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ch3 - Nucleotides and Nucleic Acids 1 Nucleotide Structure...

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Unformatted text preview: 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 CN-l, N3, N-7, and N-Q) have the potential to form hydrogen bonds (see Figs. 8—-1, 8—1 1, and 38). However, N—l is involved in Watson-Crick hydrogen bonding with a pyrimidine, and NS) is involved in the N—giycosyl linkage with deoxyribose and has very limited hydrogembonding capacity. Thus, N-3 and N? are available to form further hydrogen bonds. 2. Base Sequence of Complementary DNA Strands One strand of a donbie-helicai DNA has the se« quence {5'DGCGCAATATTTCTCAAAATATTGCGC(3’)_ Write the base sequence of the complementary strand. What special type of sequence is contained in this DNA segment? Does the doublemstranded DNA have the potentiai 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 ') CGCGTTATAAAGAGTTTTATAAGGCG (5 ’) Because this sequence is self~compiementary, 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 Caicuiate the weight in grams of a doubie—helical DNA molecule stretching from the Earth to the moon @320 000 km) The DNA doubie 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 DNAE Answer The iengtn of the DNA is (3.2 >< 105 ammo” nm/km)(10 A/nm) m 3.2 x 1018 A The number of base pairs (bp) is 3.2 >< 1018A 3.4 A/bp Thus, the weight of the DNA molecule is = 9.4 x 10” hp (9.4 x 1017 bp)(i x 10‘33 g/io3 hp) = 9.4 x 10‘4 g 2 0.00094 g 8—90 ' Chapter 8 Nucleotides and Nucieic Acids 4. DNA Bending Assume that a poly(A) tract five base pairs iong 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) i5 base pairs apart. Assume 10 base pairs per turn in the DNA double helix. Answer When bending eiements 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 haif~turn as in (b), they cancel each other out. Thus, the net bend is (a) 40°; (11) 0°. 5. Distinction between DNA Structure and RNA Structure Hairpins may form at paiindromic 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 simiiar hairpin in DNA? Answer The RNA helix assumes the A conformation; the DNA heh‘x generally assumes the B conformation. (The presence of the 2’—OH group on ribose makes it stericaily impossible for double-heiical RNA to assume the B—form heiix.) 6. Nucleotide Chemistry The cells of many eukaryotic organisms have highly specialized systems that specifically repair GMT mismatches in DNA. The mismatch is repaired to form a GEO (not AWT) 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 (3 residues of CpG sequences in eukaryotic DNA are methyiated at the 5’ position to 5—methylcytosine. (About 5% of all 0 residues are methylated.) Spontaneotis deamination of 5—methylcytosine yields thymine, T, and a Gw-T mismatch resulting from spontaneous deamination of 5mmethyicytosine in a GfiC base pair is one of the most common mismatches , in eukaryotic cells. The specialized repair mechanism to convert G—T back to GEO is directed at this common class of mismatch. 7. Spontaneous DNA Damage Hydrolysis of the N—glycosyl bond between deoxyrihose and a purine in DNA creates an A? 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—83b) and describe some chemical consequences of base toss. Answer Without the base, the ribose ring can be opened to generate the noncyclic aldehyde form. This, and the loss of basemstaching interactions, could contribute significant flexibility to the DNA backbone. 8. Nucleic Acid Strncture Explain why the absorption of UV iight by doubts-stranded DNA increases (the hyperchromic effect) when the DNA is denatured. Answer The double—helicai 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 Eight (relative to the non- stacked structure). On denaturation of DNA, the base stacking is lost and W 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 absorbence (A) of the solution at 280 nm and 260 nm and using the table that ioliows, which gives Rggojgfio, the ratio of absorbances at 280 and 260 nm; the percentage of total mass that is nucieic acid; and a factor, F, that corrects the A280 reading and gives a more accurate protein estimate. The protein concentration (in rag/mi) = F X A230 (assuming the cuvette is 1 cm wide). Calculate the protein con- centration in a solution oi‘Aggg = 0.69 and A266 3 0.94. Chapter 8 Nucleotides and Nucteic Acids 8-91 fimfikmsfimzmsxzzzzrw;fi1u xflquc'm-Lzz.’.1r.hnrqt m,1\::'::"zx.n<: mummmmemcy- -;;4:5m:».:2¢::::. Proportion of 5 fr! 2 a E; Rzgmso nucleic acid ml . E 2.75 0.00 1.115 E E 1.53 0.25 1.081 E E 1.52 0.50 1.054 E E 1.40 0.75 3.023 EE E! 1.36 1.00 0.094 E E 1.30 1.25 0.970 E E 1.25 3.50 0.944 EE l.i6 2.00 0.899 3‘ E! 2.09 2.50 0.852 E 1.03 3.00 0.814 E 0.979 3.50 0.776 E 0.939 4.00 0.743 E 0874 5.00 0.682 E 0.846 5.50 0.556 E 0.822 6.00 0.532 E 0.804 6.50 0.607 E! 0.784 7.00 0.585 E 0.767 7.50 0.565 E 0.753 8.00 0.545 . E 0.730 9.00 0.508 E 0705 10.00 0.478 E 0.671 $2.00 0.422 E 0.644 14.00 0.377 0.615 17.00 0.322 E 0.595 20.00 0.278 E Answer For this protein solution, £280,260 = 069/094 = 0.73, so (from the table) F m 0.508. The concentration of protein 131*" X A235 = (0.508 X 0.69} mg/mL = 0.85 mg/mL. Note: the tabie applies to mixtures of proteins, such as might be found in a crude celluiar 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. 