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The MOLECULES of LIFE Physical and Chemical Principles Solutions Manual Prepared by James Fraser and Samuel Leachman Chapter 2 Nucleic Acid Structure Problems and Solutions True/False and Multiple Choice Fill in the Blank 1. The tertiary structure of functional RNA molecules is easily predicted. 7. The ______________ structure of a nucleic acid is the sequence of nucleotides in the DNA or RNA molecule. Answer: primary True/False 2. Which of the following is not a stabilizing force for the structure and stability of double-stranded nucleic acids? a. b. c. d. base stacking hydrogen bonding disulfide bonds electrostatic forces 3. H-bond acceptor, H-bond donor, H-bond acceptor, methyl group is the pattern of potential interactions at the edge of which Watson-Crick base pair? a. b. c. d. e. A-T G-C A-A C-G T-A 4. Genomic DNA can become deformed from its normal B-form by DNA binding proteins, such as the histone proteins and the TATA-box binding protein. True/False 5. Which nonstandard base pair is most likely to form a wobble base pair? a. b. c. d. e. U-U A-G A-A G-U G-G 6. Classify the following RNA structural elements as secondary or tertiary structure: a. b. c. d. e. f. g. coaxial helices pseudoknot hairpin junction adenosine platform ribose zipper bulge Answer: Secondary: c, d, g; tertiary a, b, e, f 8. The modified RNA base in which two methyl groups are added to guanine is __________. Answer: N,N-dimethylguanine. The corresponding nucleoside is N,N-dimethylguanosine. 9. B-form DNA has a C2ʹ ________ sugar pucker. Answer: endo 10. Hoogsteen base pairs, where the hydrogen-bonding interactions utilize the Watson-Crick base-pairing edge on one base and the major groove edge in the other base, can be utilized to form an RNA _______ helix. Answer: triple 11. Metal ions, such as K+, Na+, and Mg2+, typically interact with the __________ group on the backbone of nucleic acids. Answer: phosphate Quantitative/Essay 12. What physical factors force RNA to adopt only the C3ʹ endo sugar configuration, but allow DNA to adopt either the C2ʹ endo or the C3ʹ endo sugar configurations? Answer: The C2ʹ endo configuration results in close contact (1.9 Å) between one of the oxygen atoms of the 3ʹ phosphate group and the hydrogen atom at the 2ʹ position of the ribose ring in DNA. Compared with DNA, RNA has the 2ʹ hydrogen replaced by hydroxyl group (OH) and the two oxygen atoms (2ʹ ribose and 3ʹ phosphate) repel each other strongly. Rather than distorting the structure away from the Watson-Crick model, it is energetically favorable for RNA to switch to the C3ʹ endo conformation. 2 Chapter 2: Nucleic Acid Structure 2Q14 Problem question 2) (na_57_v1) 13. WhatSet type(quantitative of RNA interaction is shown below? Answer: H H minor groove G N N O H N N N N H H Label the major and minor grooves of the Watson-Crick base pair and the potential hydrogen bonds between all bases. Answer: This shows an A-minor motif in which the minor groove edges of an adenine base interact with the minor groove 2Q13 of a G-C base pair. major groove U O H O N H major groove Interactions on minor groove: H-bond acceptor, H-bond donor, H-bond acceptor. Interactions on major groove: hydrogen atom, H-bond acceptor, H-bond acceptor, H-bond acceptor. The G-U base pair is not a standard Watson-Crick base pair. 15. List at least three physical features that distinguish A- and Z-form DNA. Answer: The main differences between A- and Z-form DNA are as follows: For A-form DNA, the repeating unit is one base; however, in Z-form DNA, the repeating unit is two bases (most often alternating pyrimidine and purines). minor groove A-form DNA is a right-handed double helix. Z-form DNA is a left-handed double helix. In A-form DNA, the sugar pucker of the bases is C3ʹ endo, whereas Z-form DNA alternates between C2ʹ endo and C3ʹ endo. 14. Label the atoms, the bases, the hydrogen bonds across bases, the major and minor grooves, and the interactions along the grooves for the following base pair (oxygen and nitrogen atoms are not identified explicitly): Problem set (quantitative question 2) (na_56_v1) The minor groove of A-form DNA is wide and shallow, whereas the Z-form minor grove is narrow. The major groove of A-form DNA is deep and narrow, whereas the Z-form major groove is relatively shallow. In A-form DNA the base pairs are tilted to the helical axis, whereas the Z-form base pairs are nearly perpendicular to the helical axis. 16. In water, why is base stacking relatively more important than hydrogen bonding to forming a DNA double helix? Answer: Water can make H bonds with the DNA base H-bond donor and acceptors, so the energetic gain in interactions when those H bonds are satisfied by other nucleotides is not great. However, the many small polar interactions along the nucleotide rings that are formed upon base stacking are energetically more favorable than the interactions with water. Is the base pair a standard Watson-Crick base pair? 17. How are the interactions between a nucleic acid and Mg2+ mediated by water? PROBLEMS and solutions Answer: Water molecules can form a hydration shell around Mg2+, facilitating indirect interactions between it and the phosphate and functional groups in nucleic acids. 