatograph under different conditions.
Paper electrophoresis at pH 12 will work, since asp will have a net charge of -2 while gln will have a net charge of -1. Paper chromatography will work if you use a different pH and or organic solvent than the first paper chromatographic separation. Paper electrophoresis at pH 2 will not work because both molecules would have a net charge of +1.

    2C. 20. All the carbons of aspartate are derived from glucose (the only source of C in the minimal medium). The glucose contained 30 units per mole, and had 6 carbons, all equally labeled, so there are 30/6 = 5 radioactivity units per carbon atom (per mole of each carbon atom) in glucose. Aspartate has 4 carbon atoms, or 4 moles of carbon atoms per mole of aspartate. Each carbon derived from glucose has 5 units of radioactivity. Four carbons, at 5 units each, yields 5 X 4 = 20 radioactivity units per mole of aspartate.
 

3. Hints:
    A. Compare the straight chain and ring forms.
    B. What are the products of complete hydrolysis? Will any of them isomerize?

    Answers:

     3A. Carbon #5. The C5 OH is positioned so that when it swings around and approached C1, it can form a bond to the C1 carbon.  The C1 double bond opens up and a hydroxyl is formed on the C1 oxygen. 
    Another way to reason it, without having to remember that it is the C5 OH that attacks:  Inspection of the ring structure shows that the ring oxygen atom lies between carbons 1 and 5, so the oxygen must have come from either C1 or C5.  The anomeric carbon is number 1, which has a carbonyl (C=O) in the straight chain form.  It is this double bond that is attacked by a hydroxyl to form the ring. After this attack, the carbonyl oxygen becomes a hydroxyl, so that is not the one in the ring, so it must be the C5 OH that becomes the ring oxygen.

    3B. A carbonyl, beta-D-glucose, alpha-D-glucose, carbon with tetrahedral bonds, monomers. The free D-glucose would be comprised of an equilibrium mixture of the straight chain form (containing an aldehyde, which has a carbonyl), and the alpha and beta anomer ring forms. No L-glucose would form: one would have to break and reform the bonds between the hydroxyls and carbons at positions 2, 3 and 4 for that transformation to occur. No dimers or glycosidic bonds would be left as these are what is destroyed by the hydrolysis.  The D-glucose monomers would be the products.