10. Solubility of the Components of DNA Draw the following structures and rate their relative soiubiiities in water (most 00103010 to least soiuble): deoxyribose, guanine, phosphate. How are these soiubilities consistent With the threendimensionai structure of {ioubie—stranded DNA? Answer E? HOCH o 2 OH /C\\ /N O“ H H mII (I! \CH I H H c\ 0“ / “om—IEm—mo / \ N HO H H2N N/ H OH Deoxyrihose Guanine Phosphate 8-92 Chapter 8 Nucleotides and Nucleic Acids Soiubilities: phosphate > deoxyribose > guanine. The negatively charged phosphate is the most water~soiuble; 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. Sanger Sequencing Logic In the Sanger (dideoxy) method for DNA sequencing, a small amount of a dideoxynucleotide triphosphatem—say, ddGTPmis added to the sequencing reaction along with a larger amount of the corresponding dC’I‘P. What result would be observed if the dCTP were omitted? Answer if dCTP is omitted from the reaction mixture, when the first G residue is encountered in the template, ddCTP is added and polymerization baits. Only one band will appear in the sequencing gei. 12. DNA Sequencing The foliowing DNA fragment was sequenced by the Sanger method. The asterisk indicates a fluorescent label. *5! 3r 3 ' DH ATTACGCAAGGACATTAGAC m ~ 5 ' A sample of the DNA was reacted with DNA poiymerase and each of the nucleotide inixtures (in an appropriate buffer) listed below. Bideoxynucleotides (ddN’I‘Ps) 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 etectrophoresis on an agarose gel, and the fluorescent bands on the gel were iocated. 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 ianes of the gel took like? Electrophoresis Chapter 8 Nucleotides and Nucleic Acids 8-93 Answer Mums: Electrophoresis Lane I: The reaction mixture that generated these bands inchded all the deoxynucleotides, plus dideoxythymidine. The fragments are of various lengths, all terminating where a ddTTP was substituted for a dT’I‘P. For a email portion of the strands synthesized in the experiment, dd’E‘TP 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’«pmnernTAATGCGTTCCTGTAATCTG 5’-primet—TAATGCGT’I‘CCTGTAATCT 5‘wprimenTAA’i‘GCGTTCCTGTAAT 5’~primer—TAATGGGTTCCTGT 5’-primer~TAATGCGTTCCT 5’-primer—TAATGCGTT 5’»primer~TAATGCGT 5’-primer—TAAT 5'~pritner~T Lane 2: Similarly, this lane will have four bands (top to bottom), for the foiiowing 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 d'E‘TP, every fragment was terminated immediateiy after the primer as ddTTP was inserted, to form 5’—prizner~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 dideoxynucieotide, a single labeled product formed: 5’-primer—TAATGCGTTCCTGTAATCTG. This will appear as a singie thick band at the top of the gel. 13. Snake Venom Phosphodiesterase An exonuciease is an enzyme that sequentially cleaves nu— cleotides from the end of a polynuoleotide strand. Snake venom phosphodiesterase, which hydrolyzes nucleotides from the 8’ end of any oligonucleotide with a free 3’-hydroxyl group, cleaves between the 8’ hydroxyl of the ribose or deoxyribose and the phosphoryi group of the next nucleotide, It acts on singEe—stranded DNA or RNA and has no base specificity. This enzyme was used in sequence 8-94 Chapter 8 Nucleotides and Nucleic Acids determination 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 following sequence? (5’)GCGCCAUUGC{3')—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): (sop—oooocnuuoccso—OH (5’)P—GCGCCAUUG(3’)uOI~i (5’)P—GCGCCAUU(3')—OH (5’)P—GCGCCAU(8’)—OH (5’)?»«eoeccncso—0H (5’)P~GCGCC(3’)—OH (5')?»ocoors')—on (anaconda-0H {5’)P-«GC(3’)—OH and the released nucleoside 5’—phosphates, GMP, UMP, AMP, and (3MP. 14. Preserving DNA in Bacterial Endospores Bacterial endospores form when the environment is no 15. longer conducive to active cell metabolism. The soil bacterium Bacillus subtitles, for example, begins the process of sporelation 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. B. subtilis spores are much more resistant than are 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 mutationmcausing enzymes and slows the rate of nonehzymatic depurination reactions, which are hydrolysis reactions. (b) UV light induces the condensation of adjacent pyrimidine bases to form cyclo‘outane ' pyrimidine dimers. The spores of B. subtilts, 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 {EV-induced mutation is critical to spore DNA integrity. Oligonueleotide Synthesis In the scheme of Figure 8-35, each new base to be added to the growing oiigonucleotide is modified so that its 3' hydroxyl is activated and the 5’ hydroxyl has a dimethoxytrityl (EMT) group attached. What is the function of the DMT group on the incoming base? Answer DMT is a blocking group that prevents reaction of the incoming base with itself. Chapter 8 Amino Acids, Peptides, and Proteins 5‘95 Biochemistry on the Internet 16. The Structure of DNA Elucidation of