Ch.8 Lehringer - 1 Nucleotide Structure Which positions in a purine ring of a purine nucleotide in DNA have the potential to form hydrogen bonds

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Unformatted text preview: 1. Nucleotide Structure Which positions in a 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-l, N—8, N7, and N~9) have the potential to form hydrogen bonds (see Figs. 8—1, 8—11, and 2—3). However, N-l is involved in Watson—Crick hydrogen bonding with a pyrimidine, and NE) is involved in the N—glycosidic linkage with deoxyribose and has very limited hydrogen—bonding capacity. Thus, N-3 and N7 are available to form further hydrogen bonds. 2. Base Sequence of Complementary DNA Strands One strand of a double—helical DNA has the se- quence (5’)GCGCAATATTTCTCAAAATATTGCGC(8’). 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 (~320,0001<In). The DNA double helix weighs about 1 X 10’18 g per 1,000 nucleotide pairs; each base pair extends 3.4 A. For an interesting comparison, your body contains about 0.5 g of DNA! Answer The length of the DNA is (3.2 x 1051<m)(1012 nm/kalO A/nm) = 3.2 x 1018 A The number of base pairs (bp) is 3.2 x 10% 3.4 A/bp Thus, the weight of the DNA molecule is = 9.4 x 1017 bp (9.4 x 1017 bp)(1 x 10-18 g/103 bp) = 9.4 x 10"4 g = 0.00094 g 8—64 Chapter 8 Nucleotides and Nucleic Acids 8-65 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; (1)) 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 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 form; the DNA helix generally assumes the B form. (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 GEC (not A=T) 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 GEO base pair is one of the most common mismatches in eukaryotic cells. The specialized repair mechanism to convert G-T back to GEC is directed at this common class of mismatch. 7. Nucleic Acid Structure Explain why the absorption of UV light by double~stranded DNA increases (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. 8. 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. Their respective concentrations in a mixture can be estimated by measuring the ab— sorbance (A) of the solution at 280 nm and 260 nm and using the table below, which gives [£280,260, 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 X A280 (assuming the cuvette is 1 cm wide). Calculate the protein concentration in a solution of A280 = 0.69 and A260 2 0.94. S-66 Chapter 8 Nucleotides and Nucleic Acids Proportion of 1.75 0.00 1.63 0.25 1.52 0.50 1.40 0.75 1.36 1.00 1.30 1.25 , 1.25 1.50 1.16 2.00 1.09 2.50 1.03 3.00 0.979 3.50 0.939 4.00 0.874 5.00 0.846 5.50 0.822 6.00 0.804 6.50 0.784 7.00 0.767 7.50 0.753 8.00 0.730 9.00 0.705 10.00 0.671 12.00 0.644 14.00 0.615 17.00 0.595 20.00 Answer For this protein solution, [€280,260 = 069/094 = 0.73, so (from the table) F = 0.508. The concentration of protein is F X A280 = (0.508 X 0.69) mg/mL = 0.35 mg/mL. Note: the table applies to mixtures of proteins, such as might be found in a crude cellular extract, and reflects the absorption properties of average proteins. For a purified protein, the values of F would have to be altered to reflect the unique molar extinction coefficient of that protein. 9. Base Pairing in DNA In samples of DNA isolated from two unidentified species of bacteria, X and Y, adenine makes up 32% and 17%, respectively, of the total bases. What relative proportions of adenine, guanine, thymine, and cytosine would you expect to find in the two DNA samples? What assumptions have you made? One of these species was isolated from a hot spring (64 °C). Suggest which species is this thermophilic bacterium. What is the basis for your answer? Answer For any double-helical DNA, A = T and G = C. Because the GEO base pair involves three hydrogen bonds and the A=T base pair involves only two, the higher the G + C content of a DNA molecule, the higher the melting temperature. The DNA from species X has 32% A and therefore must contain 32% T, 18% G, and 18% C. The sample from Y, with 17% A, must contain Chapter 8 Nucleotides and Nucleic Acids 8-67 17% T, 38% G, and 83% C. This calculation is based on the assumption that both DNA molecules are double—stranded. Species Y, having the DNA with the higher G + C content (66%), most likely is the thermophilic bacterium; its DNA has a higher melting temperature and thus is more stable at the temperature of the hot spring. 10. Solubility of the Components of DNA Draw the following structures and rate their relative solubilities in water (most soluble to least soluble): deoxyribose, guanine, phosphate. How are these solubilities consistent with the three-dimensional structure of double—stranded DNA? Answer O HOCH 0 ll 2 OH /C\ /N O _ H H HN C \\ I H H (‘3 (“3 PH ‘0 —P=O / \ / \N l HO H H2N N H OH Deoxyribose Guanine Phosphate Solubilities: phosphate > deoxyribose > guanine. The negatively charged phosphate is the most water—soluble; the deoxyribose, with several hydroxyl groups, is quite water—soluble; and guanine, a hydrophobic base, is relatively insoluble in water. The polar phosphate groups and sugars are on the outside of the DNA double helix, exposed to water. The hydrophobic bases are located in the interior of the double helix, away from water. 