Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 79
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It is about 1.3 megabase in length.What is the approximate physical length of this chromosome in millimeters? (Note: One megabase (Mb) equals 1 ×106 base pairs, and 1 Å = 10-7 mm.)Answer: The standard form of duplex DNA contains 10 nucleotide pairs per 34 Å, or 3.4 Å per nucleotide pair.Therefore, a 1.3-Mb molecule has a length of 1.3 × 106 × 3.4 Å = 4.4 × 106 Å = 0.44 mm; that is, the 1.3-megabasemolecule of DNA in Dp(1;f)1187 has a physical length just a little smaller than half a millimeter.Problem 2: The genome size of Drosophila melanogaster is 165,000 kilobases (kb); approximately 2/3 of thegenome is euchromatic and 1/3 is heterochromatic. In salivary gland chromosomes, the euchromatic part of thegenome becomes polytene, and the giant chromosomes exhibit approximately 5000 transverse bands used aslandmarks for the locations of genes, chromosome breakpoints, and other cytogenetic features.
What is theapproximate DNA content of an average band in the salivary gland chromosomes?Answer: The euchromatic part of the Drosophila genome consists of approximately 165,000 kb × 2/3 = 110,000 kbof DNA. This amount of DNA is distributed over 5000 bands, for an average of 110,000 kb/5000 = 22 kb per band.Problem 3: Suppose that the DNA sequences repeated at the ends of a transposable element undergo pairing andrecombination.
What are the genetic consequences if:(a) the sequences are direct repeats?(b) the sequences are inverted repeats?Draw diagrams to support your answers.Answer: The diagrams show the consequences.Page 255(a) When direct repeats pair and undergocrossing-over, the result is the deletion of oneof the repeats and the DNA segment betweenthem (part A).(b) When the repeats are inverted repeats, theresult is that the transposable element is invertedin the chromosome (part B).Analysis and Applications6.1 The wpch mutant allele of the white gene in Drosophila contains an insertion of 1.3 kb within the controllingregion of the gene.
The Wpch allele undergoes further mutation at a high rate, including reverse mutations to w+.The w+ derivatives are found to lack the 1.3-kb insert. What kind of DNA sequence is the 1.3-kb insert likely tobe?6.2 Some transposable elements in animals are active only in the germ line and not in somatic cells. How mightthis restriction of activity be advantageous to the persistence of the transposable element?6.3 Most transposable elements create a short duplication of host sequence when they insert. The duplication isalways in a direct, rather than an inverted, orientation. What does this tell you about the process of insertion?6.4 The E. coli chromosome is 4700 kb in length.
What is its length in millimeters? The haploid human genomecontains 3 × 109 base pairs of DNA. What is its length in millimeters?6.5 If you had never seen a chromosome in a microscope but had seen nuclei, what feature of DNA structure wouldtell you that DNA must exist in a highly coiled state within cells?6.6 What causes the bands in a polytene chromosome? How do polytene chromosomes undergo division?6.7 Is there an analog of a telomere in the E. coli chromosome?6.8 Many transposable elements contain a pair of repeated sequences. What is their position in the element? Arethey in direct or inverted orientation?6.9 Consider a long linear DNA molecule, one end of which is rotated four times with respect to the other end inthe unwinding direction.(a) If the two ends are joined to keep the molecule in the underwound state, how many base pairs will be broken?(b) If the underwound molecule is allowed to form a supercoil, how many twists will be present?6.10 Endonuclease S1 can break single stranded DNA but does not break double-stranded linear DNA.
However,S1 can cleave supercoiled DNA, usually making a single break. Why does this occur?6.11 Circular DNA molecules of the same size migrate at different rates in electrophoretic gels, depending onwhether they are supercoiled or relaxed circles. Why?6.12 Denaturation of DNA refers to breakdown of the double helix and ultimate separation of the individualstrands, and it can be induced by heat and other treatments. Because G-C base pairs have three hydrogen bonds andA–T base pairs have only two, the temperature required for denaturation increases with the G-C content and alsowith the length of continuous G-C tracts in the molecule. Renaturation of DNA refers to the formation of doublestranded DNA from complementary single strands.(a) Which process—denaturation or renaturation—is dependent on the concentration of DNA?PageChapter 6 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 instructor1. The centromere consensus sequences for CDE1, CDE2, and CDE3 can be compared among all yeastchromosomes because the complete sequence of the entire genome is available. This exercise concerns only thecentromeres of chromosomes IV, IX, XIII, and XVI. Find the CDE1, CDE2, and CDE3 sequences in each of thesechromosomes, and compare them with each other and with the consensus sequences given in the chapter. If assigneto do so, copy the CDE1, CDE2, and CDE3 sequences and align them.2.
