Answer Key -- nucleic acid 02s

1.         (a) I;  (b) I;  (c) 0.46;  (d) I;  (e) 0.3;  (f) I;  (g) I;  (h) 0.24;  (i) I

 2.   The nucleotide sequences of complementary strands are such that wherever an A occurs in one strand, there is a T in the other strand with which it can form a hydrogen-bonded base pair.  Wherever a C occurs in one strand, a G occurs in the other.  A is the base complementary to T, and C is the base complementary to G.

 3.   3.4Å x 10,000 = 34,000 Å = 3.4 mm.  (See p. 338.)

 4.   A-form DNA is a right-handed helix; Z-form DNA is a left-handed helix. (See Fig. 10-19, page   338.)

 5.   Any double-stranded sequence that has the form:

       1-2-3-4-5-6-6'-5'-4'-3'-2'-1'

       1'-2'-3'-4'-5'-6'-6-5-4-3-2-1

where each number and its prime represent correctly paired bases (A with T, C with G) all along the double-stranded molecule.

 6.   F; T; F

 7.   Hydrogen-bonding in regions of complementarity within an RNA chain can result in regions of double helix which are stabilized by base-stacking.  Breaks in complementary regions can result in loops and bulges which, together with the helical regions, can generate a precise three-dimensional structure.

 8.   Lower ionic strength reduces the screening of the negative charges on the phosphate groups by positive ions in the medium. The result is stronger charge-charge repulsion between the phosphate, which favors strand separation.

 9.   In general, the higher the proportion of G and C, the higher the melting temperature, t m .  More thermal energy is required to break the three hydrogen bonds holding G=C pairs than to break the two hydrogen bonds holding A=T pairs.

10. The DNA molecules are partially denatured; in each molecule approximately 50% of the DNA is single-stranded and 50% is double-helical.  The single-stranded regions, which appear as "bubbles" within the molecules, are those which denatured at lower temperatures because of their higher content of A=T base pairs.

11. In general, the more similar the sequences in two DNA molecules are, the more readily they will hybridize.  Because the evolutionary distance between mouse and yeast is greater than that between mouse and human, mouse and human DNA sequences are more similar than those of mouse and yeast.

12. UV radiation causes the formation of a dimer between adjacent thymine bases on the same DNA strand.  This results in a kink in the double helix at that site.

13. In these molecules, there is a phosphoric anhydride linkage between the phosphates.  Hydrolysis of this linkage (i.e., of ATP to ADP and phosphate) is a highly exergonic reaction that results in the release of a large amount of free energy.  Conversely, the input of a large amount of free energy is required for the synthesis of the linkage, i.e. for the phosphorylation of ADP to form ATP.  Thus, synthesis of ATP results in the storage of energy which can be released upon hydrolysis.

14. Both have a nitrogenous base and a pentose; nucleotides also have a phosphate group, which nucleosides lack.

15. See page 361; problem #3.

16. See page 361; problem #10. As an example.

 17.       (a)        See page 327; Figure 10-4

             (b)      See page 329; Figure 10-7 but you should be able to fill in the base not just the letter U

                         and G.

 18.       (a) 5'-CGACGGCGCGAAGUCAGGGGUGUUAAG-3'

            (b)  The nonsense strand of DNA would code for the RNA that is complimentary to the sequence shown above. 

 19.       (a)   The %A=%T and %G=%C 

             (b) DNA sample #2 has a greater concentration of G,C base pairs.  Since G,C base pairs form                     three H-bonds, T_m increase as G,C concentration increases.

 20.  (a) B (b) A, C (c) A (d) B, C, D (e) A (f) C (g) A (h) B (i) C, but strictly speaking, A is called a dinucleotide, not a nucleotide; (j) B (k) A, C, D

Answer Key -- lipid questions 02s

1.   For this structure, see Fig. 11-8, p. 370.  At neutral pH, there is a charge on the phosphate group, and serine is in the zwitterionic form; it has a protonated amino group and an ionized carboxyl group.

 2.   See Fig. 11-8, p. 370 for the phospholipid structure and Table 11-1, p. 364 for the structures of the fatty acids.  There are two carboxylate esters and two phosphate esters (one phosphodiester) in the molecule.

