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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Pages:
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



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