Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 91
Текст из файла (страница 91)
Abnormal oncogene expression is implicated in cancer. Viral oncogenes, found incertain cancer-causing viruses, are derived from cellular oncogenes. Cells also contain tumor-suppressor genes, theabsence of which predisposes to cancer. The gene for retinoblastoma codes for a tumor-suppressor in the retina ofthe eyes. The gene is dominant in predisposing to retinal malignancy but recessive at the cellular level. People whoinherit one copy of the gene through the germ line develop retinal tumors when the gene becomes homozygous incells in the retina. Homozygosity in the retina can result from any number of genetic events, including newmutation, deletion, chromosome loss and nondisjunction, and mitotic recombination.Page 301Key Termsacentric chromosomefragile-X chromosomeretinblastomaacrocentric chromosomefragile-X syndromerhodopsinadjacent-1 segregationhaploidring chromosomeadjacent-2 segregationhexaploidRobertsonian translocationallopolyploidimprintingsemisterilityalternate segregationinversionsingle-active-X principleamniocentesisinversion loopsubmetacentric chromosomeaneuploidkaryotypetandem duplicationautopolyploidyKlinefelter syndrometestis-determining factor (TDF)Barr bodymetacentric chromosometetraploidcellular oncogenemonoploidtranslocationchimeric genemonosomictrinucleotide repeatchromosome paintingmosaictriplicationcolchicineoctoploidtriploiddecaploidoncogenetrisomicdeficiencyparacentric inversiontrisomy-X syndromedeletionpericentric inversiontrivalentdeuteranopiaPhiladelphia chromosometumor-suppressor genedeuteranomalyposition effectTurner syndromedicentric chromosomeposition-effect variegation (PEV)uncoveringdosage compensationprotanopiaunequal crossing-overdouble-Y syndromeprotanomalyunivalentDown syndromepolyploidyvariegationduplicationpolysomyviral oncogeneembryoidpseudoautosomalX inactivationendoreduplicationreciprocal translocationeuploidred-green color blindnessReview the Basics• What is a metacentric chromosome? A submetacentric chromosome? An acrocentric chromosome?• Why does an acentric chromosome fail to align at the metaphase plate and fail to move to one of the poles inanaphase? What type of abnormal chromosome forms a chromosomal "bridge" between the daughter cells atanaphase?• How can a normal gamete have two sets of chromosomes? What possible gametes could be produced by atetraploid plant with the genotype AAaa BBbb?• Define each of the following terms: allopolyploid, aneuploid, deletion, duplication, paracentric inversion,pericentric inversion, reciprocal translocation, Robertsonian translocation.• What does it mean to say that a deletion "uncovers" a recessive mutation?• Which type of chromosome rearrangement can change a metacentric chromosome into a submetacentricchromosome? Which type of chromosome rearrangement can fuse two acrocentric chromosomes to make onemetacentric chromosome?• Inversions are often called "suppressors" of crossing-over.
Is this term literally true? If not, what is meant by theterm?• What does dosage compensation mean with reference to X-linked genes?• What are two broad classes of genes often associated with inherited cancers?Guide to Problem SolvingProblem 1: The first artificial allotetraploid was created by the Russian agronomist G. D. Karpechenko in the1930s by crossing the radish Raphanus sativus with the cabbage Brassica oleracea. Both species have a diploidchromosome number of 18. The initial F1 hybrid was virtually sterile, but among the off-spring was a rare, fullyfertile allotetraploid that he called Raphanobrassica.
What is the chromosome number of the F1 of the cross R.sativus × B. oleracea? What is the chromosome number of Raphanobrassica? How many bivalents are formed inmeiosis in Raphanobrassica?Answer: Both R. sativus and B. oleracea produce gametes with 9 chromosomes, so the F1 has 18 chromosomes.The allotetraploid results from a doubling of the F1 chromosome complement, so Raphanobrassica has 36chromosomes.
Because each chromosome has a homolog, Raphanobrassica forms 18 bivalents.Page 302Problem 2: Genes a, b, c, d, e, and f are closely linked in a chromosome, but their order is unknown. Threedeletions in the region are found to uncover recessive alleles of the genes as follows:Deletion 1 uncovers a, b, and d.Deletion 2 uncovers a, d, c, and e.Deletion 3 uncovers e and f.What is the order of the genes? In this problem, you will see that there is enough information to order most, but notall, of the genes. Suggest what experiments you might carry out to complete the ordering.Answer: Problems of this sort are worked by noting that genes uncovered by a single deletion must be contiguous.The gene order is deduced from the overlaps between the deletions. The overlaps are a and d between the first andsecond deletions and e between the second and third.
The gene order (as far as can be determined from these data)is diagrammed in part A of the accompanying illustration. The deletions are shown in red. Gene b is at the far left,then a and d (in unknown order), then c, e, and f. (Gene c must be to the left of e, because otherwise c would beuncovered by deletion 3.) Part B is a completely equivalent map with gene b at the right.
The ordering can becompleted with a three-point cross between b, a, and d or between a, d, and c or by examining additional deletions.Any deletion that uncovers either a or d (but not both) plus at least one other marker on either side would providethe information to complete the ordering.Problem 3: In each of the following cases, determine the consequences of a single crossover within the invertedregion of a pair of homologous chromosomes with the gene order A B C D in one chromosome and a c b d in theother.(a) The centromere is not included within the inversion.(b) The centromere is included within the inversion.Answer: These kinds of problems are most easily solved by drawing a diagram.
The accompanying illustrationshows how pairing within a heterozygous inversion results in a looped configuration. (Crossing-over occurs at thefourstrand stage of meiosis, but for simplicity only the chromatids participating in the crossover are diagrammed.)Part A illustrates the situation when the centromere is not included within the inversion. The crossover chromatidsconsist of a dicentric (two centromeres) and an acentric (no centromere), and the products are duplicated for theterminal region containing A and deficient for the terminal region containing D, or the other way around. Thenoncrossover chromatids (not shown) are the parental A B C D and a c b d monocentric chromosomes.Part B illustrates the situation when the centromere is included in the inversion. The duplications and deficienciesare the same as in part A, but in this case both products are monocentrics.
As before, the noncrossover chromatidsare the parental A B C D and a c b d configurations.Analysis and Applications7.1 How many Barr bodies would be present in each of the following human conditions?(a) Klinefelter syndrome(b) Turner syndrome(c) Down syndrome(d) XYY(e) XXXPagChapter 7 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 bpresented 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 makes usof the information available at the site. This report, or an alternative, may be assigned by your instructor.1. The original Bar duplication may itself have arisen from unequal crossing-over.
You can access the details bysearching for the keyword at this site. If assigned to do so, write a brief description of the proposed origin of theoriginal duplication, and draw a diagram illustrating the events in the process.2. Molecular studies of the red and green opsin genes indicate abnormalities in 15.7 percent of Caucasian men. Thifrequency is substantially smaller than the incidence of color blindness as shown by color-vision testing, whichsuggests that some chimeric genes created by unequal crossing-over may be compatible with normal color vision. Morinformation about the molecular genetics and physiology of color vision can be found by searching the keyword sit"protanopia" and "deuteranopia." You can also find a description of the "Nagel anomaloscope'' used in testing colorvision.
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