5 Структура нуклеиновых кислот (1160074), страница 11
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(1984)The chemistry and biology of left-handed Z-DNA.Annu. Rev. Biochem. 53, 791-846.Wells, R.D. & Harvey, S.C. (eds) (1988) UnusualDNA Structures, Springer-Verlag, New York.Wells, R.D. (1988) Unusual DNA structures. J.Biol. Chem. 263, 1095-1098.Minireview; a concise summary.ATP As Energy CarrierJencks, W.P.
(1987) Economics of enzyme catalysis. Cold Spring Harb. Symp. Quant. Biol. 52, 6573.A relatively short article, full of insights.HistoricalOlby, R. (1974) The Path to the Double Helix, University of Washington Press, Seattle.Sayre, A. (1978) Rosalind Franklin and DNA,W.W. Norton & Co., Inc., New York.Watson, J.D. (1968) The Double Helix, AtheneumPublishers, New York.A personal account of the human aspects of the discovery.Problems1. Determination of Protein Concentration by UVAbsorption in a Solution Containing Nucleic AcidsThe concentration of protein or nucleic acid in solutions containing both can be estimated by usingtheir light absorption properties.
Proteins have astrong absorption centered at a wavelength of280 nm, whereas nucleic acids absorb moststrongly at 260 nm. When both proteins and nucleic acids are present in a solution, their respective concentrations can be estimated by measuringthe absorbance (A) of the solution at 280 nm and260 nm and using the table at the top of page 357.^280/260 is the ratio of the absorbance at 280 and260 nm. The table indicates the percentage of totalmass that is nucleic acid, and provides a factor, F,to correct the A28o reading and give a more accurate protein estimate. The protein concentration(in mg/ml) is equal to F x A280 (assuming thecuvette is 1 cm wide). What are the protein andnucleic acid concentration if A280 = 0.69 andA260 = 0.94?Chapter 12 Nucleotides and Nucleic Acids-^280/2601.751.631.521.401.361.301.251.161.091.030.9790.9390.8740.8460.8220.8040.7840.7670.7530.7300.7050.6710.6440.6150.595Proportion ofnucleic acid(%)F0.000.250.500.751.001.251.502.002.503.003.504.005.005.506.006.507.007.508.009.0010.0012.0014.0017.0020.001.1161.0811.0541.0230.9940.9700.9440.8990.8520.8140.7760.7430.6820.6560.6320.6070.5850.5650.5450.5080.4780.4220.3770.3220.2782.
Nucleotide Structure What positions in a purine ring have the potential to form hydrogenbonds, but are not involved in the hydrogen bondsof Watson-Crick base pairs?3. Base Sequence of Complementary DNA StrandsWrite the base sequence of the complementarystrand of double-helical DNA in which one strandhas the sequence (5')ATGCCCGTATGCATTC(3;).4. DNA of the Human Body Calculate the weightin grams of a double-helical DNA molecule stretching from the earth to the moon (-320,000 km).
The357DNA double helix weighs about 1 x 10 18 g per1,000 nucleotide pairs; each base pair extends0.34 nm. For an interesting comparison, your bodycontains about 0.5 g of DNA!5. DNA Bending Assume that a poly(A) tract fivebase pairs long produces a bend of about 20°. Calculate the total (net) bend produced in the DNA ifthe center base pairs (the third of five) of two successive (dA)5 tracts are located (a) 10 or (b) 15 basepairs apart.
Assume that there are 10 base pairsper turn in the DNA double helix.6. Distinction between DNA Structure and RNAStructure Hairpins may form at palindromic sequences in single strands of either RNA or DNA.How is the helical structure of a hairpin in RNAdifferent from that of a hairpin in DNA?7. Nucleotide Chemistry In the cells of many eukaryotic organisms, there are highly specializedsystems that specifically repair G-T mismatchesin DNA. The mismatch is repaired to form a G=Cbase pair (not A=T).
This G-T mismatch repairsystem occurs in addition to a more general systemthat repairs virtually all mismatches. Can youthink of a reason why cells require a specializedsystem to repair G-T mismatches?8. Nucleic Acid Structure Explain why there isan increase in the absorption of UV light (hyperchromic effect) when double-stranded DNA isdenatured.9. Base Pairing in DNA In samples of DNA isolated from two unidentified species of bacteria,adenine makes up 32 and 17%, respectively, of thetotal bases. What relative proportions of adenine,guanine, thymine, and cytosine would you expectto find in the two DNA samples? What assumptions have you made? One of these bacteria wasisolated from a hot spring (64 °C). Which DNAcame from this thermophilic bacterium? What isthe basis for your answer?P A R TInformation Pathways789790Part IV Information PathwaysReplicationDNATranscriptionfRNATranslationfProteinThe central dogma of molecular genetics, showingthe general pathways of information flow via theprocesses of replication, transcription, and translation.
The term "dogma" is a misnomer here. It wasintroduced by Francis Crick at a time when littleevidence supported these ideas. The "dogma" is nowa well-established principle.DNA molecules having identical nucleotide sequences. The second istranscription, the process by which parts of the coded genetic message in DNA are copied precisely in the form of RNA. The third istranslation, in which the genetic message coded in messenger RNA istranslated on the ribosomes into a protein with a specific sequence ofamino acids.Part IV is devoted to an explanation of these and related processes.First (Chapter 23) we will examine the structure, topology, and packaging of chromosomes and genes.
The processes that make up the central dogma will be elaborated in Chapters 24 through 26. Then, as wehave done for biosynthetic pathways, we will turn to regulation andexamine how the expression of genetic information is controlled (Chapter 27).A major theme running through these chapters is the added complexity encountered in the biosynthesis of a macromolecule when thatmacromolecule contains information. Assembling nucleic acids andproteins with the correct sequences of nucleotides and amino acids,respectively, represents nothing less than preserving the faithful expression of the template upon which life itself is based.
The formationof phosphodiester bonds in DNA or peptide bonds in proteins might beexpected to be a trivial feat for cells, given the arsenal of enzymatic andchemical tools described in Part III of this book. Nevertheless, theframework of patterns and rules established in the examination ofmetabolic pathways must be enlarged considerably when informationis added to the equation.
Forming specific bonds and preventing sequence errors in these polymers has an enormous impact on the thermodynamics, chemistry, and enzymology of the synthetic processes.For example, formation of a peptide bond should require an input ofonly about 21 kJ, and relatively simple enzymes that catalyze comparable reactions are known.
To synthesize the correct peptide bond between two specific amino acids at a given point in a protein, however,the cell invests about 125 kJ in chemical energy and makes use of thecombined activities of over 200 RNA molecules, enzymes, and specialized proteins. Information is expensive.The dynamic interaction between nucleic acids and proteins is another central theme of Part IV.
With the important exception of a fewcatalytic RNA molecules (discussed in Chapter 25), the processes thatmake up the pathways of cellular information flow are catalyzed andregulated by proteins. An understanding of these enzymes and proteins can have practical as well as intellectual rewards because theyform the basis of the development of recombinant DNA technology.This technology is making possible the prenatal diagnosis of geneticdisease; the production of a wide range of potent new pharmaceuticalagents; the sequencing of the entire human genome; the introductionof new traits into bacteria, plants, and animals for industry and agriculture; human gene therapy; and many other advances.
We finish ourtour of the information pathways, and indeed the entire book, in Chapter 28 with a look at this technology and its implications for the future..
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