UMD BCHM 461 - 228904467-Solutions-Ch-08 (10 pages)

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228904467-Solutions-Ch-08



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228904467-Solutions-Ch-08

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10
School:
University of Maryland, College Park
Course:
Bchm 461 - Biochemistry I
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2608T ch08sm S89 S98 2 1 08 5 09PM Page 89 ntt 102 WHQY028 Solutions Manual Ch 08 chapter 8 Nucleotides and Nucleic Acids 1 Nucleotide Structure Which positions in the purine ring of a purine nucleotide in DNA have the potential 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 hydrogen bonds see Figs 8 1 8 11 and 2 3 However N 1 is involved in Watson Crick hydrogen bonding with a pyrimidine and N 9 is involved in the N glycosyl linkage with deoxyribose and has very limited hydrogen bonding capacity Thus N 3 and N 7 are available to form further hydrogen bonds 2 Base Sequence of Complementary DNA Strands One strand of a double helical DNA has the sequence 5 GCGCAATATTTCTCAAAATATTGCGC 3 Write the base sequence of the complementary strand What special type of sequence is contained in this DNA segment Does the double stranded DNA 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 This sequence 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 form hairpin 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 stretching from the Earth to the moon 320 000 km The DNA double helix weighs about 1 10 18 g per 1 000 nucleotide pairs each base pair extends 3 4 For an interesting comparison your body contains about 0 5 g of DNA Answer The length of the DNA is 3 2 105 km 1012 nm km 10 nm 3 2 1018 The number of base pairs bp is 3 2 1018 9 4 1017 bp 3 4 bp Thus the weight of the DNA molecule is 9 4 1017 bp 1 10 18 g 103 bp 9 4 10 4 g 0 00094 g S 89 2608T ch08sm S89 S98 S 90 2 1 08 5 09PM Page 90 ntt 102 WHQY028 Solutions Manual Ch 08 Chapter 8 Nucleotides and Nucleic Acids 4 DNA Bending Assume that a poly A tract five base pairs long produces a 20 bend in a DNA strand Calculate the total net bend produced in a DNA if the center base pairs the third of five of two successive dA 5 tracts 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 base pairs as in a the total bend is additive when bending elements are repeated out of phase by 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 palindromic sequences in single strands of either RNA or DNA How is the helical structure of a long and fully basepaired 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 the B conformation The presence of the 2 OH group on ribose makes it sterically impossible for double helical RNA to assume the B form helix 6 Nucleotide Chemistry The cells of many eukaryotic organisms have highly specialized systems that specifically repair G T mismatches in DNA The mismatch is repaired to form a GqC not AUT base pair This G T mismatch repair mechanism occurs in addition to a more general system that repairs virtually all mismatches Can you suggest why cells might require a specialized system to repair G T mismatches Answer Many C residues of CpG sequences in eukaryotic DNA are methylated at the 5 position to 5 methylcytosine About 5 of all C residues are methylated Spontaneous deamination of 5 methylcytosine yields thymine T and a G T mismatch resulting from spontaneous deamination of 5 methylcytosine in a GqC base pair is one of the most common mismatches in eukaryotic cells The specialized repair mechanism to convert G T back to GqC is directed at this common class of mismatch 7 Spontaneous DNA Damage Hydrolysis of the N glycosyl bond between deoxyribose and a purine in DNA creates an AP site An AP site generates a thermodynamic destabilization greater than that created by any DNA mismatched base pair This effect is not completely understood Examine the structure 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 aldehyde form This and the loss of base stacking interactions could contribute significant flexibility to the 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 complementary bases on opposite strands and by base stacking between adjacent bases on the same strand Base stacking in nucleic acids causes a decrease in the absorption of UV light relative to the nonstacked 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 Nucleic Acids The concentration of protein or nucleic acid in a solution containing both can be estimated by using their different light absorption properties proteins absorb most strongly at 280 nm and nucleic acids at 260 nm Estimates of their respective concentrations in a mixture can be made by measuring the absorbance A of the solution at 280 nm and 260 nm and using the table that follows which gives R280 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 A280 reading and gives a more accurate protein estimate The protein concentration in mg ml F A280 assuming the cuvette is 1 cm wide Calculate the protein concentration in a solution of A280 0 69 and A260 0 94 2608T ch08sm S89 S98 2 1 08 5 09PM Page 91 ntt 102 WHQY028 Solutions Manual Ch 08 Chapter 8 Nucleotides and Nucleic Acids R280 260 Proportion of nucleic acid F 1 75 0 00 1 116 1 63 0 25 1 081 1 52 0 50 1 054 1 40 0 75 1 023 1 36 1 00 0 994 1 30 1 25 0 970 1 25 1 50 0 944 1 16 2 00 0 899 1 09 2 50 0 852 1 03 3 00 0 814 0 979 3 50 0 776 0 939 4 00 0 743 0 874 5 00 0 682 0 846 5 50 0 656 0 822 …


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