11. DNA Sequencing The following DNA fragment was sequenced by the Sanger method. The asterisk indicates a fluorescent label. 3 ’—OH ATTACGCAAGGACATTAGAC—--5' 1:5: 3: A sample of the DNA was reacted with DNA polymerase and each of the nucleotide mixtures (in an appropriate buffer) listed below. Dideoxynucleotides (ddNTPs) were added in relatively small amounts. 1. dATP, dTTP, dCTP, dGTP, ddTTP ' 2. dATP, dTTP, dCTP, dGTP, ddGTP 3. dATP, dCTP, dGTP, ddTTP 4. dATP, dTTP, dCTP, dGTP The resulting DNA was separated by electrophoresis on an agarose gel, and the fluorescent bands on the gel were located. The band pattern resulting from nucleotide mixture 1 is shown below. Assuming that all mixtures were run on the same gel, what did the remaining lanes of the gel look like? 5-68 Chapter 8 Nucleotides and Nucleic Acids Electrophoresis Answer Lane 1: The reaction mixture that generated these bands included all the deoxynucleotides, plus dideoxythymidine. The fragments are of various lengths, all terminating where a ddTTP was substituted for a dTTP. For a small portion of the strands synthesized in the experiment, ddTTP would not be inserted and the strand would thus extend to the final G. Thus, the nine products are (from top to bottom of the gel) 5’—primer~TAATGCGTTCCTGTAATCTG 5’-primer—TAATGCGTTCCTGTAATCT 5’~piimer~TAATGCGTTCCTGTAAT 5’—primer-TAATGCGTTCCTGT 5’-primer-TAATGCGTTCCT 5’-primer—TAATGCGTT 5’—primer-TAATGCGT 5’-primer—TAAT 5"primer—T Lane 2: Similarly, this lane will have four bands (top to bottom), for the follong fragments, each terminating Where ddGTP was inserted in place of dGTP: 5’-primer—TAATGCGTTCCTGTAATCTG 5’—primer—TAATGCGTTCCTG 5’—primer-TAATGCG 5’~primer~TAATG Lane 3: Because mixture 3 lacked dTTP, every fragment was terminated inunediately after the primer as ddTTP was inserted, to form 5’sprimer—T, The result will be a single thick band near the bottom of the gel. Lane 4: When all the deoxynucleotides were provided, but no dideoxynucleotide, a single labeled product formed: 5’-primer—TAATGCGTTCCTGTAATCTG. This will appear as a single thick band at the top of the gel. 12. Snake Venom Phosphodiesterase An exonuclease is an enzyme that sequentially cleaves nu— cleotides from the end of a polynucleotide strand. Snake venom phosphodiesterase, which hydrolyzes nucleotides from the 3’ end of any oligonucleotide with a free 3’-hydroxyl group, cleaves between the 3’ hydroxyl of the ribose or deoxyribose and the phosphoryl group of the next nucleotide. It acts on single—stranded DNA or RNA and has no base specificity, This enzyme was used in sequence determi— Chapter 8 Nucleotides and Nucleic Acids 8-69 nation experiments before the development of modern nucleic acid sequencing techniques. What are the products of partial digestion by snake venom phosphodiesterase of an oligonucleotide with the fol- lowing sequence? (5’)GCGCCAUUGC (BU—OH Answer When snake venom phosphodiesterase cleaves a nucleotide from a nucleic acid strand, it leaves the phosphoryl group attached to the 5’ position of the released nucleotide and a free 3’-OH group on the remaining strand. Partial digestion of the oligonucleotide gives a mixture of fragments of all lengths, as well as some of the original, undigested strand, so the products are (P represents the phosphate group) ' (5')P—GCGCCAUUGC (8’)—OH (5’)P~GCGCCAUUG (3’)—OH (5’)P—GCGCCAUU (BO—OH (5’)P—GCGCCAU(3’)—OH (5’)P—GCGCCA(3’)——OH (5’)P—GCGCC(3’)—OH (5’)P—GCGC(8’)—OH (5’)P.GCG(8’)~OH (5’)P—GC(8’)-OH and the released nucleoside 5’-phosphates, GMP, UMP, AMP, and CMP. \ 13. Preserving DNA in Bacterial Endospores Bacterial endospores form when the environment is no longer conducive to active cell metabolism. The soil bacterium Bacillus subtllls, for example, begins the process of sporulation when one or more nutrients are depleted. The end product is a small, meta» bolically dormant structure that can survive almost indefinitely with no detectable metabolism. Spores have mechanisms to prevent accumulation of potentially lethal mutations in their DNA over periods of dormancy that can exceed 1,000 years. Bacillus subtll'ls spores are much more resistant than the organism’s growing cells to heat, UV radiation, and oxidizing agents, all of which promote mutations. (a) One factor that prevents potential DNA damage in spores is their greatly decreased water con— tent. How would this affect some types of mutations? (b) Endospores have a category of proteins called small acid—soluble proteins (SASPS) that bind to their DNA, preventing formation of cyclobutane—type dimers. What causes cyclobutane dimers, and why do bacterial endospores need mechanisms to prevent their formation? Answer (a) Water is a participant in most biological reactions, including those that cause mutations. The low water content in endospores reduces the activity of mutation'causing enzymes and slows the rate of nonenzymatic depurination reactions, which are hydrolysis reactions. (b) UV light induces the condensation of adjacent pyrimidine bases to form cyclobutane pyrimidine dimers. The spores of B. submits, a soil organism, are at constant risk of being lofted to the top of the soil or into the air, where they are subject to UV exposure, possibly for prolonged periods. Protection from UV—induced mutation is critical to spore DNA integrity. Biochemistry on the internet 14. The Structure of DNA Elucidation of the three-dimensional structure of DNA helped researchers understand how this molecule conveys information that can be faithfully replicated from one gen— eration to the next. To see the secondary structure of double—stranded DNA, go to the Protein Data Bank website (W). Use the PDB identifiers listed below to retrieve the data pages for the tim forms of DNA. Open the structures using Chime, and use the different viewing options to complete the following exercises. 