the three-dimensional structure of DNA heiped researchers understand how this molecule conveys information that can be faithfully repiicated from one genera- tion to the next. To see the secondary structure of double-stranded DNA, go to the i’rotein Data Bank website (wwwrcsborg). Use the PDB identifiers listed below to retrieve the structure summaries for the two forms of DNA. Open the structures using Jmol (linked under the Display Options), and use the controls in the Jmol menu (accessed with a control-click or by clicking on the Jmol logo in the lower right corner of the image screen) to complete the following exercises. Refer to the Jmol help links as needed. (3.) Obtain the file for 1411?), a highly conserved, repeated DNA sequence from the end of the Elli/«1 (the virus that causes AIDS) genome. Display the molecule as a ball-and-stick structure (in the control menu, choose Seiect > All, then Render > Scheme > Ball and Stick). identify the sugarmphosphate backbone for each strand of the DNA dupiex. Locate and identify individual bases. identify the 5’ end of each strand. Locate the major and minor grooves. Is this a right- or left-handed helix? (1}) Obtain the file for 145D, a DNA with the Z conformation. Display the molecule as a bail-andmstick structure. Identify the sugar-phosphate backbone for each strand of the DNA duplex. is this a right— or left-handed helix? (c) To fully appreciate the secondary structure of DNA, View the molecules in stereo. 0n the control menu, Seiect > All, then Render > Stereographic > Crossveyed or Wall-eyed. You will see two images of the DNA molecule. Sit with your nose approximately 20 inches from the monitor and focus on the tip of your nose (cross-eyed) or the Opposite edges of the screen (wali~eyed). In the background you should see three images of the DNA helix. Shift your focus to the middle image, which should appear three-dimensional. (Note that only one of the two authors can make this work.) Answer (a) The DNA fragment modeled in tile 141D, from the human immunodeficiency virus, is the B form, the standard Watson-Crick structure (although this particular structure is a bent B—form DNA). This fragment has an adenine at the 5’ end and a guanine at the 3’ 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 minor groove is in the center of the model. Rotating the modei so that the 5' adenine is at the upper righbhand 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. (1)) The model of DNA in the Z conformation includes a shell of water moiecules around the helix. ri‘he water molecules are visible when the complex is viewed in ball-and-sticix mode. ri‘urn off the display of the water molecules using the console controls Select > Nucleic > DNA. Then Select > Display Selected Only. The backbone of DNA in the Z confonnation is very different from that in the B conformation. The helix spiral runs up- ward in a ciockwise direction, so this is a ieftwhanded helix. (c) Viewing the structures in stereo takes a bit of practice, but perseverance wili be re— warded? Here are some tips for successful threendimensional viewing: (1) Turn off or lower the room lighting. (2) Sit directly in front of the screen. (3) Use a ruler to make sure you are it) to El inches from the screen. (4) 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). (5) Move your head slightly closer to or farther away from the screen to bring the mid- dle image into focus. Don’t look directly at the middle image as you try to bring it into focus. S~96 Chapter 8 Amino Acids, Peptides, and Proteins (6) If you find it uncomfortabie to focus on the tip of your nose1 try using the tip of a fin— ger {positioned just beyond the tip of your nose) instead. (7) 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! Data Anaiysis Prohiem 1’7. Chargaff’s Studies of DNA Structure The chapter section “DNA Is a Double Helix that Stores Ge~ netic lnforrnation”i11ciudes a summary of the main findings of Erwin Chargaff and his coworkers, listed as four conclusions {“Chargaff’s rules"; 9. 278). In this problem, you will examine the data Char~ gaff collected in support of these conclusions. In one paper, Chargaff (1950) described his analytical methods and some eariy results. Briefly, he treated DNA samples with acid to remove the bases, separated the bases by paper chromatography, and measured the amount of each base with UV spectroscopy. His results are shown in the three tabies be— low. The molar ratio is the ratio of the number of moles of each base in the sample to the number of moles of phosphate in the sampie—this gives the fraction of the total number of bases represented by each particuiar base. The recovery is the sum of all four bases {the sum of the molar ratios); full recovu ery of all bases in the EDNA would give a recovery of 1.0. ficu mamvxumnui-awwfl wm:m=m~mcm-Km:mz maxi-.1". minnow firs/1.“; "Ev 2:, mxwww m , ,~ mm 22:2: .N-w" emu-L: 11...: Adenine E Guanine E Cytosine Thymine Recovery Adenine t! . a Guanine 0:8 32; Cytosine 0&5 ? Thymioe 0.27 ’ 214;:st :.:.: _*.:x:_:o'..e:. r..::_~...
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