18. Why is the R (purine) in GNRA tetraloops required rather than a pyrimidine? Answer: The R position, adjacent to the A in sequence and space, is functionally significant because it provides a large surface area (the base has the two rings of a purine, as opposed to one ring for a pyrimidine) against which the A can base stack. 19. Why do most transcription factors interact with the major groove of B-form DNA rather than the minor groove? 3 22. The structure of the large ribosomal subunit from Haloarcula marismortui (PDB code: 1FFK) has been solved by x-ray crystallography. The 23S RNA contains 2922 nucleotides (758 A, 889 G, 739 C, and 536 U). a. Assuming a random distribution of nucleotides, how many four-mer sequences with the sequence G-N (any base)-R (purine)-A are possible? b. There are actually 21 GNRA tetraloops in the structure. What percentage of possible GNRA sequences actually formed tetraloops in the structure? Answer: a. Probability of G = total G/total nucleotides = 889/2922 Probability of N = 1. Probability of R = (total A + total G)/total nucleotides = (758 + 889)/2922. Answer: Probability of A = total A/total nucleotides = 758/2922. The major groove has more distinguishing interactions than the minor groove. Each base pair has a unique pattern of molecular interactions along the major, but not the minor, groove. Since transcription factors need to bind specifically to sequences of DNA, the best potential for distinguishing different sequences is to interact with the major groove. Further, the minor groove is narrow compared with the major groove. The depth of the major groove is compatible with the interaction with protein structural elements such as an α helix. Each four-mer has a GRNA probability = (probability of G) × (probability of N) × (probability of R) × (probability of A) = 0.0445. 20. Which of the following DNA sequences is most likely to adopt the Z-form? Why? a. b. GCGCGCGCATATGCGCGCGCC AGAGAGCTCTCTCTCTAAAAT Answer: Sequence a is more likely to adopt Z-form because it alternates purine and pyrimidines in a GC-rich sequence. This alternating pattern of 2ʹ endo and 3ʹ endo puckers yields the zig-zag pattern, where the smallest repeating unit is two base pairs, characteristic of Z-form DNA. 21. Consider a relaxed, closed-circular DNA plasmid that has 1040 base pairs with writhe = 0. An intercalator is added, such that there is one intercalator per 104 base pairs. The effect of the intercalator is to cause the twist between the base pairs that flank it to be reduced to zero. Will the resulting intercalator-bound DNA be positively or negatively supercoiled? Answer: Positively supercoiled. The relaxed plasmid contains 100 turns (1040 bp × 1 turn/10.4 bp) and therefore has linking number L = 100. Because it is not initially supercoiled, it has writhe W = 0. By the relation W = L – T, initial twist T is 100, with each base pair contributing ~35°. The cumulative effect of binding 10 intercalators is a reduction of the twist by a full turn (~35° × 10 ≈ 360°) to T = 99. As L is constant, positive supercoiling with W = 1 results. Total four-mers = 2922 – 3 = 2919. Probable GRNA four-mers = probability × total fourmers = 0.0445 × 2919 = 130. b. If only 21 GNRA tetraloops formed out of a possible 130, then16% of the sequences form a stable tertiary GNRA tetraloop motif. 23. A bacterial DNA polymerase moves at approximately 1000 base pairs per second when replicating DNA. The polymerase holoenzyme is approximately 110 Å long. By what multiple of its length does the polymerase move forward along the axis of the DNA double helix in 3 seconds? Assume the DNA stays fixed and ignore the rotational component of the motion of the polymerase along the DNA. Answer: Each bp of DNA ~3.4 Å rise per bp. In 3 s, the polymerase moves 3000 bp. Length = 3000 bp × 3.4 Å•bp–1 = 10,200 Å. Length/polymerase length = 10,200 Å/110 Å = 92.7 times the total length of the polymerase in 3 sec.