5-70 (a) (b) (c) (1) (2) (3) C4) (5) (5) (7) Chapter 8 Nucleotides and Nucleic Acids Obtain the file for 141D, a highly conserved, repeated DNA sequence from the end of the HIV-1 (the virus that causes AIDS) genome. Display the molecule as a stick or ball—and-stick structure. Identify the sugar-phosphate backbone for each strand of the DNA duplex. Locate and identify individual bases. Which is the 5’ end of this molecule? Locate the major and minor grooves. is this a right- or left~handed helix? Obtain the file for 145D, a DNA with the Z conformation. Display the molecule as a stick or ball- and-stick structure. Identify the sugar—phosphate backbone for each strand of the DNA duplex. ls this a right— or left-handed helix? To fully appreciate the secondary structure of DNA, select “Stereo” in the Options menu in the viewer. You will see two images of the DNA molecule. Sit with your nose approrn'mately 10 inches from the monitor and focus on the tip of your nose. 1n the background you should see three im- ages of the DNA helix. Shift your focus from the tip of your nose to the middle image, which should appear three~dimensional (Note that only one of the two authors can make this work.) For additional tips, see the Study Guide or the textbook website (wwwvvhfreemancom/lehninger) . Answer (a) The DNA fragment modeled in file 141D, from the human irrununodeficiency virus, is the B form, the standard Watson-Crick structure (although this particular structure is a bent B—form DNA). To identify the sugar—phosphate backbone, turn on the “Backbone” dis- play. To identify individual bases, return to the ball—"indestick structure and click on the relevant portion of the molecule; the corresponding base is identified at the bottom of the frame. Alternatively, go to the “Options” menu and click on “Labels” to display the names and numbers of every base in the structure. This fragment has an adenine at the 5’ end and a guanine at the 8’ end; click on the bases at each end of the helix to identify which is the 5' end. When the helix is oriented with the 5’ adenine at the upper left— hand side of the model, the mirror groove is in the center of the model. Rotating the model so that the 5’ adenine is at the upper right—hand side positions the major groove in the center. The spiral of this helix runs upward in a counterclockwise direction, so this is a right-handed helix. The model of DNA in the Z conformation includes a number of water molecules and sev- eral ligands. Note the shell of water molecules around the helix. These can be viewed by using the Rasmol command line, entering “select water” and then “spacefill” on the next line. If you use the Chime program for viewing instead of Rasmol, you can turn off the display of these water molecules by toggling the “Display Hetero Atoms” tool. it is also helpful to toggle off the display of the hydrogen atoms. The backbone of DNA in the Z conformation is very different from that in the B conformation. Turn on the “Backbone” display to fully appreciate why this is referred to as the Z conformation. The spiral of the helix runs upward in a clockwise direction, so this is a left-handed helix. Viewing the structures in stereo takes a bit of practice, but perseverance will be re- warded! Here are some tips for successful three-dimensional viewing: (b) (0) Turn off or lower the room lighting. Sit directly in front of the screen. Use a ruler to make sure you are 10 to 11 inches from the screen. Position your head so that when you focus on the tip of your nose, the screen images are on either side of the tip {i.e., look down your nose at the structures). Move your head slightly closer to or farther away from the screen to bring the middle image into focus. Don’t look directly at the middle image as you try to bring it into focus. If you find it uncomfortable to focus on the tip of your nose, try using the tip of a finger (positioned just beyond the tip of your nose) instead. Relax as you attempt to View the three—dimensional image. Note that many people, including one of the text authors, have some trouble making this work! ...
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This homework help was uploaded on 02/13/2008 for the course BIOBM 3310 taught by Professor Feigenson,gw during the Fall '07 term at Cornell University (Engineering School).

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Ch.8 Lehringer - 1 Nucleotide Structure Which positions in a purine ring of a purine nucleotide in DNA have the potential to form hydrogen bonds

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