 3.   For this structure, see Fig. 11-8, p. 370.  At neutral pH, there is a negative charge on the phosphate group, and the quaternary amino group of choline carries a fixed positive charge; this entire phosphorylcholine moiety is polar.  The acyl chains attached to glycerol are the nonpolar part of the molecule.

 4.   All glycerophospholipids have two fatty acids in ester linkage with C-1 and C-2 of glycerol; often the fatty acid at C-1 is saturated, and that at C-2 is unsaturated.  C-3 of glycerol is joined to an alcohol-containing head group through a phosphodiester linkage, which is negatively charged at neutral pH.  See Fig. 11-8, p. 370.

 5.   A wax consists of a long-chain fatty acid in ester linkage with a long-chain fatty alcohol.  See Fig. 11-5, p. 368.

 6.   Triacylglycerols are nonpolar hydrophobic molecules that can be stored in specialized nonaqueous cellular compartments.  Glycerophospholipids are amphipathic molecules that can serve as structural components of membranes, which have hydrophilic and hydrophobic regions.

 7.   (1) The long-chain acyl group attached to C-1 of glycerol is ether-linked in a plasmalogen, but is an ester-linked fatty acyl group in typical glycerophospholipids. (2) There is a double bond between C-1 and C-2 of this fatty acyl chain in plasmalogens, but not in other phospholipids.

 8.   The structure of sphingosine is shown in Fig. 11-10, p. 372, which also shows that the attachment of a fatty acyl group to sphingosine in amide linkage converts it to ceramide.

 9.   A cerebroside has a single sugar residue joined to ceramide; a ganglioside has an oligosaccharide joined to ceramide. See Fig. 11-10, p. 372.

 10.       c; d; a; b

 11.       K; A; D; E

 12.       They are all lipids with potent biological activities derived from isoprenoid precursors.

 13. Triacylglycerols provide mammals with (1) stored fuel, (2) insulation, and (3) padding.  In some animals, such as camels and desert rats, the oxidation of stored lipids provides water; in hibernating animals, oxidation of stored lipids generates heat to maintain body temperature (see Chapter 4).  In plants, oxidation of the triacylglycerols stored in seeds provides the energy and precursors for biosynthetic processes during germination, before photosynthetic mechanisms become functional.

 14. See the structures shown in question #15:

            (a) B is sphingomyelin      (b) A is phosphatidylcholine

 15. (a) A, D   (b) C   (c) B, C   (d) A, B   (e) A, D   (f) A, B, C, D

            A is phosphatidylcholine; B is sphingomyelin; C is cerebroside; D is phosphatidyl glycerol

 16. c

 17. c

 18.       e, c, a, b, d       The lipid bilayer is even more permeable to O₂ than to water because O₂ lacks an appreciable dipole.  Ile is more hydrophobic than Tyr.

 19. See outline for polysaccharides.

Answer Key -- carbohydrates practice 02s

 1.   (a) Any of the hexoses drawn with a six-membered ring, as shown in Fig. 9-6 on p. 298, is correct.  The hydroxyls at C-2, C-3, and C-4 can point either up or down.  (b) For the anomer, the structure should be identical to the first, except that the hydroxyl group at C-1 should point up if it pointed down in your first structure, and vice versa.  (c) The number of chiral centers is 5; all carbons except C-6.  (d) The number of possible stereoisomers for a compound with n chiral centers is 2 n ; in this case, 2 5 , or 32 possible isomers.

 2.   (a) The anomeric carbon is the carbonyl carbon atom of a sugar, which is involved in ring formation.  (b) Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.  (c) Furanose is a sugar with a five-membered ring; pyranose is a sugar with a six-membered ring.  (d) A glycoside is an acetal formed between a sugar anomeric carbon hemi-acetal and an alcohol, which may be part of a second sugar.  (e) Epimers are stereoisomers differing in configuration at only one asymmetric carbon.  (f) An aldose is a sugar with an aldehyde carbonyl group; a ketose is a sugar with a ketone carbonyl group.

 3.   (a) 2 pyranoses;  (b) b1 ÿ 4;  (c) Yes.  There is a free anomeric carbon on one of the monosaccharide units that can undergo oxidation.

