Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 68
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Semiconservative replication was first demonstrated in the Meselson-Stahlexperiment, which used equilibrium density-gradient centrifugation to separate DNA molecules containing two15N-labeled strands, two 14N-labeled strands, or one of each.
Replication proceeds by a DNA polymerase (1)bringing in a nucleotide triphosphate with a base capable of hydrogenbonding with the corresponding base in thetemplate strand and (2) joining the 5'-P group of the nucleotide to the free 3'-OH group of the growing strand. (Theterminal P–P from the nucleotide triphosphate is cleaved off and released.) Because double-stranded DNA isantiparallel, only one strand (the leading strand) grows in the direction of movement of the replication fork.
Theother strand (the lagging strand) is synthesized in the opposite direction as short fragments (Okazaki fragments)that are subsequently joined together. DNA polymerases cannot initiate synthesis, so a primer is always needed.The primer is an RNA fragment made by an RNA-polymerizing enzyme called primase; the RNA primer isremoved at later stages of replication. DNA molecules of prokaryotes usually have a single replication origin;eukaryotic DNA molecules usually have many origins.Restriction enzymes cleave DNA molecules at the positions of specific sequences (restriction sites) of usually fouror six nucleotides.
Each restriction enzyme produces a unique set of fragments for any particular DNA molecule.These fragments can be separated by electrophoresis and used for purposes such as DNA sequencing. The positionsof particular restriction fragments in a gel can be visualized by means of a Southern blot, in which radioactiveprobe DNA is mixed with denatured DNA made up of single-stranded restriction fragments that have beentransferred to a filter membrane after electrophoresis. The probe DNA will form stable duplexes (anneal orrenature) with whatever fragments contain sufficiently complementary base sequences, and the positions of theseduplexes can be determined by autoradiography of the filter. Particular DNA sequences can also be amplifiedwithout cloning by means of the polymerase chain reaction (PCR), in which short, synthetic oligonucleotides areused as primers to replicate repeatedly and amplify the sequence between them.The base sequence of a DNA molecule can be determined by dideoxynucleotide sequencing.
In this method, theDNA is isolated in discrete fragments containing several hundred nucleotide pairs. Complementary strands of eachfragment are sequenced, and the sequences of overlapping fragments are combined to yield the complete sequence.The dideoxy sequencing method uses dideoxynucleotides to terminate daughter strand synthesis and reveal theidentity of the base present in the daughter strand at the site of termination.Key TermsanalogDNA polymerasemajor grooveannealingDNA polymerase I (Pol I)melting curveantiparallelDNA polymerase III (Pol III)melting temperatureB formediting functionminor groovebandendonucleasenickbase compositionequilibrium density-gradientnucleasebidirectional replicationcentrifugationnucleic acid hybridizationblunt endsexonucleasenucleosideChargaff's rules5'-P groupnucleotidedaughter strand5'Okazaki fragmentsdenaturationgel electrophoresisoligonucleotide primersdideoxyribosehelicasepalindromedideoxy sequencing methodinitiationparent strandDNA cloninglagging strandpercent G + CDNA ligaseleading strandphosphodiester bond3' exonucleasePage 215polarityreplication forktemplatepolymerase αreplication originthermophilespolymerase chain reaction (PCR) restriction endonucleaseθ replicationpolymerase γrestriction enzyme3'-OH grouppolynucleotide chainrestriction fragmenttopoisomeraseprecursor fragmentrestriction mapultracentrifugeprimaserestriction siteZ-form DNAprimerriboseprimosomeRNA polymeraseprobe DNArolling-circle replicationproofreading functionsemiconservative replicationpurinesingle-strand binding proteinspyrimidineSouthern blotrenaturationsticky endsReview the Basics• What are the four bases commonly found in DNA? Which form base pairs? What five-carbon sugar is found inDNA? What is the difference between a nucleoside and a nucleotide?• How many phosphate groups are there per base in DNA, and how many phosphates are there in each precursorfor DNA synthesis?• Which chemical groups are at the ends of a single polynucleotide strand?• What is the relationship between the amount of DNA in a somatic cell and the amount in a gamete?• Name four requirements for initiation of DNA synthesis.
To what chemical group in a DNA chain is an incomingnucleotide added, and what group in the nucleotide reacts with the DNA terminus?• In what sense are the two strands of DNA antiparallel?• How does the polymerase chain reaction work? What is it used for? What information about the target sequencemust be known in advance?Guide to Problem SolvingProblem 1: A technique is used for determining the base composition of double-stranded DNA. Rather than givingthe relative amounts of each of the four bases—[A], [T], [G], and [C]—it yields the value of the ratio [A] [C].
Ifthis ratio is 1/3, what are the relative amounts of each of the four bases?Answer: In double-stranded DNA, [A] = [T] and [G] = [C] because of the base pairing between complementarystrands. Therefore, if [A]/[C] = 1/3, then [C] = [G] = 3 × [A]. Because [A] + [T] + [G] + [C] = 1, everything can beput in terms of [A] asorThis makesIn other words, the DNA is 12.5 percent A, 12.5 percent T, 37.5 percent G, and 37.5 percent C.Problem 2: The restriction enzyme EcoRI cleaves double-stranded DNA at the sequence 5'-GAATTC-3', and therestriction enzyme HindIII cleaves at 5'-AAGCTT-3'.