Another view of the telomerase at work can be found at this keyword site. Although the 5'-to-3' polarities of theDNA and RNA strands are not indicated, you should be able to deduce them. If assigned to do so, make a sketch ofthe telomerase, and indicate the polarity of each of the DNA and RNA strands.3. You will find up-to-date descriptions of the transposable element that Barbara McClintock first described bysearching the keyword site for the terms Activator and Dissociation. Use the hot links to locate the nucleotidesequence of Ac, and identify the(text box continued on next page)(b) Which of the two illustrated DNA molecules (1 or 2) would have the lower temperature for strand separation?Why?6.13 DNA from species A, labeled with 14N and randomly fragmented, is renatured with an equal concentration ofDNA from species B, labeled with 15N and randomly fragmented, and then centrifuged to equilibrium in CsCl.
Fivepercent of the total renatured DNA has a hybrid density. What fraction of the base sequences is common to the twospecies?6.14 What is meant by the terms ''direct repeat" and "inverted repeat"? Using the base sequence illustrated, diagramduplex DNA molecule, showing the sequence as a direct repeat and as an inverted repeat.6.15 Mutations have not been observed that result in non-functional histones. Why should this be expected?6.16 A sequence of middle-repetitive DNA from Drosophila is isolated and purified.
It is used as a template in apolymerization reaction with DNA polymerase and radioactive substrates, and highly radioactive probe DNA isprepared. The probe DNA is then used in an in situ hybridization experiment with cells containing polytenechromosomes obtained from ten different flies of the same species. Autoradiography indicates that the radioactivematerial is localized to about 20 sites in the genome, but they are in different sites in each fly examined. What does thiobservation suggest about the DNA sequence being studied?Challenge Problem6.17 A sample of identical DNA molecules, each containing about 3000 base pairs per molecule, is mixed with histonoctamers under conditions that allow formation of chromatin.
The reconstituted chromatin is then treated with anuclease, and enzymatic digestion is allowed to occur. The histones are removed, and the positions of the cuts in theDNA are identified by sequencing the fragments. It is found that the breaks have been made at random positions and,asPage(text box continued from previous page)inverted repeats at its ends.
If assigned to do so, write a 150-word report on these transposable elements, payingparticular attention to the DNA sequence relations between the Ac and Ds elements and specifying the element thatencodes the active transposase.MUTABLE SITE EXERCISESThe Mutable Site Exercise changes frequently. Each new update includes a different exercise that makes use ofgenetics resources available on the World Wide Web.
Select the Mutable Site for Chapter 6, and you will be linkedto the current exercise that relates to the material presented in this chapter.PIC SITEThe Pic Site showcases some of the most visually appealing genetics sites on the World Wide Web. To visit theshowcase genetics site, select the Pic Site for Chapter 6.expected, at about 200-base-pair intervals. The experiment is then repeated with a single variation. A protein known tbind to DNA is added to the DNA sample before addition of the histone octamers. Again, reconstituted chromatin isformed and digested with nuclease. In this experiment, it is found that the breaks are again at intervals of about 200base pairs, but they are localized at particular positions in the base sequence. At each position, the site of breakage cavary over only a 2-to-3-base range.
However, examination of the base sequences in which the breaks have occurreddoes not indicate that breakage occurs in a particular sequence; in other words, each 2-to-3-base region in whichcutting occurs has a different sequence. Explain the difference between the two experiments.Further ReadingBerg, D.
E., and M. M. Howe. 1989. Mobile DNA. American Association for Microbiology. Washington, DC.Blackburn, E. H. 1990. Telomeres and their synthesis. Science 249: 489.Comfort, N. C. 1995. Two genes, no enzyme: A second look at Barbara McClintock and the 1951 Cold Spring HarboSymposium. Genetics 140: 1161.Curtis, B. C., and D. R. Johnson. 1969. Hybrid wheat.
Scientific American, May.Elgin, S. C. R. 1996. Heterochromatin and gene regulation in Drosophila. Current Opinion in Genetics &Development 6: 193.Engels, W. R. 1997. Invasions of P elements. Genetics 145: 11.Federoff, N. 1984. Transposable genetic elements in maize. Scientific American, June.Green, M. M. 1980. Transposable elements in Drosophila and other Diptera. Annual Review of Genetics 14: 109.Greider, C. W., and E.
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