 4.   (a) A reducing sugar is one with a free carbonyl carbon that can be oxidized by Cu 2+ or Fe 3+ .  (b) The carbonyl carbon is C-1 of glucose and C-2 of fructose.  When the carbonyl carbon is involved in a glycosidic linkage, it is no longer accessible to oxidizing agents.  In sucrose (Glc(a1ÿ2)Fru), both oxidizable carbons are involved in the glycosidic linkage.

 5.   The enzymes that act on these polymers to mobilize glucose for metabolism act only on their nonreducing ends.  With extensive branching, there are more such ends for enzymatic attack than would be present in the same quantity of glucose stored in a linear polymer.  In effect, branched polymers increase the substrate concentration for these enzymes.

    6.   The molecule is branched, with each branch ending in a nonreducing end.  (See Fig. 9-15c, p. 305.)

 7.   The polymers are essentially insoluble and contribute little to the osmolarity of the cell, thereby avoiding the influx of water that would occur with the glucose in solution.  They also make the uptake of glucose energetically more feasible than it would be with free glucose in the cell.

 8.   (a) For the structure of amylose, see Fig. 9-15a, p. 305. The repeating unit is a-D-glucose linked to a-D-glucose; the glycosidic bond is therefore (a1 ÿ 4).          (b) Cellulose has the same structure as amylose, except that the repeating units are b-D-glucose and the glycosidic bond is (b1ÿ 4). (See Fig. 9-17a, p. 307.)

 9.   The ruminant animals have in their rumens microorganisms that produce the enzyme cellulase, which splits the (b1ÿ 4) linkages in cellulose, releasing glucose.  Humans do not produce an a enzyme with this activity; their digestive enzyme a-amylase can split only (a1 ÿ 4) linkages (such as those in glycogen and starch).

10. Uronic acids such as glucuronic acid and sulfated hydroxyl groups such as GalNAc4SO₃G  and GlcNAc6SO₃G  (see Fig. 9-20, p. 309).

 11. g; b; c; d; f; e; h; a

 12. A typical proteoglycan consists of a core protein with covalently attached glycosaminoglycan polysaccharides, such as chondroitin sulfate and keratin sulfate.  The polysaccharides generally attach to a serine residue in the protein via a trisaccharide (gal-gal-xyl) (see Fig. 9-21, p. 311).

 13. Both are made up of proteins and polysaccharides.  In proteoglycans, the carbohydrate moiety dominates, constituting 95% or more of the mass of the complex.  In glycoproteins, the protein constitutes a larger fraction, generally 50% or more of the total mass.

 14. A proteoglycan aggregate is a supramolecular assembly of proteoglycan monomers.  Each monomer consists of a core protein with multiple, covalently linked polysaccharide chains.  Hundreds of these monomers can bind noncovalently to a single extended molecule of hyaluronic acid to form very large structures.

 15. Hydrophilic carbohydrates can alter the polarity and solubility of the proteins.  Steric and charge interactions may influence the conformation of regions of the polypeptide and protect it from proteolysis.

 16. Newly synthesized serum glycoproteins bear oligosaccharide chains that end in sialic acid.  With time, the sialic acid is removed.  Glycoproteins that lack the terminal sialic acid are recognized by asialoglycoprotein receptors in the liver, internalized, and destroyed.

 17. Lectins are proteins that bind to specific oligosaccharides.  They interact with specific cell-surface glycoproteins thus mediating cell-cell recognition and adhesion.  Several microbial toxins and viral capsid proteins, which interact with cell surface receptors, are lectins.

 18.       (a)  2, 5   (b)  3, 6, 7   (c)  1, 2, 4, 8   (d)  3, 6, 7, 9   (e)  1, 2, 4, 5     

            Note that #3 should read:   (3) contains b(1-4) glucosidic bond

 19. a-D-ribofuranose; See page 326, Figure 10-3.  Note that this is the b-anomer.

 20. b-D-glucuronate; See page 300, Figure 9-9. 

21. a, d

22. c is false because keratan sulfate and chondroitin sulfate chains are bound to the core protein.

23. (a)  A  (b) B, C  (c) B and C contain fructose: B, D and E contain or are glucose.  Note that the glucose is in the a-anomer form in B and D and is in the b-anomer form in sugar E; (d) A, B, D  (e)  C, D, E  (f)  B and E.  In structure B the fructose ring is flipped over.  (g)  C, D, E  (h)  B    (i)  D and E.  Although E is in the b-anomer form, remember that in solution it can mutarotate.