A 20-kilobase (kb) circular plasmid is digested with eachenzyme individually and then in combination, and the resulting fragment sizes are determined by means ofelectrophoresis. The results are as follows:EcoRI alone: fragments of 6 kb and 14 kbHindIII alone: fragments of 7 kb and 13 kbEcoRI and HindIII: fragments of 2 kb, 4 kb, 5 kb and 9 kbHow many possible restriction maps are compatible with these data? For each possible restriction map, make adiagram of the circular molecule and indicate the relative positions of the EcoRI and HindIII restriction sites.Page 216Answer: Because the single-enzyme digests give two bands each, there must be two restriction sites for eachenzyme in the molecule.
Furthermore, because digestion with HindIII makes both the 6-kb and the 14-kbrestriction fragments disappear, each of these fragments must contain one HindIII site. Considering the sizes of thefragments in the double digest, the 6-kb EcoRI fragment must be cleaved into 2-kb and 4-kb fragments, and the 14kb EcoRI fragment must be cleaved into 5-kb and 9-kb fragments. Two restriction maps are compatible with thedata, depending on which end of the 6-kb EcoRI fragment the HindIII site is nearest.
The position of the remainingHindIII site is determined by the fact that the 2-kb and 5-kb fragments in the double digest must be adjacent in theintact molecule in order for a 13-kb fragment to be produced by HindIII digestion alone.The accompanying figure shows the relative positions of the EcoRI sites (part A). Parts B and C are the twopossible restriction maps, which differ according to whether the EcoRI site at the top generates the 2-kb or the 4-kbfragment in the double digest.Problem 3: The DNA sequencing gels diagrammed in parts A and B of the accompanying illustration wereobtained from human DNA by the dideoxy sequencing method.
Part A comes from a person homozygous for thenormal allele of the cystic fibrosis gene, part B from an affected person homozygous for a mutant allele. Use thebands in the gel to answer the following questions.(a) Why are there discrete bands in the gel?(b) Which end of the gel (top or bottom) corresponds to the 5' end of the DNA sequence?(c) What nucleotide sequence is indicated by the bands in the gels?(d) What is the sequence of the complementary strand in duplex DNA?(e) Compare the sequence in parts A and B.
What is special about the three nucleotides that give the bandsindicated in red?Answer:(a) Each band results from a fragment of DNA whose replication was terminated by the incorporation of a dideoxynucleotide. The fragments differ by one nucleotide in length, and they are arranged by size with the smallestfragments at the bottom, the largest at the top.(b) Because DNA replication elongates at the 3' end, each fragment of increasing length has an additional 3'nucleotide. Therefore, the DNA sequence is oriented with the 5' end at the bottom of the gel and the 3' end at thetop.(c) The sequences, read directly from the gel from bottom to top, are (part A)5'-ATTAAAGAAAATATCATCTTTGGTGTTTCCTATGATGAATAT-3'and (part B),5'-ATTAAAGAAAATATCATTGGTGTTTCCTATGATGAATATAGA-3'(d) The complementary strands in duplex DNA are antiparallel and have A paired with T and G paired with C.
Forpart A, the complementary strand is3'-TAATTTCTTTTATAGTAGAAACCACAAAGGATACTACTTATA-5'and, for part B, the complementary strand is3'-TAATTTCTTTTATAGTAACCACAAAGGATACTACTTATATCT-5'(e) The three nucleotides in red in the wildtype gene are missing in the mutant gene. This three-base deletion isfound in about 70 percent of the mutations in cystic fibrosis patients, and it results in a missing amino acid atposition 508 in the corresponding protein.Page 217Chapter 5 GeNETics on the webGeNETics on the web will introduce you to some of the most important sites for finding genetic information on theInternet. To complete the exercises below, visit the Jones and Bartlett home page athttp://WWW.jbpub.com/geneticsSelect the link to Genetics: Principles and Analysis and then choose the link to GeNETics on the web. You will bepresented with a chapter-by-chapter list of highlighted keywords.GeNETics EXERCISESSelect the highlighted keyword in any of the exercises below, and you will be linked to a web site containing thegenetic information necessary to complete the exercise.
Each exercise suggests a specific, written report that makesuse of the information available at the site. This report, or an alternative, may be assigned by your instructor.1. The keyword DNA will connect you with a site that has an excellent collection of DNA structural models,including some animations.
Browse through some of these and you will get a better intuitive feel for the structureand the manner in which the two strands are paired. If assigned to do so, take a close look at DNA model Number24. There is no legend, but you should be able to deduce the color coding of the spheres. What type of atomcorresponds to red? To green? To blue? What is the color of the phosphorous atoms?2. One of the principal tools of the molecular geneticist is the polymerase chain reaction, which is unique in itsability to bring about exponential amplification of a particular DNA sequence present in a large background ofother sequences.
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