S-89Nucleotides andNucleic Acidschapter 81. Nucleotide Structure Which positions in the purine ring of a purine nucleotide in DNA have thepotential to form hydrogen bonds but are not involved in Watson-Crick base pairing?Answer All purine ring nitrogens (N-1, N-3, N-7, and N-9) have the potential to form hydrogenbonds (see Figs. 8–1, 8–11, and 2–3). However, N-1 is involved in Watson-Crick hydrogenbonding with a pyrimidine, and N-9 is involved in the N-glycosyl linkage with deoxyribose andhas very limited hydrogen-bonding capacity. Thus, N-3 and N-7 are available to form furtherhydrogen bonds.2. Base Sequence of Complementary DNA Strands One strand of a double-helical DNA has the se-quence (5)GCGCAATATTTCTCAAAATATTGCGC(3). Write the base sequence of the complementarystrand. What special type of sequence is contained in this DNA segment? Does the double-strandedDNA have the potential to form any alternative structures?Answer The complementary strand is(5)GCGCAATATTTTGAGAAATATTGCGC(3) (Note that the sequence of a single strand is always written in the 5→3 direction.) Thissequence has a palindrome, an inverted repeat with twofold symmetry:(5)GCGCAATATTTCTCAAAATATTGCGC(3)(3)CGCGTTATAAAGAGTTTTATAACGCG(5)Because this sequence is self-complementary, the individual strands have the potential to formhairpin structures. The two strands together may also form a cruciform.3. DNA of the Human Body Calculate the weight in grams of a double-helical DNA molecule stretchingfrom the Earth to the moon (~320,000 km). The DNA double helix weighs about 1 1018g per 1,000nucleotide pairs; each base pair extends 3.4 Å. For an interesting comparison, your body containsabout 0.5 g of DNA!AnswerThe length of the DNA is(3.2 105km)(1012nm/km)(10 Å/nm)  3.2 1018ÅThe number of base pairs (bp) is 9.4 1017bpThus, the weight of the DNA molecule is(9.4 1017bp)(1 1018g/103bp)  9.4 104g  0.00094 g3.2 1018Å3.4 Å/bp2608T_ch08sm_S89-S98 2/1/08 5:09PM Page 89 ntt 102:WHQY028:Solutions Manual:Ch-08:S-90 Chapter 8 Nucleotides and Nucleic Acids4. DNA Bending Assume that a poly(A) tract five base pairs long produces a 20 bend in a DNAstrand. Calculate the total (net) bend produced in a DNA if the center base pairs (the third of five)of two successive (dA)5tracts are located (a) 10 base pairs apart; (b) 15 base pairs apart. Assume 10 base pairs per turn in the DNA double helix.Answer When bending elements are repeated in phase with the helix turn (i.e., every 10 basepairs) as in (a), the total bend is additive; when bending elements are repeated out of phaseby one half-turn as in (b), they cancel each other out. Thus, the net bend is (a) 40; (b) 0.5. Distinction between DNA Structure and RNA Structure Hairpins may form at palindromicsequences in single strands of either RNA or DNA. How is the helical structure of a long and fully base-paired (except at the end) hairpin in RNA different from that of a similar hairpin in DNA?Answer The RNA helix assumes the A conformation; the DNA helix generally assumes theB conformation. (The presence of the 2-OH group on ribose makes it sterically impossible fordouble-helical RNA to assume the B-form helix.)6. Nucleotide Chemistry The cells of many eukaryotic organisms have highly specialized systems thatspecifically repair G–T mismatches in DNA. The mismatch is repaired to form a GqC (not AUT) basepair. This G–T mismatch repair mechanism occurs in addition to a more general system that repairsvirtually all mismatches. Can you suggest why cells might require a specialized system to repair G–Tmismatches?Answer Many C residues of CpG sequences in eukaryotic DNA are methylated at the 5 positionto 5-methylcytosine. (About 5% of all C residues are methylated.) Spontaneous deaminationof 5-methylcytosine yields thymine, T, and a G–T mismatch resulting from spontaneousdeamination of 5-methylcytosine in a GqC base pair is one of the most common mismatchesin eukaryotic cells. The specialized repair mechanism to convert G–T back to GqC is directedat this common class of mismatch.7. Spontaneous DNA Damage Hydrolysis of the N-glycosyl bond between deoxyribose and a purinein DNA creates an AP site. An AP site generates a thermodynamic destabilization greater than thatcreated by any DNA mismatched base pair. This effect is not completely understood. Examine thestructure of an AP site (see Fig. 8–33b) and describe some chemical consequences of base loss.Answer Without the base, the ribose ring can be opened to generate the noncyclic aldehydeform. This, and the loss of base-stacking interactions, could contribute significant flexibility tothe DNA backbone.8. Nucleic Acid Structure Explain why the absorption of UV light by double-stranded DNA increases(the hyperchromic effect) when the DNA is denatured.Answer The double-helical structure is stabilized by hydrogen bonding between complementarybases on opposite strands and by base stacking between adjacent bases on the same strand. Basestacking in nucleic acids causes a decrease in the absorption of UV light (relative to the non-stacked structure). On denaturation of DNA, the base stacking is lost and UV absorption increases.9. Determination of Protein Concentration in a Solution Containing Proteins and NucleicAcids The concentration of protein or nucleic acid in a solution containing both can be estimated byusing their different light absorption properties: proteins absorb most strongly at 280 nm and nucleicacids at 260 nm. Estimates of their respective concentrations in a mixture can be made by measuringthe absorbance (A) of the solution at 280 nm and 260 nm and using the table that follows, which givesR280/260, the ratio of absorbances at 280 and 260 nm; the percentage of total mass that is nucleic acid;and a factor, F, that corrects the A280reading and gives a more accurate protein estimate. The proteinconcentration (in mg/ml)  F A280(assuming the cuvette is 1 cm wide). Calculate the protein con-centration in a solution of A280 0.69 and A260 0.94.2608T_ch08sm_S89-S98 2/1/08 5:09PM Page 90 ntt 102:WHQY028:Solutions Manual:Ch-08:Answer For this protein solution, R280/260 0.69/0.94  0.73, so (from the table) F  0.508. The concentration of protein is F A280 (0.508 0.69) mg/mL  0.35 mg/mL.Note: the table applies to mixtures of proteins, such as might be found in a crude cellularextract, and reflects the absorption